Deck 16: Trigonometric Models

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Question
Evaluate the integral. (10x+25)sin(x2+5x)dx\int ( 10 x + 25 ) \sin \left( x ^ { 2 } + 5 x \right) \mathrm { d } x

A) 5xsin(x2+5x)+C- 5 x \sin \left( x ^ { 2 } + 5 x \right) + C
B) 5cos(x2+5x)+C5 \cos \left( x ^ { 2 } + 5 x \right) + C
C) 5xcos(x2+5x)+C5 x \cos \left( x ^ { 2 } + 5 x \right) + C
D) 5cos(x2+5x)+C- 5 \cos \left( x ^ { 2 } + 5 x \right) + C
E) 5sin(x2+5x)+C5 \sin \left( x ^ { 2 } + 5 x \right) + C
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Question
Recall that the average of a function f(x)f ( x ) on an interval [a,b][ a , b ] is fˉ=1baabf(x)dx\bar { f } = \frac { 1 } { b - a } \int _ { a } ^ { b } f ( x ) \mathrm { d } x
Find the average of the given function.
f(x)=sin(5x)f ( x ) = \sin ( 5 x ) over [0,5π][ 0,5 \pi ]

A)Average = 225π\frac { 2 } { 25 \pi }
B) Average = 25π2\frac { 25 \pi } { 2 }
C) Average = π5\frac { \pi } { 5 }
D) Average = 25π4\frac { 25 \pi } { 4 }
E) Average = 5π\frac { 5 } { \pi }
Question
Use geometry to compute the given integral. π6π62sinxdx\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x d x

A) π6π62sinxdx=6\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x d x = 6
B) π6π62sinx dx=4\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x \mathrm {~d} x = - 4
C) π6π62sinxdx=0\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x d x = 0
D) π6π62sinxdx=8\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x d x = 8
E) none of these
Question
Evaluate the integral. π3π4sinx dx\int _ { - \frac { \pi } { 3 } } ^ { \frac { \pi } { 4 } } \sin x \mathrm {~d} x

A) cosπ4cosπ\cos \frac { \pi } { 4 } - \cos \pi
B) sinπ3+sinπ4\sin \frac { \pi } { 3 } + \sin \frac { \pi } { 4 }
C) sinπ3cosπ5\sin \frac { \pi } { 3 } - \cos \frac { \pi } { 5 }
D) cosπ3cosπ4\cos \frac { \pi } { 3 } - \cos \frac { \pi } { 4 }
E) none of these
Question
Evaluate the integral. ? (8cosx4.1sinx9.3)dx\int ( 8 \cos x - 4.1 \sin x - 9.3 ) \mathrm { d } x ?

A) 8sinx+4.1cosx9.3x+C8 \sin x + 4.1 \cos x - 9.3 x + C
B) 8sinx+4.1cosx+C- 8 \sin x + 4.1 \cos x + C
C) 8sinx4.1cosx9.3x+C8 \sin x - 4.1 \cos x - 9.3 x + C
D) 8sinx4.1cosx+C- 8 \sin x - 4.1 \cos x + C
E) 8sinx4.1cosx+9.3x+C8 \sin x - 4.1 \cos x + 9.3 x + C
Question
Recall that the total income received from time t=at = a to time t=bt = b from a continuous income stream of R(t)R ( t ) dollars per year is
Total value = TV = abR(t)dt\int _ { a } ^ { b } R ( t ) \mathrm { d } t
Find the total value of the given income stream and also find its future value (at the end of the given interval) using the given interest rate.
R(t)=800,000sin(2πt)R ( t ) = 800,000 \sin ( 2 \pi t ) , 0t50 \leq t \leq 5 , at 9%

A)TV = $0, FV = $72,344.91
B) TV = $0, FV = $327,074.77
C) TV = $1,600,000, FV = $834,722.23
D) TV = $0, FV = $256,372.45
E) none of these
Question
Evaluate the integral. (3.4sec2x+cosx1.43.6ex)dx\int \left( 3.4 \sec ^ { 2 } x + \frac { \cos x } { 1.4 } - 3.6 e ^ { x } \right) \mathrm { d } x

A) 3.4tanx+sinx1.43.6ex+C3.4 \tan x + \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
B) 3.4tanx+sinx1.43.6ex+C\frac { 3.4 } { \tan x } + \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
C) 3.4tanxsinx1.43.6ex+C3.4 \tan x - \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
D) 3.4tanxsinx1.43.6ex+C- \frac { 3.4 } { \tan x } - \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
E) 3.4tanx+sinx1.43.6ex+C- \frac { 3.4 } { \tan x } + \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
Question
Evaluate the integral. (x3+x4)sec2(5x4+4x5)dx\int \left( x ^ { 3 } + x ^ { 4 } \right) \sec ^ { 2 } \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) \mathrm { d } x

A) 15tan(5x4+4x5)+C\frac { 1 } { 5 } \tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
B) 120tan(5x4+4x5)+C\frac { 1 } { 20 } \tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
C) 14tan(5x4+4x5)+C\frac { 1 } { 4 } \tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
D) tan(5x4+4x5)+C\tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
E) 20tan(5x4+4x5)+C20 \tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
Question
Evaluate the integral.
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Question
Decide whether the integral converges. If the integral converges, compute its value. 0+e3xcos(3x)dx\int _ { 0 } ^ { + \infty } e ^ { - 3 x } \cos ( 3 x ) \mathrm { d } x

A) 15\frac { 1 } { 5 }
B) 12\frac { 1 } { 2 }
C) 16\frac { 1 } { 6 }
D) 13\frac { 1 } { 3 }
E) diverges
Question
Evaluate the integral. π4π4tanx dx\int _ { - \frac { \pi } { 4 } } ^ { \frac { \pi } { 4 } } \tan x \mathrm {~d} x

A) lnsec(π4)lnsec(π4)\ln \left| \sec \left( \frac { \pi } { 4 } \right) \right| - \ln \left| \sec \left( - \frac { \pi } { 4 } \right) \right|
B) lnsin(π4)lnsin(π4)\ln \left| \sin \left( \frac { \pi } { 4 } \right) \right| - \ln \left| \sin \left( \frac { \pi } { 4 } \right) \right|
C) sin(π4)sin(π4)\sin \left( - \frac { \pi } { 4 } \right) - \sin \left( \frac { \pi } { 4 } \right)
D) cos(π4)cos(π4)\cos \left( \frac { \pi } { 4 } \right) - \cos \left( - \frac { \pi } { 4 } \right)
E) none of these
Question
Recall that the average of a function f(x)f ( x ) on an interval [a,b][ a , b ] is fˉ=1baabf(x)dx\bar { f } = \frac { 1 } { b - a } \int _ { a } ^ { b } f ( x ) \mathrm { d } x
Calculate the 9-unit moving average of the function.
f(x)=cos(πx18)f ( x ) = \cos \left( \frac { \pi x } { 18 } \right)

A) fˉ(x)=2π(sin(πx18)cos(πx18))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) - \cos \left( \frac { \pi x } { 18 } \right) \right)
B) fˉ(x)=2π(sin(πx18)cos(πx3))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) - \cos \left( \frac { \pi x } { 3 } \right) \right)
C) fˉ(x)=2π(sin(πx18)+cos(πx2))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) + \cos \left( \frac { \pi x } { 2 } \right) \right)
D) fˉ(x)=2π(cos(πx18)sin(πx18))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \cos \left( \frac { \pi x } { 18 } \right) - \sin \left( \frac { \pi x } { 18 } \right) \right)
E) fˉ(x)=2π(sin(πx18)+cos(πx18))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) + \cos \left( \frac { \pi x } { 18 } \right) \right)
Question
Evaluate the integral. sin(4x7)dx\int - \sin ( - 4 x - 7 ) \mathrm { d } x

A) cos(4x+7)4+C\frac { \cos ( 4 x + 7 ) } { 4 } + C
B) 4cos(4x7)+C4 \cos ( 4 x - 7 ) + C
C) cos(4x+7)4+C- \frac { \cos ( 4 x + 7 ) } { 4 } + C
D) 4cos(4x+7)+C- 4 \cos ( 4 x + 7 ) + C
E) cos(4x7)4+C- \frac { \cos ( 4 x - 7 ) } { 4 } + C
Question
Evaluate the integral.
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Use the symbol C to write the constant.
Question
Evaluate the integral.
Evaluate the integral. ​   ​ Use the symbol C to write the constant.<div style=padding-top: 35px>
Use the symbol C to write the constant.
Question
Evaluate the integral 1π3π9sin(1x)x2 dx\int _ { \frac { 1 } { \pi } } ^ { \frac { 3 } { \pi } } 9 \frac { \sin \left( \frac { 1 } { x } \right) } { x ^ { 2 } } \mathrm {~d} x

A)9.5
B) 9
C) 31.5
D) 22.5
E) 13.5
Question
Evaluate the integral. 6sec(3x7)dx\int 6 \sec ( 3 x - 7 ) \mathrm { d } x

A) 2lntan(3x7)+C2 \ln | \tan ( 3 x - 7 ) | + C
B) 2lnsec(3x7)+tan(3x7)+C2 \ln | \sec ( 3 x - 7 ) + \tan ( 3 x - 7 ) | + C
C) 6lnsec(3x7)+tan(3x7)+C6 \ln | \sec ( 3 x - 7 ) + \tan ( 3 x - 7 ) | + C
D) lnsec(3x7)+tan(3x7)+C\ln | \sec ( 3 x - 7 ) + \tan ( 3 x - 7 ) | + C
E) 2lnsec(3x7)+C2 \ln | \sec ( 3 x - 7 ) | + C
Question
Evaluate the integral. 10.8cos(6x1)dx\int 10.8 \cos ( 6 x - 1 ) \mathrm { d } x

A) 10.8sin(6x1)+C10.8 \sin ( 6 x - 1 ) + C
B) 1.8cos(6x1)+C1.8 \cos ( 6 x - 1 ) + C
C) 1.8sin(6x1)+C1.8 \sin ( 6 x - 1 ) + C
D) 1.8sin(6x1)+C- 1.8 \sin ( 6 x - 1 ) + C
E) 1.8sin(6x2x)+C1.8 \sin \left( 6 x ^ { 2 } - x \right) + C
Question
Evaluate the integral. 1π2+14xcos(x21)dx\int _ { 1 } ^ { \sqrt { \frac { \pi } { 2 } + 1 } } 4 x \cos \left( x ^ { 2 } - 1 \right) \mathrm { d } x

A)2
B) 10
C) 1
D) 6
E) 4
Question
Use geometry to compute the given integral. 05π6cosxdx\int _ { 0 } ^ { 5 \pi } 6 \cos x d x

A) 05π6cosxdx=5\int _ { 0 } ^ { 5 \pi } 6 \cos x d x = 5
B) 05π6cosxdx=12\int _ { 0 } ^ { 5 \pi } 6 \cos x d x = 12
C) 05π6cosxdx=1\int _ { 0 } ^ { 5 \pi } 6 \cos x d x = 1
D) 05π6cosxdx=0\int _ { 0 } ^ { 5 \pi } 6 \cos x d x = 0
E) none of these
Question
Calculate the derivative. ddx([lnx][cot(6x+5)])\frac { d } { d x } ( [ \ln | x | ] [ \cot ( 6 x + 5 ) ] )

A) cot(6x5)x6lnxcsc2(6x5)\frac { \cot ( 6 x - 5 ) } { x } - 6 \ln | x | \csc ^ { 2 } ( 6 x - 5 )
B) cot(6x+5)x6lnxcsc2(6x+5)\frac { \cot ( 6 x + 5 ) } { x } - 6 \ln | x | \csc ^ { 2 } ( 6 x + 5 )
C) 6lnxcsc2(6x+5)- 6 \ln | x | \csc ^ { 2 } ( 6 x + 5 )
D) cot(6x+5)x\frac { \cot ( 6 x + 5 ) } { x }
E) tan(6x+5)x6lnxcsc2(6x5)\frac { \tan ( 6 x + 5 ) } { x } - 6 \ln | x | \csc ^ { 2 } ( 6 x - 5 )
Question
Find the derivative of the function. w(x)=2sec(x)tan(x22)w ( x ) = 2 \sec ( x ) \cdot \tan \left( x ^ { 2 } - 2 \right)

A) w(x)=2sec2xsinxtan(x22)+4xsecxsec2(x22)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } - 2 \right) + 4 x \cdot \sec x \cdot \sec ^ { 2 } \left( x ^ { 2 } - 2 \right)
B) w(x)=2sec2xsinxtan(x22)+4xsecxsec2(2x2)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } - 2 \right) + 4 x \cdot \sec x \cdot \sec ^ { 2 } ( 2 x - 2 )
C) w(x)=2sec2xsinxtan(x2+2)+4xsecxsec2(2x2)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } + 2 \right) + 4 x \cdot \sec x \cdot \sec ^ { 2 } ( 2 x - 2 )
D) w(x)=2sec2xsinxtan(x22)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } - 2 \right)
E) w(x)=2sec2xsinxtan(x22)+4xsecxsec2(x24)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } - 2 \right) + 4 x \cdot \sec x \cdot \sec ^ { 2 } \left( x ^ { 2 } - 4 \right)
Question
Calculate the derivative. ddx[sec(6x3x21)]\frac { d } { d x } \left[ \sec \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \right]

A) 6x418x2(x21)2csc(6x3x21)tan(6x3x21)\frac { 6 x ^ { 4 } - 18 x ^ { 2 } } { \left( x ^ { 2 } - 1 \right) ^ { 2 } } \csc \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right)
B) 6x3x21csc(6x3x21)tan(6x3x21)\frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \csc \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right)
C) 6x418x2(x21)2sec(6x3x21)tan(6x3x21)\frac { 6 x ^ { 4 } - 18 x ^ { 2 } } { \left( x ^ { 2 } - 1 \right) ^ { 2 } } \sec \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right)
D) 6x318x2x21sec(6x3x21)tan(6x3x21)\frac { 6 x ^ { 3 } - 18 x ^ { 2 } } { x ^ { 2 } - 1 } \sec \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right)
E) 6x418x2(x2+1)2sec(6x3x21)tan(6x3x2+1)\frac { 6 x ^ { 4 } - 18 x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 2 } } \sec \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } + 1 } \right)
Question
Find the derivative of the function. t(x)=cotx4+8secxt ( x ) = \frac { \cot x } { 4 + 8 \sec x }

A) t(x)=(48secx)csc2x+8secx(48secx)2t ^ { \prime } ( x ) = - \frac { ( 4 - 8 \sec x ) \csc ^ { 2 } x + 8 \sec x } { ( 4 - 8 \sec x ) ^ { 2 } }
B) t(x)=(4+8secx)csc2x+8secx(4+8secx)2t ^ { \prime } ( x ) = - \frac { ( 4 + 8 \sec x ) \csc ^ { 2 } x + 8 \sec x } { ( 4 + 8 \sec x ) ^ { 2 } }
C) t(x)=(4+8secx)csc2x+8secx(4+8secx)2t ^ { \prime } ( x ) = \frac { ( 4 + 8 \sec x ) \csc ^ { 2 } x + 8 \sec x } { ( 4 + 8 \sec x ) ^ { 2 } }
D) t(x)=(4+8secx)csc2x+8secx(48secx)2t ^ { \prime } ( x ) = - \frac { ( 4 + 8 \sec x ) \csc ^ { 2 } x + 8 \sec x } { ( 4 - 8 \sec x ) ^ { 2 } }
E) t(x)=(48secx)csc2x8secx(48secx)2t ^ { \prime } ( x ) = - \frac { ( 4 - 8 \sec x ) \csc ^ { 2 } x - 8 \sec x } { ( 4 - 8 \sec x ) ^ { 2 } }
Question
Calculate the derivative. ddx[sin(7x)]0.5\frac { d } { d x } [ \sin ( 7 x ) ] ^ { 0.5 }

A) 3.5sin(7x)cos(7x)3.5 \sin ( 7 x ) \cdot \cos ( 7 x )
B) 3.5(sin(7x))0.5cos(7x)3.5 ( \sin ( 7 x ) ) ^ { - 0.5 } \cdot \cos ( 7 x )
C) (sin(7x))0.5cos(7x)( \sin ( 7 x ) ) ^ { - 0.5 } \cdot \cos ( 7 x )
D) 7(sin(7x))0.5cos(7x)7 ( \sin ( 7 x ) ) ^ { - 0.5 } \cdot \cos ( 7 x )
E) 7(sin(7x))0.5cos(7x)7 ( \sin ( 7 x ) ) ^ { 0.5 } \cdot \cos ( 7 x )
Question
Find the derivative of the function. g(x)=9sin(x)tanxg ( x ) = 9 \sin ( x ) \cdot \tan x

A) g(x)=9cos(x)tanx+9sin(x)csc2xg' ( x ) = 9 \cos ( x ) \cdot \tan x + 9 \sin ( x ) \cdot \csc ^ { 2 } x
B) g(x)=9cos(x)tanx+9sin(x)sec2xg '( x ) = 9 \cos ( x ) \cdot \tan x + 9 \sin ( x ) \cdot \sec ^ { 2 } x
C) g(x)=9cos(x)tanx9sin(x)cotxg '( x ) = 9 \cos ( x ) \cdot \tan x - 9 \sin ( x ) \cdot \cot x
D) g(x)=9cos(x)tanx+9cos(x)cotxg' ( x ) = 9 \cos ( x ) \cdot \tan x + 9 \cos ( x ) \cdot \cot x
E) g(x)=9cos(x)tanx9sin(x)sec2xg '( x ) = 9 \cos ( x ) \cdot \tan x - 9 \sin ( x ) \cdot \sec ^ { 2 } x
Question
Calculate the derivative. ddx[e3xsin(3πx)]\frac { d } { d x } \left[ e ^ { - 3 x } \sin ( 3 \pi x ) \right]

A) e3x(3sin(3πx)3πcos(3πx))e ^ { 3 x } ( - 3 \sin ( 3 \pi x ) - 3 \pi \cos ( 3 \pi x ) )
B) e3x(3sin(3πx)+3πcos(3πx))e ^ { - 3 x } ( - 3 \sin ( 3 \pi x ) + 3 \pi \cos ( 3 \pi x ) )
C) e3x(3sin(3πx)3πcos(3πx))e ^ { - 3 x } ( - 3 \sin ( 3 \pi x ) - 3 \pi \cos ( 3 \pi x ) )
D) e3x(sin(3πx)+cos(3πx))e ^ { - 3 x } ( \sin ( 3 \pi x ) + \cos ( 3 \pi x ) )
E) e3x(sin(3πx)3πcos(3πx))e ^ { - 3 x } ( \sin ( 3 \pi x ) - 3 \pi \cos ( 3 \pi x ) )
Question
Find the derivative of the function. u(x)=sec(x5.2+2.8x9)u ( x ) = \sec \left( x ^ { 5.2 } + 2.8 x - 9 \right)

A) u(x)=(5.2x4.2+2.8)sec(x5.22.8x9)tan(x5.22.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } + 2.8 \right) \sec \left( x ^ { 5.2 } - 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } - 2.8 x - 9 \right)
B) u(x)=(5.2x4.22.8)sec(x5.2+2.8x9)tan(x5.2+2.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } - 2.8 \right) \sec \left( x ^ { 5.2 } + 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } + 2.8 x - 9 \right)
C) u(x)=(5.2x4.22.8)sec(x5.22.8x9)tan(x5.2+2.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } - 2.8 \right) \sec \left( x ^ { 5.2 } - 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } + 2.8 x - 9 \right)
D) u(x)=(5.2x4.2+2.8)sec(x5.2+2.8x9)tan(x5.2+2.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } + 2.8 \right) \sec \left( x ^ { 5.2 } + 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } + 2.8 x - 9 \right)
E) u(x)=(5.2x4.22.8)sec(x5.22.8x9)tan(x5.22.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } - 2.8 \right) \sec \left( x ^ { 5.2 } - 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } - 2.8 x - 9 \right)
Question
Find the derivative of the function. s(x)=2cos2xs ( x ) = 2 \cos ^ { 2 } x

A) s(x)=4cos(x)cosxs '( x ) = - 4 \cos ( x ) \cdot \cos x
B) s(x)=4cos(x)sinxs '( x ) = - 4 \cos ( x ) \cdot \sin x
C) s(x)=2cos2(x)sinxs ^ { \prime } ( x ) = - 2 \cos ^ { 2 } ( x ) \cdot \sin x
D) s(x)=2cos(x)sinxs' ( x ) = - 2 \cos ( x ) \cdot \sin x
E) s(x)=4cscxs' ( x ) = - 4 \csc x
Question
Find the derivative of the function. y(x)=7cos(ex)+9excosxy ( x ) = 7 \cos \left( e ^ { x } \right) + 9 e ^ { x } \cos x

A) y(x)=ex(7sin(ex)+9cosx9sinx)y' ( x ) = e ^ { x } \left( - 7 \sin \left( e ^ { x } \right) + 9 \cos x - 9 \sin x \right)
B) y(x)=7sin(ex)+9sinx9cosxy' ( x ) = - 7 \sin \left( e ^ { x } \right) + 9 \sin x - 9 \cos x
C) y(x)=7sin(ex)+9cosx9sinxy' ( x ) = - 7 \sin \left( e ^ { x } \right) + 9 \cos x - 9 \sin x
D) y(x)=ex(7sin(ex)9cosx+9sinx)y '( x ) = e ^ { x } \left( - 7 \sin \left( e ^ { x } \right) - 9 \cos x + 9 \sin x \right)
E) y(x)=ex2(7sin(ex)9cosx+9sinx)y ^ { \prime } ( x ) = e ^ { x ^ { 2 } } \left( - 7 \sin \left( e ^ { x } \right) - 9 \cos x + 9 \sin x \right)
Question
Find the derivative of the function. z(x)=3ln4(secx+tanx)z ( x ) = 3 \ln \mid 4 ( \sec x + \tan x )

A) z(x)=9secxz ( x ) = 9 \sec x
B) z(x)=3secx4z ^ { \prime } ( x ) = \frac { 3 \sec x } { 4 }
C) z(x)=3(1+tan2x)z ^ { \prime } ( x ) = 3 \left( 1 + \tan ^ { 2 } x \right)
D) z(x)=secxz ^ { \prime } ( x ) = \sec x
E) z(x)=3secxz ( x ) = 3 \sec x
Question
Find the derivative of the function. r(x)=4xcosx+9x2+8r ( x ) = 4 x \cos x + 9 x ^ { 2 } + 8

A) r(x)=4xcosx+4cosx+18xr'( x ) = - 4 x \cos x + 4 \cos x + 18 x
B) r(x)=4xsinx+4cosx+18xr' ( x ) = - 4 x \sin x + 4 \cos x + 18 x
C) r(x)=4xsinx+4cosx18xr '( x ) = - 4 x \sin x + 4 \cos x - 18 x
D) r(x)=4xsinx+4sinx+18xr '( x ) = - 4 x \sin x + 4 \sin x + 18 x
E) r(x)=4xsinx4cosx+18xr '( x ) = 4 x \sin x - 4 \cos x + 18 x
Question
Find the derivative of the function. f(x)=9sinx8cosxf ( x ) = 9 \sin x - 8 \cos x

A) f(x)=9cosx+8sinxf ^ { \prime } ( x ) = 9 \cos x + 8 \sin x
B) f(x)=8cosx9sinxf ^ { \prime } ( x ) = 8 \cos x - 9 \sin x
C) f(x)=9cosx8sinxf ^ { \prime } ( x ) = 9 \cos x - 8 \sin x
D) f(x)=9cosx8sinxf ^ { \prime } ( x ) = - 9 \cos x - 8 \sin x
E) f(x)=8cosx+9sinxf ^ { \prime } ( x ) = 8 \cos x + 9 \sin x
Question
Decide whether each integral converges. If the integral converges, compute its value.
Choose the correct letter for each question.


-converges to 14\frac { 1 } { 4 }

A) 0+sin(2x)dx\int _ { 0 } ^ { + \infty } \sin ( 2 x ) \mathrm { d } x
B) 0+e2xsin(2x)dx\int _ { 0 } ^ { + \infty } e ^ { - 2 x } \sin ( 2 x ) \mathrm { d } x
Question
Decide whether each integral converges. If the integral converges, compute its value.
Choose the correct letter for each question.


-diverges

A) 0+sin(2x)dx\int _ { 0 } ^ { + \infty } \sin ( 2 x ) \mathrm { d } x
B) 0+e2xsin(2x)dx\int _ { 0 } ^ { + \infty } e ^ { - 2 x } \sin ( 2 x ) \mathrm { d } x
Question
Find the derivative of the function. u(x)=cos(x22x)u ( x ) = \cos \left( x ^ { 2 } - 2 x \right)

A) u(x)=(2x2)cos(x22x)u ^ { \prime } ( x ) = ( 2 x - 2 ) \cos \left( x ^ { 2 } - 2 x \right)
B) u(x)=(2x+2)sin(x22x)u ^ { \prime } ( x ) = - ( 2 x + 2 ) \sin \left( x ^ { 2 } - 2 x \right)
C) u(x)=(2x2)cos(x22x)u ^ { \prime } ( x ) = - ( 2 x - 2 ) \cos \left( x ^ { 2 } - 2 x \right)
D) u(x)=(2x2)sin(x22x)u ^ { \prime } ( x ) = - ( 2 x - 2 ) \sin \left( x ^ { 2 } - 2 x \right)
E) u(x)=(2x+2)sin(x22x)u ^ { \prime } ( x ) = ( 2 x + 2 ) \sin \left( x ^ { 2 } - 2 x \right)
Question
Find the derivative of the function. j(x)=7sec2xj ( x ) = 7 \sec ^ { 2 } x

A) j(x)=14sec2xtanxj '( x ) = 14 \sec ^ { 2 } x \cdot \tan x
B) j(x)=7sec2xtanxj '( x ) = 7 \sec ^ { 2 } x \cdot \tan x
C) j(x)=14sec2xtan2xj '( x ) = 14 \sec ^ { 2 } x \cdot \tan ^ { 2 } x
D) j(x)=14sec2xtanxj ^ { \prime } ( x ) = - 14 \sec ^ { 2 } x \cdot \tan x
E) j(x)=14secxtanxj '( x ) = 14 \sec x \cdot \tan x
Question
Find the derivative of the function. s(x)=(4x22x+2)tanxs ( x ) = \left( 4 x ^ { 2 } - 2 x + 2 \right) \tan x

A) s(x)=(8x2)tanx+(4x22x2)sec2xs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x + \left( 4 x ^ { 2 } - 2 x - 2 \right) \sec ^ { 2 } x
B) s(x)=(8x2)tanx+(4x22x+2)secxs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x + \left( 4 x ^ { 2 } - 2 x + 2 \right) \sec x
C) s(x)=(8x2)tanx+(4x2+2x+2)sec2xs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x + \left( 4 x ^ { 2 } + 2 x + 2 \right) \sec ^ { 2 } x
D) s(x)=(8x2)tanx+(4x22x+2)sec2xs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x + \left( 4 x ^ { 2 } - 2 x + 2 \right) \sec ^ { 2 } x
E) s(x)=(8x2)tanx(4x22x+2)sec2xs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x - \left( 4 x ^ { 2 } - 2 x + 2 \right) \sec ^ { 2 } x
Question
Find the derivative of the function. p(x)=3+2sin(π2(x1))p ( x ) = 3 + 2 \sin \left( \frac { \pi } { 2 } ( x - 1 ) \right)

A) p(x)=πcos(π2(x1))p ^ { \prime } ( x ) = \pi \cos \left( \frac { \pi } { 2 } ( x - 1 ) \right)
B) p(x)=2cos(π2(x1))p ^ { \prime } ( x ) = 2 \cos \left( \frac { \pi } { 2 } ( x - 1 ) \right)
C) p(x)=πcos(π2(x+1))p ^ { \prime } ( x ) = \pi \cos \left( \frac { \pi } { 2 } ( x + 1 ) \right)
D) p(x)=πsin(π2(x1))p ^ { \prime } ( x ) = \pi \sin \left( \frac { \pi } { 2 } ( x - 1 ) \right)
E) p(x)=3+πcos(π2(x1))p ^ { \prime } ( x ) = 3 + \pi \cos \left( \frac { \pi } { 2 } ( x - 1 ) \right)
Question
Find the derivative of the function. y(x)=sec(e4x)y ( x ) = \sec \left( e ^ { 4 x } \right)

A) y(x)=4e4xcsc(e4x)tan2(e4x)y ^ { \prime } ( x ) = 4 e ^ { 4 x } \csc \left( e ^ { 4 x } \right) \tan ^ { 2 } \left( e ^ { 4 x } \right)
B) y(x)=16e4xsec(e4x)tan(e4x)y ^ { \prime } ( x ) = 16 e ^ { 4 x } \sec \left( e ^ { 4 x } \right) \tan \left( e ^ { 4 x } \right)
C) y(x)=4e4xsec(e4x)tan(e4x)y ^ { \prime } ( x ) = 4 e ^ { 4 x } \sec \left( e ^ { 4 x } \right) \tan \left( e ^ { 4 x } \right)
D) y(x)=4e4xcsc2(e4x)tan2(e4x)y ^ { \prime } ( x ) = 4 e ^ { 4 x } \csc ^ { 2 } \left( e ^ { 4 x } \right) \tan ^ { 2 } \left( e ^ { 4 x } \right)
E) y(x)=4e4xcsc(e4x)tan(e4x)y ^ { \prime } ( x ) = 4 e ^ { 4 x } \csc \left( e ^ { 4 x } \right) \tan \left( e ^ { 4 x } \right)
Question
Model the curve with a cosine function.  <strong>Model the curve with a cosine function.    Note that the period of the curve is  P = \frac { 1 } { 6 }  and its range is  [ - 1,1 ]  . </strong> A)  f ( x ) = \cos ( 12 x )  B)  f ( x ) = \cos \left( \frac { \pi x } { 12 } \right)  C)  f ( x ) = \cos ( 12 \pi x )  D)  f ( x ) = 12 \cos ( \pi x )  E)  f ( x ) = \cos \left( \frac { x } { 12 } \right)  <div style=padding-top: 35px>
Note that the period of the curve is P=16P = \frac { 1 } { 6 } and its range is [1,1][ - 1,1 ] .

A) f(x)=cos(12x)f ( x ) = \cos ( 12 x )
B) f(x)=cos(πx12)f ( x ) = \cos \left( \frac { \pi x } { 12 } \right)
C) f(x)=cos(12πx)f ( x ) = \cos ( 12 \pi x )
D) f(x)=12cos(πx)f ( x ) = 12 \cos ( \pi x )
E) f(x)=cos(x12)f ( x ) = \cos \left( \frac { x } { 12 } \right)
Question
The cost of Dig-It brand snow shovels is given by c(t)=3sin(2π(t0.75))c ( t ) = 3 \sin ( 2 \pi ( t - 0.75 ) )
Where t is time in years since January 1, 1997. How fast, in dollars per year, is the cost increasing on October 30, 1997

A)$21.85 per year
B) $18.85 per year
C) $9.42 per year
D) $20.85 per year
E) $6.00 per year
Question
Starting with the identity sin2x+cos2x=1\sin ^ { 2 } x + \cos ^ { 2 } x = 1 , choose the right trigonometric identity.

A) sec2x=1+tan2x\sec ^ { 2 } x = 1 + \tan ^ { 2 } x
B) sec2x=1tan2x\sec ^ { 2 } x = 1 - \tan ^ { 2 } x
C) cot2x=1tan2x\cot ^ { 2 } x = 1 - \tan ^ { 2 } x
D) sin2x=1+tan2x\sin ^ { 2 } x = 1 + \tan ^ { 2 } x
E) sec2x=1+csc2x\sec ^ { 2 } x = 1 + \csc ^ { 2 } x
Question
Sketch the curves without any technological help. f(t)=2costf ( t ) = 2 \cos t ; g(t)=3.3cos(2t)g ( t ) = 3.3 \cos ( 2 t )

A)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)     <div style=padding-top: 35px>
B)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)     <div style=padding-top: 35px>
C)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)     <div style=padding-top: 35px>
D)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)     <div style=padding-top: 35px>
E)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)     <div style=padding-top: 35px>
Question
Sketch the curves without any technological help. f(t)=costf ( t ) = \cos t ; g(t)=cos(t+π)g ( t ) = \cos ( t + \pi )

A)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)    <div style=padding-top: 35px>
B)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)    <div style=padding-top: 35px>
C)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)    <div style=padding-top: 35px>
D)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)    <div style=padding-top: 35px>
E)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)    <div style=padding-top: 35px>
Question
Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function s(t)=0.468sin(1.25t+1.73)+0.79(1t6)s ( t ) = 0.468 \sin ( 1.25 t + 1.73 ) + 0.79 ( 1 \leq t \leq 6 )
Where t is time in quarters ( t=1t = 1 represents the end of the first quarter of 1995) and s(t)s ( t ) is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales.

A) smax=1.526s _ { \max } = 1.526 , smin=0.074s _ { \min } = 0.074
B) smax=0.79s _ { \max } = 0.79 , smin=0.468s _ { \min } = 0.468
C) smax=2.048s _ { \max } = 2.048 , smin=1.112s _ { \min } = 1.112
D) smax=2.194s _ { \max } = 2.194 , smin=0.146s _ { \min } = 0.146
E) smax=1.258s _ { \max } = 1.258 , smin=0.322s _ { \min } = 0.322
Question
Model the curve with a sine function.  <strong>Model the curve with a sine function.    Note that the period of the curve is  P = 0.4  and its range is  [ - 3 , - 1 ]  . </strong> A)  f ( x ) = 2 - \sin x  B)  f ( x ) = - 2 + 5 \sin x  C)   f ( x ) = - 2 + \sin ( 5 \pi x )  D)  f ( x ) = 2 - \sin ( 5 \pi x )  E)  f ( x ) = - 2 + \sin ( \pi x )  <div style=padding-top: 35px>
Note that the period of the curve is P=0.4P = 0.4 and its range is [3,1][ - 3 , - 1 ] .

A) f(x)=2sinxf ( x ) = 2 - \sin x
B) f(x)=2+5sinxf ( x ) = - 2 + 5 \sin x
C) f(x)=2+sin(5πx)f ( x ) = - 2 + \sin ( 5 \pi x )
D) f(x)=2sin(5πx)f ( x ) = 2 - \sin ( 5 \pi x )
E) f(x)=2+sin(πx)f ( x ) = - 2 + \sin ( \pi x )
Question
Use the addition formulas: sin(x+y)=sinxcosy+cosxsinysin(xy)=sinxcosycosxsinycos(x+y)=cosxcosysinxsinycos(xy)=cosxcosy+sinxsiny\begin{array} { l } \sin ( x + y ) = \sin x \cdot \cos y + \cos x \cdot \sin y \\\sin ( x - y ) = \sin x \cdot \cos y - \cos x \cdot \sin y \\\cos ( x + y ) = \cos x \cdot \cos y - \sin x \cdot \sin y \\\cos ( x - y ) = \cos x \cdot \cos y + \sin x \cdot \sin y\end{array}
To express tan(x+23π)\tan ( x + 23 \pi ) in terms of tan(x)\tan ( x ) .

A) tan(x)23π\frac { \tan ( x ) } { 23 \pi }
B) tan(x)+23π\tan ( x ) + 23 \pi
C) 23πtan(x)23 \pi \tan ( x )
D) tan(x)\tan ( x )
E) tan(x)23π\tan ( x ) - 23 \pi
Question
Use the formula for sin(x+y)\sin ( x + y ) to simplify the expression sin(t+17π2)\sin \left( t + \frac { 17 \pi } { 2 } \right) .

A) cost17π\cos t - 17 \pi
B) cost+17π\cos t + 17 \pi
C) cost17π\frac { \cos t } { 17 \pi }
D) 17πcost17 \pi \cos t
E) cost\cos t
Question
Use the addition formulas: sin(x+y)=sinxcosy+cosxsinysin(xy)=sinxcosycosxsinycos(x+y)=cosxcosysinxsinycos(xy)=cosxcosy+sinxsiny\begin{array} { l } \sin ( x + y ) = \sin x \cdot \cos y + \cos x \cdot \sin y \\\sin ( x - y ) = \sin x \cdot \cos y - \cos x \cdot \sin y \\\cos ( x + y ) = \cos x \cdot \cos y - \sin x \cdot \sin y \\\cos ( x - y ) = \cos x \cdot \cos y + \sin x \cdot \sin y\end{array}
To calculate cos(π3)\cos \left( \frac { \pi } { 3 } \right) , given that sin(π6)=12\sin \left( \frac { \pi } { 6 } \right) = \frac { 1 } { 2 } and cos(π6)=32\cos \left( \frac { \pi } { 6 } \right) = \frac { \sqrt { 3 } } { 2 } .

A) cos(π3)=0\cos \left( \frac { \pi } { 3 } \right) = 0
B) cos(π3)=32\cos \left( \frac { \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 2 }
C) cos(π3)=12\cos \left( \frac { \pi } { 3 } \right) = \frac { 1 } { 2 }
D) cos(π3)=32\cos \left( \frac { \pi } { 3 } \right) = - \frac { \sqrt { 3 } } { 2 }
E) cos(π3)=12\cos \left( \frac { \pi } { 3 } \right) = - \frac { 1 } { 2 }
Question
Use the conversion formula cosx=sin(π2x)\cos x = \sin \left( \frac { \pi } { 2 } - x \right) to replace the expression g(x)=25cos[2π(3x1)]+9g ( x ) = 25 \cos [ 2 \pi ( 3 x - 1 ) ] + 9
By a sine function.

A) g(x)=25sin(6πx+5π2)9+π2g ( x ) = 25 \sin \left( 6 \pi x + \frac { 5 \pi } { 2 } \right) - 9 + \frac { \pi } { 2 }
B) g(x)=25sin(6πx5π2)+9g ( x ) = 25 \sin \left( 6 \pi x - \frac { 5 \pi } { 2 } \right) + 9
C) g(x)=25sin(6πx5π2)+9g ( x ) = 25 \sin \left( - 6 \pi x - \frac { 5 \pi } { 2 } \right) + 9
D) g(x)=25sin(6πx+5π2)+9g ( x ) = 25 \sin \left( - 6 \pi x + \frac { 5 \pi } { 2 } \right) + 9
E) g(x)=25sin(12πx+3π2)+9g ( x ) = 25 \sin \left( 12 \pi x + \frac { 3 \pi } { 2 } \right) + 9
Question
Use the conversion formula cosx=sin(π2x)\cos x = \sin \left( \frac { \pi } { 2 } - x \right) to replace the expression g(t)=45cos(t5)g ( t ) = 45 - \cos ( t - 5 )
By a sine function.

A) g(t)=45sin(π2t+5)g ( t ) = 45 - \sin \left( \frac { \pi } { 2 } - t + 5 \right)
B) g(t)=45sin(π2t5)g ( t ) = 45 - \sin \left( \frac { \pi } { 2 } - t - 5 \right)
C) g(t)=45sin(πt+5)g ( t ) = 45 - \sin ( \pi - t + 5 )
D) g(t)=45sin(π2+t+5)g ( t ) = 45 - \sin \left( \frac { \pi } { 2 } + t + 5 \right)
E) g(t)=π245sin(t+5)g ( t ) = \frac { \pi } { 2 } - 45 - \sin ( t + 5 )
Question
Model the curve with a cosine function.  <strong>Model the curve with a cosine function.    Note that the period of the curve is  P = 14  , its range is  [ 0,120 ]  the graph of the cosine function is shifted upward 60 units and shifted to the right 7 units. </strong> A)  f ( x ) = 120 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) + 7  B)  f ( x ) = 120 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) - 7  C)  f ( x ) = 60 \cos \left( \frac { \pi ( x + 7 ) } { 7 } \right) + 60  D)  f ( x ) = 7 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) + 7  E)  f ( x ) = 60 \cos \left( \frac { \pi ( x - 7 ) } { 7 } \right) + 60  <div style=padding-top: 35px>
Note that the period of the curve is P=14P = 14 , its range is [0,120][ 0,120 ] the graph of the cosine function is shifted upward 60 units and shifted to the right 7 units.

A) f(x)=120cos(π(x60)60)+7f ( x ) = 120 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) + 7
B) f(x)=120cos(π(x60)60)7f ( x ) = 120 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) - 7
C) f(x)=60cos(π(x+7)7)+60f ( x ) = 60 \cos \left( \frac { \pi ( x + 7 ) } { 7 } \right) + 60
D) f(x)=7cos(π(x60)60)+7f ( x ) = 7 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) + 7
E) f(x)=60cos(π(x7)7)+60f ( x ) = 60 \cos \left( \frac { \pi ( x - 7 ) } { 7 } \right) + 60
Question
Use the conversion formula cosx=sin(π2x)\cos x = \sin \left( \frac { \pi } { 2 } - x \right) to replace the expression f(t)=5.2cos(6πt)+10f ( t ) = 5.2 \cos ( 6 \pi t ) + 10
By a sine function.

A) f(t)=5.2sin(π26πt)+10f ( t ) = 5.2 \sin \left( \frac { \pi } { 2 } - 6 \pi t \right) + 10
B) f(t)=5.2sin(π6πt2)+10f ( t ) = 5.2 \sin \left( \frac { \pi - 6 \pi t } { 2 } \right) + 10
C) f(t)=5.2sin(π26t)+10f ( t ) = 5.2 \sin \left( \frac { \pi } { 2 } - 6 t \right) + 10
D) f(t)=6sin(π25.2πt)+10f ( t ) = 6 \sin \left( \frac { \pi } { 2 } - 5.2 \pi t \right) + 10
E) f(t)=10sin(π26πt)+5.2f ( t ) = 10 \sin \left( \frac { \pi } { 2 } - 6 \pi t \right) + 5.2
Question
Model the curve with a sine function.  <strong>Model the curve with a sine function.    Note that the period of the curve is  P = 32  and its range is  [ - 40,0 ]  , the graph of the sine function is shifted to the right 7 units. </strong> A)  f ( x ) = 20 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) + 20  B)  f ( x ) = - 20 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) + 20  C)  f ( x ) = 40 \sin \left( \frac { \pi ( x - 7 ) } { 16 } \right) - 40  D)  f ( x ) = 40 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) - 40  E)  f ( x ) = 20 \sin \left( \frac { \pi ( x - 7 ) } { 16 } \right) - 20  <div style=padding-top: 35px>
Note that the period of the curve is P=32P = 32 and its range is [40,0][ - 40,0 ] , the graph of the sine function is shifted to the right 7 units.

A) f(x)=20sin(π(x+7)16)+20f ( x ) = 20 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) + 20
B) f(x)=20sin(π(x+7)16)+20f ( x ) = - 20 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) + 20
C) f(x)=40sin(π(x7)16)40f ( x ) = 40 \sin \left( \frac { \pi ( x - 7 ) } { 16 } \right) - 40
D) f(x)=40sin(π(x+7)16)40f ( x ) = 40 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) - 40
E) f(x)=20sin(π(x7)16)20f ( x ) = 20 \sin \left( \frac { \pi ( x - 7 ) } { 16 } \right) - 20
Question
Model the curve with a sine function.
 <strong>Model the curve with a sine function.   Note that the period of the curve is  P = \frac { 1 } { 5 }  and its range is  [ - 2.2,2.2 ]  and the graph of the sine function is shifted to the left 0.55 units. </strong> A)  f ( x ) = 2.2 \sin ( 10 \pi ( x + 0.55 ) )  B)  f ( x ) = 2.2 \sin ( 10 \pi ( x - 0.55 ) )  C)  f ( x ) = 2.2 \sin ( 10 \pi x + 0.55 )  D)  f ( x ) = 2.2 \sin ( 10 \pi ( 2 x + 0.55 ) )  E)  f ( x ) = 4.4 \sin ( 5 \pi ( x + 0.55 ) )  <div style=padding-top: 35px>
Note that the period of the curve is P=15P = \frac { 1 } { 5 } and its range is [2.2,2.2][ - 2.2,2.2 ] and the graph of the sine function is shifted to the left 0.55 units.

A) f(x)=2.2sin(10π(x+0.55))f ( x ) = 2.2 \sin ( 10 \pi ( x + 0.55 ) )
B) f(x)=2.2sin(10π(x0.55))f ( x ) = 2.2 \sin ( 10 \pi ( x - 0.55 ) )
C) f(x)=2.2sin(10πx+0.55)f ( x ) = 2.2 \sin ( 10 \pi x + 0.55 )
D) f(x)=2.2sin(10π(2x+0.55))f ( x ) = 2.2 \sin ( 10 \pi ( 2 x + 0.55 ) )
E) f(x)=4.4sin(5π(x+0.55))f ( x ) = 4.4 \sin ( 5 \pi ( x + 0.55 ) )
Question
Calculate the derivative.
Calculate the derivative. ​  <div style=padding-top: 35px>
Question
The depth of water d(t)d ( t ) at my favorite surfing spot varies from 8 to 20 feet, depending on the time. Last Sunday high tide occurred at 5:00 A.M. and the next high tide occurred at 6:30 P.M. Use a sine function to model the depth of water as a function of time t in hours since midnight on Sunday morning.

A) d(t)=10sin(2π(t1.625)13.5)+4d ( t ) = 10 \sin \left( \frac { 2 \pi ( t - 1.625 ) } { 13.5 } \right) + 4
B) d(t)=6sin(2π(t+1.625)11.5)14d ( t ) = 6 \sin \left( \frac { 2 \pi ( t + 1.625 ) } { 11.5 } \right) - 14
C) d(t)=6sin(2π(t+1.625)13.5)+14d ( t ) = - 6 \sin \left( \frac { - 2 \pi ( t + 1.625 ) } { 13.5 } \right) + 14
D) d(t)=6sin(2π(t1.625)13.5)+14d ( t ) = 6 \sin \left( \frac { 2 \pi ( t - 1.625 ) } { 13.5 } \right) + 14
E) d(t)=14sin(2π(t1.625)13.5)+6d ( t ) = 14 \sin \left( \frac { 2 \pi ( t - 1.625 ) } { 13.5 } \right) + 6
Question
Model the curve with a cosine function.  <strong>Model the curve with a cosine function.     Note that the period of the curve is  P = \frac { 1 } { 5 }  , its range is  [ - 3.3,3.3 ]  and the graph of the cosine function is shifted to the right 0.35 units. </strong> A)  f ( x ) = 6.6 \cos ( 20 \pi ( 2 x - 0.35 ) )  B)  f ( x ) = 3.3 \cos ( 10 ( x - 0.35 ) )  C)  f ( x ) = 6.6 \cos ( 20 \pi ( 2 x + 0.35 ) )  D)  f ( x ) = 3.3 \cos ( 10 \pi ( x - 0.35 ) )  E)  f ( x ) = 3.3 \cos ( 10 \pi ( x + 0.35 ) )  <div style=padding-top: 35px>
Note that the period of the curve is P=15P = \frac { 1 } { 5 } , its range is [3.3,3.3][ - 3.3,3.3 ] and the graph of the cosine function is shifted to the right 0.35 units.

A) f(x)=6.6cos(20π(2x0.35))f ( x ) = 6.6 \cos ( 20 \pi ( 2 x - 0.35 ) )
B) f(x)=3.3cos(10(x0.35))f ( x ) = 3.3 \cos ( 10 ( x - 0.35 ) )
C) f(x)=6.6cos(20π(2x+0.35))f ( x ) = 6.6 \cos ( 20 \pi ( 2 x + 0.35 ) )
D) f(x)=3.3cos(10π(x0.35))f ( x ) = 3.3 \cos ( 10 \pi ( x - 0.35 ) )
E) f(x)=3.3cos(10π(x+0.35))f ( x ) = 3.3 \cos ( 10 \pi ( x + 0.35 ) )
Question
Model the curve with a cosine function.  <strong>Model the curve with a cosine function.    Note that the period of the curve is  P = \frac { 1 } { 3 }  and its range is  [ - 1,1 ]  . </strong> A)  f ( x ) = \cos ( 6 x )  B)  f ( x ) = \cos ( 6 \pi x )  C)  f ( x ) = \cos \left( \frac { x } { 6 } \right)  D)  f ( x ) = \cos \left( \frac { \pi x } { 6 } \right)  E)  f ( x ) = 6 \cos ( \pi x )  <div style=padding-top: 35px>
Note that the period of the curve is P=13P = \frac { 1 } { 3 } and its range is [1,1][ - 1,1 ] .

A) f(x)=cos(6x)f ( x ) = \cos ( 6 x )
B) f(x)=cos(6πx)f ( x ) = \cos ( 6 \pi x )
C) f(x)=cos(x6)f ( x ) = \cos \left( \frac { x } { 6 } \right)
D) f(x)=cos(πx6)f ( x ) = \cos \left( \frac { \pi x } { 6 } \right)
E) f(x)=6cos(πx)f ( x ) = 6 \cos ( \pi x )
Question
Starting with the identity Starting with the identity   and then dividing both sides of the equation by a suitable trigonometric function, derive the trigonometric identity. ​  <div style=padding-top: 35px> and then dividing both sides of the equation by a suitable trigonometric function, derive the trigonometric identity.
Starting with the identity   and then dividing both sides of the equation by a suitable trigonometric function, derive the trigonometric identity. ​  <div style=padding-top: 35px>
Question
The uninflated cost of Dugout brand snow shovels currently varies from a high of $30 on January 1 (t=0)( t = 0 ) to a low of $6 on July 1 (t=0.5)( t = 0.5 ) . Assuming this trend were to continue indefinitely, calculate the uninflated cost u(t)u ( t ) of Dugout snow shovels as a function of time t in years. (Use a sine function.)

A) u(t)=15sin(2π(t0.75))+3u ( t ) = 15 \sin ( 2 \pi ( t - 0.75 ) ) + 3
B) u(t)=12sin(2π(t0.75))+18u ( t ) = 12 \sin ( 2 \pi ( t - 0.75 ) ) + 18
C) u(t)=18sin(2π(t0.75))+12u ( t ) = 18 \sin ( 2 \pi ( t - 0.75 ) ) + 12
D) u(t)=12sin(2π(t0.75))18u ( t ) = - 12 \sin ( 2 \pi ( t - 0.75 ) ) - 18
E) u(t)=12sin(2π(t+0.75))18u ( t ) = 12 \sin ( 2 \pi ( t + 0.75 ) ) - 18
Question
The depth of water The depth of water   at my favorite surfing spot varies from 5 to 15 feet, depending on the time. Last Sunday high tide occurred at 5:00 A.M. and the next high tide occurred at 6:30 P.M. Use a sine function to model to the depth of water as a function of time t in hours since midnight in Sunday morning.<div style=padding-top: 35px> at my favorite surfing spot varies from 5 to 15 feet, depending on the time. Last Sunday high tide occurred at 5:00 A.M. and the next high tide occurred at 6:30 P.M. Use a sine function to model to the depth of water as a function of time t in hours since midnight in Sunday morning.
Question
Model the curve with a sine function.
Model the curve with a sine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the sine function is shifted to the left 0.9 units. Write the model function as a function of x and π.<div style=padding-top: 35px>
Note that the period of the curve is Model the curve with a sine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the sine function is shifted to the left 0.9 units. Write the model function as a function of x and π.<div style=padding-top: 35px> , its range is Model the curve with a sine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the sine function is shifted to the left 0.9 units. Write the model function as a function of x and π.<div style=padding-top: 35px> and the graph of the sine function is shifted to the left 0.9 units. Write the model function as a function of x and π.
Question
Model the curve with a cosine function.
Model the curve with a cosine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the cosine function is shifted upward 55 units and shifted to the right 14 units. Write the model function as a function of x and π.<div style=padding-top: 35px>
Note that the period of the curve is Model the curve with a cosine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the cosine function is shifted upward 55 units and shifted to the right 14 units. Write the model function as a function of x and π.<div style=padding-top: 35px> , its range is Model the curve with a cosine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the cosine function is shifted upward 55 units and shifted to the right 14 units. Write the model function as a function of x and π.<div style=padding-top: 35px> and the graph of the cosine function is shifted upward 55 units and shifted to the right 14 units. Write the model function as a function of x and π.
Question
Model the curve with a sine function.

Model the curve with a sine function. ​ ​   ​ Note that the period of the curve is   and its range is   . Write the model function as a function of x and π.<div style=padding-top: 35px>
Note that the period of the curve is Model the curve with a sine function. ​ ​   ​ Note that the period of the curve is   and its range is   . Write the model function as a function of x and π.<div style=padding-top: 35px> and its range is Model the curve with a sine function. ​ ​   ​ Note that the period of the curve is   and its range is   . Write the model function as a function of x and π.<div style=padding-top: 35px> . Write the model function as a function of x and π.
Question
Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function
Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function ​   ​ where ​t is time in quarters (   represents the end of the first quarter of 1995) and   is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales. ​ Maximum sales __________ billions of dollars ​ Minimum sales __________ billions of dollars<div style=padding-top: 35px>
where ​t is time in quarters ( Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function ​   ​ where ​t is time in quarters (   represents the end of the first quarter of 1995) and   is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales. ​ Maximum sales __________ billions of dollars ​ Minimum sales __________ billions of dollars<div style=padding-top: 35px> represents the end of the first quarter of 1995) and Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function ​   ​ where ​t is time in quarters (   represents the end of the first quarter of 1995) and   is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales. ​ Maximum sales __________ billions of dollars ​ Minimum sales __________ billions of dollars<div style=padding-top: 35px> is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales.

Maximum sales __________ billions of dollars

Minimum sales __________ billions of dollars
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Deck 16: Trigonometric Models
1
Evaluate the integral. (10x+25)sin(x2+5x)dx\int ( 10 x + 25 ) \sin \left( x ^ { 2 } + 5 x \right) \mathrm { d } x

A) 5xsin(x2+5x)+C- 5 x \sin \left( x ^ { 2 } + 5 x \right) + C
B) 5cos(x2+5x)+C5 \cos \left( x ^ { 2 } + 5 x \right) + C
C) 5xcos(x2+5x)+C5 x \cos \left( x ^ { 2 } + 5 x \right) + C
D) 5cos(x2+5x)+C- 5 \cos \left( x ^ { 2 } + 5 x \right) + C
E) 5sin(x2+5x)+C5 \sin \left( x ^ { 2 } + 5 x \right) + C
5cos(x2+5x)+C- 5 \cos \left( x ^ { 2 } + 5 x \right) + C
2
Recall that the average of a function f(x)f ( x ) on an interval [a,b][ a , b ] is fˉ=1baabf(x)dx\bar { f } = \frac { 1 } { b - a } \int _ { a } ^ { b } f ( x ) \mathrm { d } x
Find the average of the given function.
f(x)=sin(5x)f ( x ) = \sin ( 5 x ) over [0,5π][ 0,5 \pi ]

A)Average = 225π\frac { 2 } { 25 \pi }
B) Average = 25π2\frac { 25 \pi } { 2 }
C) Average = π5\frac { \pi } { 5 }
D) Average = 25π4\frac { 25 \pi } { 4 }
E) Average = 5π\frac { 5 } { \pi }
Average = 225π\frac { 2 } { 25 \pi }
3
Use geometry to compute the given integral. π6π62sinxdx\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x d x

A) π6π62sinxdx=6\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x d x = 6
B) π6π62sinx dx=4\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x \mathrm {~d} x = - 4
C) π6π62sinxdx=0\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x d x = 0
D) π6π62sinxdx=8\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x d x = 8
E) none of these
π6π62sinxdx=0\int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 6 } } 2 \sin x d x = 0
4
Evaluate the integral. π3π4sinx dx\int _ { - \frac { \pi } { 3 } } ^ { \frac { \pi } { 4 } } \sin x \mathrm {~d} x

A) cosπ4cosπ\cos \frac { \pi } { 4 } - \cos \pi
B) sinπ3+sinπ4\sin \frac { \pi } { 3 } + \sin \frac { \pi } { 4 }
C) sinπ3cosπ5\sin \frac { \pi } { 3 } - \cos \frac { \pi } { 5 }
D) cosπ3cosπ4\cos \frac { \pi } { 3 } - \cos \frac { \pi } { 4 }
E) none of these
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5
Evaluate the integral. ? (8cosx4.1sinx9.3)dx\int ( 8 \cos x - 4.1 \sin x - 9.3 ) \mathrm { d } x ?

A) 8sinx+4.1cosx9.3x+C8 \sin x + 4.1 \cos x - 9.3 x + C
B) 8sinx+4.1cosx+C- 8 \sin x + 4.1 \cos x + C
C) 8sinx4.1cosx9.3x+C8 \sin x - 4.1 \cos x - 9.3 x + C
D) 8sinx4.1cosx+C- 8 \sin x - 4.1 \cos x + C
E) 8sinx4.1cosx+9.3x+C8 \sin x - 4.1 \cos x + 9.3 x + C
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6
Recall that the total income received from time t=at = a to time t=bt = b from a continuous income stream of R(t)R ( t ) dollars per year is
Total value = TV = abR(t)dt\int _ { a } ^ { b } R ( t ) \mathrm { d } t
Find the total value of the given income stream and also find its future value (at the end of the given interval) using the given interest rate.
R(t)=800,000sin(2πt)R ( t ) = 800,000 \sin ( 2 \pi t ) , 0t50 \leq t \leq 5 , at 9%

A)TV = $0, FV = $72,344.91
B) TV = $0, FV = $327,074.77
C) TV = $1,600,000, FV = $834,722.23
D) TV = $0, FV = $256,372.45
E) none of these
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7
Evaluate the integral. (3.4sec2x+cosx1.43.6ex)dx\int \left( 3.4 \sec ^ { 2 } x + \frac { \cos x } { 1.4 } - 3.6 e ^ { x } \right) \mathrm { d } x

A) 3.4tanx+sinx1.43.6ex+C3.4 \tan x + \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
B) 3.4tanx+sinx1.43.6ex+C\frac { 3.4 } { \tan x } + \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
C) 3.4tanxsinx1.43.6ex+C3.4 \tan x - \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
D) 3.4tanxsinx1.43.6ex+C- \frac { 3.4 } { \tan x } - \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
E) 3.4tanx+sinx1.43.6ex+C- \frac { 3.4 } { \tan x } + \frac { \sin x } { 1.4 } - 3.6 e ^ { x } + C
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8
Evaluate the integral. (x3+x4)sec2(5x4+4x5)dx\int \left( x ^ { 3 } + x ^ { 4 } \right) \sec ^ { 2 } \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) \mathrm { d } x

A) 15tan(5x4+4x5)+C\frac { 1 } { 5 } \tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
B) 120tan(5x4+4x5)+C\frac { 1 } { 20 } \tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
C) 14tan(5x4+4x5)+C\frac { 1 } { 4 } \tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
D) tan(5x4+4x5)+C\tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
E) 20tan(5x4+4x5)+C20 \tan \left( 5 x ^ { 4 } + 4 x ^ { 5 } \right) + C
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9
Evaluate the integral.
Evaluate the integral. ​
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10
Decide whether the integral converges. If the integral converges, compute its value. 0+e3xcos(3x)dx\int _ { 0 } ^ { + \infty } e ^ { - 3 x } \cos ( 3 x ) \mathrm { d } x

A) 15\frac { 1 } { 5 }
B) 12\frac { 1 } { 2 }
C) 16\frac { 1 } { 6 }
D) 13\frac { 1 } { 3 }
E) diverges
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11
Evaluate the integral. π4π4tanx dx\int _ { - \frac { \pi } { 4 } } ^ { \frac { \pi } { 4 } } \tan x \mathrm {~d} x

A) lnsec(π4)lnsec(π4)\ln \left| \sec \left( \frac { \pi } { 4 } \right) \right| - \ln \left| \sec \left( - \frac { \pi } { 4 } \right) \right|
B) lnsin(π4)lnsin(π4)\ln \left| \sin \left( \frac { \pi } { 4 } \right) \right| - \ln \left| \sin \left( \frac { \pi } { 4 } \right) \right|
C) sin(π4)sin(π4)\sin \left( - \frac { \pi } { 4 } \right) - \sin \left( \frac { \pi } { 4 } \right)
D) cos(π4)cos(π4)\cos \left( \frac { \pi } { 4 } \right) - \cos \left( - \frac { \pi } { 4 } \right)
E) none of these
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12
Recall that the average of a function f(x)f ( x ) on an interval [a,b][ a , b ] is fˉ=1baabf(x)dx\bar { f } = \frac { 1 } { b - a } \int _ { a } ^ { b } f ( x ) \mathrm { d } x
Calculate the 9-unit moving average of the function.
f(x)=cos(πx18)f ( x ) = \cos \left( \frac { \pi x } { 18 } \right)

A) fˉ(x)=2π(sin(πx18)cos(πx18))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) - \cos \left( \frac { \pi x } { 18 } \right) \right)
B) fˉ(x)=2π(sin(πx18)cos(πx3))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) - \cos \left( \frac { \pi x } { 3 } \right) \right)
C) fˉ(x)=2π(sin(πx18)+cos(πx2))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) + \cos \left( \frac { \pi x } { 2 } \right) \right)
D) fˉ(x)=2π(cos(πx18)sin(πx18))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \cos \left( \frac { \pi x } { 18 } \right) - \sin \left( \frac { \pi x } { 18 } \right) \right)
E) fˉ(x)=2π(sin(πx18)+cos(πx18))\bar { f } ( x ) = \frac { 2 } { \pi } \left( \sin \left( \frac { \pi x } { 18 } \right) + \cos \left( \frac { \pi x } { 18 } \right) \right)
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13
Evaluate the integral. sin(4x7)dx\int - \sin ( - 4 x - 7 ) \mathrm { d } x

A) cos(4x+7)4+C\frac { \cos ( 4 x + 7 ) } { 4 } + C
B) 4cos(4x7)+C4 \cos ( 4 x - 7 ) + C
C) cos(4x+7)4+C- \frac { \cos ( 4 x + 7 ) } { 4 } + C
D) 4cos(4x+7)+C- 4 \cos ( 4 x + 7 ) + C
E) cos(4x7)4+C- \frac { \cos ( 4 x - 7 ) } { 4 } + C
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14
Evaluate the integral.
Evaluate the integral. ​   ​ Use the symbol C to write the constant.
Use the symbol C to write the constant.
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15
Evaluate the integral.
Evaluate the integral. ​   ​ Use the symbol C to write the constant.
Use the symbol C to write the constant.
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16
Evaluate the integral 1π3π9sin(1x)x2 dx\int _ { \frac { 1 } { \pi } } ^ { \frac { 3 } { \pi } } 9 \frac { \sin \left( \frac { 1 } { x } \right) } { x ^ { 2 } } \mathrm {~d} x

A)9.5
B) 9
C) 31.5
D) 22.5
E) 13.5
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17
Evaluate the integral. 6sec(3x7)dx\int 6 \sec ( 3 x - 7 ) \mathrm { d } x

A) 2lntan(3x7)+C2 \ln | \tan ( 3 x - 7 ) | + C
B) 2lnsec(3x7)+tan(3x7)+C2 \ln | \sec ( 3 x - 7 ) + \tan ( 3 x - 7 ) | + C
C) 6lnsec(3x7)+tan(3x7)+C6 \ln | \sec ( 3 x - 7 ) + \tan ( 3 x - 7 ) | + C
D) lnsec(3x7)+tan(3x7)+C\ln | \sec ( 3 x - 7 ) + \tan ( 3 x - 7 ) | + C
E) 2lnsec(3x7)+C2 \ln | \sec ( 3 x - 7 ) | + C
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18
Evaluate the integral. 10.8cos(6x1)dx\int 10.8 \cos ( 6 x - 1 ) \mathrm { d } x

A) 10.8sin(6x1)+C10.8 \sin ( 6 x - 1 ) + C
B) 1.8cos(6x1)+C1.8 \cos ( 6 x - 1 ) + C
C) 1.8sin(6x1)+C1.8 \sin ( 6 x - 1 ) + C
D) 1.8sin(6x1)+C- 1.8 \sin ( 6 x - 1 ) + C
E) 1.8sin(6x2x)+C1.8 \sin \left( 6 x ^ { 2 } - x \right) + C
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19
Evaluate the integral. 1π2+14xcos(x21)dx\int _ { 1 } ^ { \sqrt { \frac { \pi } { 2 } + 1 } } 4 x \cos \left( x ^ { 2 } - 1 \right) \mathrm { d } x

A)2
B) 10
C) 1
D) 6
E) 4
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20
Use geometry to compute the given integral. 05π6cosxdx\int _ { 0 } ^ { 5 \pi } 6 \cos x d x

A) 05π6cosxdx=5\int _ { 0 } ^ { 5 \pi } 6 \cos x d x = 5
B) 05π6cosxdx=12\int _ { 0 } ^ { 5 \pi } 6 \cos x d x = 12
C) 05π6cosxdx=1\int _ { 0 } ^ { 5 \pi } 6 \cos x d x = 1
D) 05π6cosxdx=0\int _ { 0 } ^ { 5 \pi } 6 \cos x d x = 0
E) none of these
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21
Calculate the derivative. ddx([lnx][cot(6x+5)])\frac { d } { d x } ( [ \ln | x | ] [ \cot ( 6 x + 5 ) ] )

A) cot(6x5)x6lnxcsc2(6x5)\frac { \cot ( 6 x - 5 ) } { x } - 6 \ln | x | \csc ^ { 2 } ( 6 x - 5 )
B) cot(6x+5)x6lnxcsc2(6x+5)\frac { \cot ( 6 x + 5 ) } { x } - 6 \ln | x | \csc ^ { 2 } ( 6 x + 5 )
C) 6lnxcsc2(6x+5)- 6 \ln | x | \csc ^ { 2 } ( 6 x + 5 )
D) cot(6x+5)x\frac { \cot ( 6 x + 5 ) } { x }
E) tan(6x+5)x6lnxcsc2(6x5)\frac { \tan ( 6 x + 5 ) } { x } - 6 \ln | x | \csc ^ { 2 } ( 6 x - 5 )
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22
Find the derivative of the function. w(x)=2sec(x)tan(x22)w ( x ) = 2 \sec ( x ) \cdot \tan \left( x ^ { 2 } - 2 \right)

A) w(x)=2sec2xsinxtan(x22)+4xsecxsec2(x22)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } - 2 \right) + 4 x \cdot \sec x \cdot \sec ^ { 2 } \left( x ^ { 2 } - 2 \right)
B) w(x)=2sec2xsinxtan(x22)+4xsecxsec2(2x2)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } - 2 \right) + 4 x \cdot \sec x \cdot \sec ^ { 2 } ( 2 x - 2 )
C) w(x)=2sec2xsinxtan(x2+2)+4xsecxsec2(2x2)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } + 2 \right) + 4 x \cdot \sec x \cdot \sec ^ { 2 } ( 2 x - 2 )
D) w(x)=2sec2xsinxtan(x22)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } - 2 \right)
E) w(x)=2sec2xsinxtan(x22)+4xsecxsec2(x24)w ^ { \prime } ( x ) = 2 \sec ^ { 2 } x \cdot \sin x \cdot \tan \left( x ^ { 2 } - 2 \right) + 4 x \cdot \sec x \cdot \sec ^ { 2 } \left( x ^ { 2 } - 4 \right)
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23
Calculate the derivative. ddx[sec(6x3x21)]\frac { d } { d x } \left[ \sec \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \right]

A) 6x418x2(x21)2csc(6x3x21)tan(6x3x21)\frac { 6 x ^ { 4 } - 18 x ^ { 2 } } { \left( x ^ { 2 } - 1 \right) ^ { 2 } } \csc \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right)
B) 6x3x21csc(6x3x21)tan(6x3x21)\frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \csc \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right)
C) 6x418x2(x21)2sec(6x3x21)tan(6x3x21)\frac { 6 x ^ { 4 } - 18 x ^ { 2 } } { \left( x ^ { 2 } - 1 \right) ^ { 2 } } \sec \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right)
D) 6x318x2x21sec(6x3x21)tan(6x3x21)\frac { 6 x ^ { 3 } - 18 x ^ { 2 } } { x ^ { 2 } - 1 } \sec \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right)
E) 6x418x2(x2+1)2sec(6x3x21)tan(6x3x2+1)\frac { 6 x ^ { 4 } - 18 x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 2 } } \sec \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } - 1 } \right) \tan \left( \frac { 6 x ^ { 3 } } { x ^ { 2 } + 1 } \right)
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24
Find the derivative of the function. t(x)=cotx4+8secxt ( x ) = \frac { \cot x } { 4 + 8 \sec x }

A) t(x)=(48secx)csc2x+8secx(48secx)2t ^ { \prime } ( x ) = - \frac { ( 4 - 8 \sec x ) \csc ^ { 2 } x + 8 \sec x } { ( 4 - 8 \sec x ) ^ { 2 } }
B) t(x)=(4+8secx)csc2x+8secx(4+8secx)2t ^ { \prime } ( x ) = - \frac { ( 4 + 8 \sec x ) \csc ^ { 2 } x + 8 \sec x } { ( 4 + 8 \sec x ) ^ { 2 } }
C) t(x)=(4+8secx)csc2x+8secx(4+8secx)2t ^ { \prime } ( x ) = \frac { ( 4 + 8 \sec x ) \csc ^ { 2 } x + 8 \sec x } { ( 4 + 8 \sec x ) ^ { 2 } }
D) t(x)=(4+8secx)csc2x+8secx(48secx)2t ^ { \prime } ( x ) = - \frac { ( 4 + 8 \sec x ) \csc ^ { 2 } x + 8 \sec x } { ( 4 - 8 \sec x ) ^ { 2 } }
E) t(x)=(48secx)csc2x8secx(48secx)2t ^ { \prime } ( x ) = - \frac { ( 4 - 8 \sec x ) \csc ^ { 2 } x - 8 \sec x } { ( 4 - 8 \sec x ) ^ { 2 } }
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25
Calculate the derivative. ddx[sin(7x)]0.5\frac { d } { d x } [ \sin ( 7 x ) ] ^ { 0.5 }

A) 3.5sin(7x)cos(7x)3.5 \sin ( 7 x ) \cdot \cos ( 7 x )
B) 3.5(sin(7x))0.5cos(7x)3.5 ( \sin ( 7 x ) ) ^ { - 0.5 } \cdot \cos ( 7 x )
C) (sin(7x))0.5cos(7x)( \sin ( 7 x ) ) ^ { - 0.5 } \cdot \cos ( 7 x )
D) 7(sin(7x))0.5cos(7x)7 ( \sin ( 7 x ) ) ^ { - 0.5 } \cdot \cos ( 7 x )
E) 7(sin(7x))0.5cos(7x)7 ( \sin ( 7 x ) ) ^ { 0.5 } \cdot \cos ( 7 x )
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26
Find the derivative of the function. g(x)=9sin(x)tanxg ( x ) = 9 \sin ( x ) \cdot \tan x

A) g(x)=9cos(x)tanx+9sin(x)csc2xg' ( x ) = 9 \cos ( x ) \cdot \tan x + 9 \sin ( x ) \cdot \csc ^ { 2 } x
B) g(x)=9cos(x)tanx+9sin(x)sec2xg '( x ) = 9 \cos ( x ) \cdot \tan x + 9 \sin ( x ) \cdot \sec ^ { 2 } x
C) g(x)=9cos(x)tanx9sin(x)cotxg '( x ) = 9 \cos ( x ) \cdot \tan x - 9 \sin ( x ) \cdot \cot x
D) g(x)=9cos(x)tanx+9cos(x)cotxg' ( x ) = 9 \cos ( x ) \cdot \tan x + 9 \cos ( x ) \cdot \cot x
E) g(x)=9cos(x)tanx9sin(x)sec2xg '( x ) = 9 \cos ( x ) \cdot \tan x - 9 \sin ( x ) \cdot \sec ^ { 2 } x
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27
Calculate the derivative. ddx[e3xsin(3πx)]\frac { d } { d x } \left[ e ^ { - 3 x } \sin ( 3 \pi x ) \right]

A) e3x(3sin(3πx)3πcos(3πx))e ^ { 3 x } ( - 3 \sin ( 3 \pi x ) - 3 \pi \cos ( 3 \pi x ) )
B) e3x(3sin(3πx)+3πcos(3πx))e ^ { - 3 x } ( - 3 \sin ( 3 \pi x ) + 3 \pi \cos ( 3 \pi x ) )
C) e3x(3sin(3πx)3πcos(3πx))e ^ { - 3 x } ( - 3 \sin ( 3 \pi x ) - 3 \pi \cos ( 3 \pi x ) )
D) e3x(sin(3πx)+cos(3πx))e ^ { - 3 x } ( \sin ( 3 \pi x ) + \cos ( 3 \pi x ) )
E) e3x(sin(3πx)3πcos(3πx))e ^ { - 3 x } ( \sin ( 3 \pi x ) - 3 \pi \cos ( 3 \pi x ) )
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28
Find the derivative of the function. u(x)=sec(x5.2+2.8x9)u ( x ) = \sec \left( x ^ { 5.2 } + 2.8 x - 9 \right)

A) u(x)=(5.2x4.2+2.8)sec(x5.22.8x9)tan(x5.22.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } + 2.8 \right) \sec \left( x ^ { 5.2 } - 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } - 2.8 x - 9 \right)
B) u(x)=(5.2x4.22.8)sec(x5.2+2.8x9)tan(x5.2+2.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } - 2.8 \right) \sec \left( x ^ { 5.2 } + 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } + 2.8 x - 9 \right)
C) u(x)=(5.2x4.22.8)sec(x5.22.8x9)tan(x5.2+2.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } - 2.8 \right) \sec \left( x ^ { 5.2 } - 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } + 2.8 x - 9 \right)
D) u(x)=(5.2x4.2+2.8)sec(x5.2+2.8x9)tan(x5.2+2.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } + 2.8 \right) \sec \left( x ^ { 5.2 } + 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } + 2.8 x - 9 \right)
E) u(x)=(5.2x4.22.8)sec(x5.22.8x9)tan(x5.22.8x9)u ^ { \prime } ( x ) = \left( 5.2 x ^ { 4.2 } - 2.8 \right) \sec \left( x ^ { 5.2 } - 2.8 x - 9 \right) \tan \left( x ^ { 5.2 } - 2.8 x - 9 \right)
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29
Find the derivative of the function. s(x)=2cos2xs ( x ) = 2 \cos ^ { 2 } x

A) s(x)=4cos(x)cosxs '( x ) = - 4 \cos ( x ) \cdot \cos x
B) s(x)=4cos(x)sinxs '( x ) = - 4 \cos ( x ) \cdot \sin x
C) s(x)=2cos2(x)sinxs ^ { \prime } ( x ) = - 2 \cos ^ { 2 } ( x ) \cdot \sin x
D) s(x)=2cos(x)sinxs' ( x ) = - 2 \cos ( x ) \cdot \sin x
E) s(x)=4cscxs' ( x ) = - 4 \csc x
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30
Find the derivative of the function. y(x)=7cos(ex)+9excosxy ( x ) = 7 \cos \left( e ^ { x } \right) + 9 e ^ { x } \cos x

A) y(x)=ex(7sin(ex)+9cosx9sinx)y' ( x ) = e ^ { x } \left( - 7 \sin \left( e ^ { x } \right) + 9 \cos x - 9 \sin x \right)
B) y(x)=7sin(ex)+9sinx9cosxy' ( x ) = - 7 \sin \left( e ^ { x } \right) + 9 \sin x - 9 \cos x
C) y(x)=7sin(ex)+9cosx9sinxy' ( x ) = - 7 \sin \left( e ^ { x } \right) + 9 \cos x - 9 \sin x
D) y(x)=ex(7sin(ex)9cosx+9sinx)y '( x ) = e ^ { x } \left( - 7 \sin \left( e ^ { x } \right) - 9 \cos x + 9 \sin x \right)
E) y(x)=ex2(7sin(ex)9cosx+9sinx)y ^ { \prime } ( x ) = e ^ { x ^ { 2 } } \left( - 7 \sin \left( e ^ { x } \right) - 9 \cos x + 9 \sin x \right)
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31
Find the derivative of the function. z(x)=3ln4(secx+tanx)z ( x ) = 3 \ln \mid 4 ( \sec x + \tan x )

A) z(x)=9secxz ( x ) = 9 \sec x
B) z(x)=3secx4z ^ { \prime } ( x ) = \frac { 3 \sec x } { 4 }
C) z(x)=3(1+tan2x)z ^ { \prime } ( x ) = 3 \left( 1 + \tan ^ { 2 } x \right)
D) z(x)=secxz ^ { \prime } ( x ) = \sec x
E) z(x)=3secxz ( x ) = 3 \sec x
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32
Find the derivative of the function. r(x)=4xcosx+9x2+8r ( x ) = 4 x \cos x + 9 x ^ { 2 } + 8

A) r(x)=4xcosx+4cosx+18xr'( x ) = - 4 x \cos x + 4 \cos x + 18 x
B) r(x)=4xsinx+4cosx+18xr' ( x ) = - 4 x \sin x + 4 \cos x + 18 x
C) r(x)=4xsinx+4cosx18xr '( x ) = - 4 x \sin x + 4 \cos x - 18 x
D) r(x)=4xsinx+4sinx+18xr '( x ) = - 4 x \sin x + 4 \sin x + 18 x
E) r(x)=4xsinx4cosx+18xr '( x ) = 4 x \sin x - 4 \cos x + 18 x
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33
Find the derivative of the function. f(x)=9sinx8cosxf ( x ) = 9 \sin x - 8 \cos x

A) f(x)=9cosx+8sinxf ^ { \prime } ( x ) = 9 \cos x + 8 \sin x
B) f(x)=8cosx9sinxf ^ { \prime } ( x ) = 8 \cos x - 9 \sin x
C) f(x)=9cosx8sinxf ^ { \prime } ( x ) = 9 \cos x - 8 \sin x
D) f(x)=9cosx8sinxf ^ { \prime } ( x ) = - 9 \cos x - 8 \sin x
E) f(x)=8cosx+9sinxf ^ { \prime } ( x ) = 8 \cos x + 9 \sin x
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34
Decide whether each integral converges. If the integral converges, compute its value.
Choose the correct letter for each question.


-converges to 14\frac { 1 } { 4 }

A) 0+sin(2x)dx\int _ { 0 } ^ { + \infty } \sin ( 2 x ) \mathrm { d } x
B) 0+e2xsin(2x)dx\int _ { 0 } ^ { + \infty } e ^ { - 2 x } \sin ( 2 x ) \mathrm { d } x
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35
Decide whether each integral converges. If the integral converges, compute its value.
Choose the correct letter for each question.


-diverges

A) 0+sin(2x)dx\int _ { 0 } ^ { + \infty } \sin ( 2 x ) \mathrm { d } x
B) 0+e2xsin(2x)dx\int _ { 0 } ^ { + \infty } e ^ { - 2 x } \sin ( 2 x ) \mathrm { d } x
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36
Find the derivative of the function. u(x)=cos(x22x)u ( x ) = \cos \left( x ^ { 2 } - 2 x \right)

A) u(x)=(2x2)cos(x22x)u ^ { \prime } ( x ) = ( 2 x - 2 ) \cos \left( x ^ { 2 } - 2 x \right)
B) u(x)=(2x+2)sin(x22x)u ^ { \prime } ( x ) = - ( 2 x + 2 ) \sin \left( x ^ { 2 } - 2 x \right)
C) u(x)=(2x2)cos(x22x)u ^ { \prime } ( x ) = - ( 2 x - 2 ) \cos \left( x ^ { 2 } - 2 x \right)
D) u(x)=(2x2)sin(x22x)u ^ { \prime } ( x ) = - ( 2 x - 2 ) \sin \left( x ^ { 2 } - 2 x \right)
E) u(x)=(2x+2)sin(x22x)u ^ { \prime } ( x ) = ( 2 x + 2 ) \sin \left( x ^ { 2 } - 2 x \right)
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37
Find the derivative of the function. j(x)=7sec2xj ( x ) = 7 \sec ^ { 2 } x

A) j(x)=14sec2xtanxj '( x ) = 14 \sec ^ { 2 } x \cdot \tan x
B) j(x)=7sec2xtanxj '( x ) = 7 \sec ^ { 2 } x \cdot \tan x
C) j(x)=14sec2xtan2xj '( x ) = 14 \sec ^ { 2 } x \cdot \tan ^ { 2 } x
D) j(x)=14sec2xtanxj ^ { \prime } ( x ) = - 14 \sec ^ { 2 } x \cdot \tan x
E) j(x)=14secxtanxj '( x ) = 14 \sec x \cdot \tan x
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38
Find the derivative of the function. s(x)=(4x22x+2)tanxs ( x ) = \left( 4 x ^ { 2 } - 2 x + 2 \right) \tan x

A) s(x)=(8x2)tanx+(4x22x2)sec2xs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x + \left( 4 x ^ { 2 } - 2 x - 2 \right) \sec ^ { 2 } x
B) s(x)=(8x2)tanx+(4x22x+2)secxs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x + \left( 4 x ^ { 2 } - 2 x + 2 \right) \sec x
C) s(x)=(8x2)tanx+(4x2+2x+2)sec2xs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x + \left( 4 x ^ { 2 } + 2 x + 2 \right) \sec ^ { 2 } x
D) s(x)=(8x2)tanx+(4x22x+2)sec2xs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x + \left( 4 x ^ { 2 } - 2 x + 2 \right) \sec ^ { 2 } x
E) s(x)=(8x2)tanx(4x22x+2)sec2xs ^ { \prime } ( x ) = ( 8 x - 2 ) \tan x - \left( 4 x ^ { 2 } - 2 x + 2 \right) \sec ^ { 2 } x
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39
Find the derivative of the function. p(x)=3+2sin(π2(x1))p ( x ) = 3 + 2 \sin \left( \frac { \pi } { 2 } ( x - 1 ) \right)

A) p(x)=πcos(π2(x1))p ^ { \prime } ( x ) = \pi \cos \left( \frac { \pi } { 2 } ( x - 1 ) \right)
B) p(x)=2cos(π2(x1))p ^ { \prime } ( x ) = 2 \cos \left( \frac { \pi } { 2 } ( x - 1 ) \right)
C) p(x)=πcos(π2(x+1))p ^ { \prime } ( x ) = \pi \cos \left( \frac { \pi } { 2 } ( x + 1 ) \right)
D) p(x)=πsin(π2(x1))p ^ { \prime } ( x ) = \pi \sin \left( \frac { \pi } { 2 } ( x - 1 ) \right)
E) p(x)=3+πcos(π2(x1))p ^ { \prime } ( x ) = 3 + \pi \cos \left( \frac { \pi } { 2 } ( x - 1 ) \right)
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40
Find the derivative of the function. y(x)=sec(e4x)y ( x ) = \sec \left( e ^ { 4 x } \right)

A) y(x)=4e4xcsc(e4x)tan2(e4x)y ^ { \prime } ( x ) = 4 e ^ { 4 x } \csc \left( e ^ { 4 x } \right) \tan ^ { 2 } \left( e ^ { 4 x } \right)
B) y(x)=16e4xsec(e4x)tan(e4x)y ^ { \prime } ( x ) = 16 e ^ { 4 x } \sec \left( e ^ { 4 x } \right) \tan \left( e ^ { 4 x } \right)
C) y(x)=4e4xsec(e4x)tan(e4x)y ^ { \prime } ( x ) = 4 e ^ { 4 x } \sec \left( e ^ { 4 x } \right) \tan \left( e ^ { 4 x } \right)
D) y(x)=4e4xcsc2(e4x)tan2(e4x)y ^ { \prime } ( x ) = 4 e ^ { 4 x } \csc ^ { 2 } \left( e ^ { 4 x } \right) \tan ^ { 2 } \left( e ^ { 4 x } \right)
E) y(x)=4e4xcsc(e4x)tan(e4x)y ^ { \prime } ( x ) = 4 e ^ { 4 x } \csc \left( e ^ { 4 x } \right) \tan \left( e ^ { 4 x } \right)
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41
Model the curve with a cosine function.  <strong>Model the curve with a cosine function.    Note that the period of the curve is  P = \frac { 1 } { 6 }  and its range is  [ - 1,1 ]  . </strong> A)  f ( x ) = \cos ( 12 x )  B)  f ( x ) = \cos \left( \frac { \pi x } { 12 } \right)  C)  f ( x ) = \cos ( 12 \pi x )  D)  f ( x ) = 12 \cos ( \pi x )  E)  f ( x ) = \cos \left( \frac { x } { 12 } \right)
Note that the period of the curve is P=16P = \frac { 1 } { 6 } and its range is [1,1][ - 1,1 ] .

A) f(x)=cos(12x)f ( x ) = \cos ( 12 x )
B) f(x)=cos(πx12)f ( x ) = \cos \left( \frac { \pi x } { 12 } \right)
C) f(x)=cos(12πx)f ( x ) = \cos ( 12 \pi x )
D) f(x)=12cos(πx)f ( x ) = 12 \cos ( \pi x )
E) f(x)=cos(x12)f ( x ) = \cos \left( \frac { x } { 12 } \right)
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42
The cost of Dig-It brand snow shovels is given by c(t)=3sin(2π(t0.75))c ( t ) = 3 \sin ( 2 \pi ( t - 0.75 ) )
Where t is time in years since January 1, 1997. How fast, in dollars per year, is the cost increasing on October 30, 1997

A)$21.85 per year
B) $18.85 per year
C) $9.42 per year
D) $20.85 per year
E) $6.00 per year
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43
Starting with the identity sin2x+cos2x=1\sin ^ { 2 } x + \cos ^ { 2 } x = 1 , choose the right trigonometric identity.

A) sec2x=1+tan2x\sec ^ { 2 } x = 1 + \tan ^ { 2 } x
B) sec2x=1tan2x\sec ^ { 2 } x = 1 - \tan ^ { 2 } x
C) cot2x=1tan2x\cot ^ { 2 } x = 1 - \tan ^ { 2 } x
D) sin2x=1+tan2x\sin ^ { 2 } x = 1 + \tan ^ { 2 } x
E) sec2x=1+csc2x\sec ^ { 2 } x = 1 + \csc ^ { 2 } x
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44
Sketch the curves without any technological help. f(t)=2costf ( t ) = 2 \cos t ; g(t)=3.3cos(2t)g ( t ) = 3.3 \cos ( 2 t )

A)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)
B)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)
C)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)
D)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)
E)  <strong>Sketch the curves without any technological help.  f ( t ) = 2 \cos t  ;  g ( t ) = 3.3 \cos ( 2 t )   </strong> A)   B)     C)    D)     E)
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45
Sketch the curves without any technological help. f(t)=costf ( t ) = \cos t ; g(t)=cos(t+π)g ( t ) = \cos ( t + \pi )

A)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)
B)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)
C)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)
D)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)
E)  <strong>Sketch the curves without any technological help.    f ( t ) = \cos t  ;  g ( t ) = \cos ( t + \pi )   </strong> A)    B)     C)     D)     E)
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46
Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function s(t)=0.468sin(1.25t+1.73)+0.79(1t6)s ( t ) = 0.468 \sin ( 1.25 t + 1.73 ) + 0.79 ( 1 \leq t \leq 6 )
Where t is time in quarters ( t=1t = 1 represents the end of the first quarter of 1995) and s(t)s ( t ) is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales.

A) smax=1.526s _ { \max } = 1.526 , smin=0.074s _ { \min } = 0.074
B) smax=0.79s _ { \max } = 0.79 , smin=0.468s _ { \min } = 0.468
C) smax=2.048s _ { \max } = 2.048 , smin=1.112s _ { \min } = 1.112
D) smax=2.194s _ { \max } = 2.194 , smin=0.146s _ { \min } = 0.146
E) smax=1.258s _ { \max } = 1.258 , smin=0.322s _ { \min } = 0.322
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47
Model the curve with a sine function.  <strong>Model the curve with a sine function.    Note that the period of the curve is  P = 0.4  and its range is  [ - 3 , - 1 ]  . </strong> A)  f ( x ) = 2 - \sin x  B)  f ( x ) = - 2 + 5 \sin x  C)   f ( x ) = - 2 + \sin ( 5 \pi x )  D)  f ( x ) = 2 - \sin ( 5 \pi x )  E)  f ( x ) = - 2 + \sin ( \pi x )
Note that the period of the curve is P=0.4P = 0.4 and its range is [3,1][ - 3 , - 1 ] .

A) f(x)=2sinxf ( x ) = 2 - \sin x
B) f(x)=2+5sinxf ( x ) = - 2 + 5 \sin x
C) f(x)=2+sin(5πx)f ( x ) = - 2 + \sin ( 5 \pi x )
D) f(x)=2sin(5πx)f ( x ) = 2 - \sin ( 5 \pi x )
E) f(x)=2+sin(πx)f ( x ) = - 2 + \sin ( \pi x )
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48
Use the addition formulas: sin(x+y)=sinxcosy+cosxsinysin(xy)=sinxcosycosxsinycos(x+y)=cosxcosysinxsinycos(xy)=cosxcosy+sinxsiny\begin{array} { l } \sin ( x + y ) = \sin x \cdot \cos y + \cos x \cdot \sin y \\\sin ( x - y ) = \sin x \cdot \cos y - \cos x \cdot \sin y \\\cos ( x + y ) = \cos x \cdot \cos y - \sin x \cdot \sin y \\\cos ( x - y ) = \cos x \cdot \cos y + \sin x \cdot \sin y\end{array}
To express tan(x+23π)\tan ( x + 23 \pi ) in terms of tan(x)\tan ( x ) .

A) tan(x)23π\frac { \tan ( x ) } { 23 \pi }
B) tan(x)+23π\tan ( x ) + 23 \pi
C) 23πtan(x)23 \pi \tan ( x )
D) tan(x)\tan ( x )
E) tan(x)23π\tan ( x ) - 23 \pi
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49
Use the formula for sin(x+y)\sin ( x + y ) to simplify the expression sin(t+17π2)\sin \left( t + \frac { 17 \pi } { 2 } \right) .

A) cost17π\cos t - 17 \pi
B) cost+17π\cos t + 17 \pi
C) cost17π\frac { \cos t } { 17 \pi }
D) 17πcost17 \pi \cos t
E) cost\cos t
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50
Use the addition formulas: sin(x+y)=sinxcosy+cosxsinysin(xy)=sinxcosycosxsinycos(x+y)=cosxcosysinxsinycos(xy)=cosxcosy+sinxsiny\begin{array} { l } \sin ( x + y ) = \sin x \cdot \cos y + \cos x \cdot \sin y \\\sin ( x - y ) = \sin x \cdot \cos y - \cos x \cdot \sin y \\\cos ( x + y ) = \cos x \cdot \cos y - \sin x \cdot \sin y \\\cos ( x - y ) = \cos x \cdot \cos y + \sin x \cdot \sin y\end{array}
To calculate cos(π3)\cos \left( \frac { \pi } { 3 } \right) , given that sin(π6)=12\sin \left( \frac { \pi } { 6 } \right) = \frac { 1 } { 2 } and cos(π6)=32\cos \left( \frac { \pi } { 6 } \right) = \frac { \sqrt { 3 } } { 2 } .

A) cos(π3)=0\cos \left( \frac { \pi } { 3 } \right) = 0
B) cos(π3)=32\cos \left( \frac { \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 2 }
C) cos(π3)=12\cos \left( \frac { \pi } { 3 } \right) = \frac { 1 } { 2 }
D) cos(π3)=32\cos \left( \frac { \pi } { 3 } \right) = - \frac { \sqrt { 3 } } { 2 }
E) cos(π3)=12\cos \left( \frac { \pi } { 3 } \right) = - \frac { 1 } { 2 }
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51
Use the conversion formula cosx=sin(π2x)\cos x = \sin \left( \frac { \pi } { 2 } - x \right) to replace the expression g(x)=25cos[2π(3x1)]+9g ( x ) = 25 \cos [ 2 \pi ( 3 x - 1 ) ] + 9
By a sine function.

A) g(x)=25sin(6πx+5π2)9+π2g ( x ) = 25 \sin \left( 6 \pi x + \frac { 5 \pi } { 2 } \right) - 9 + \frac { \pi } { 2 }
B) g(x)=25sin(6πx5π2)+9g ( x ) = 25 \sin \left( 6 \pi x - \frac { 5 \pi } { 2 } \right) + 9
C) g(x)=25sin(6πx5π2)+9g ( x ) = 25 \sin \left( - 6 \pi x - \frac { 5 \pi } { 2 } \right) + 9
D) g(x)=25sin(6πx+5π2)+9g ( x ) = 25 \sin \left( - 6 \pi x + \frac { 5 \pi } { 2 } \right) + 9
E) g(x)=25sin(12πx+3π2)+9g ( x ) = 25 \sin \left( 12 \pi x + \frac { 3 \pi } { 2 } \right) + 9
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52
Use the conversion formula cosx=sin(π2x)\cos x = \sin \left( \frac { \pi } { 2 } - x \right) to replace the expression g(t)=45cos(t5)g ( t ) = 45 - \cos ( t - 5 )
By a sine function.

A) g(t)=45sin(π2t+5)g ( t ) = 45 - \sin \left( \frac { \pi } { 2 } - t + 5 \right)
B) g(t)=45sin(π2t5)g ( t ) = 45 - \sin \left( \frac { \pi } { 2 } - t - 5 \right)
C) g(t)=45sin(πt+5)g ( t ) = 45 - \sin ( \pi - t + 5 )
D) g(t)=45sin(π2+t+5)g ( t ) = 45 - \sin \left( \frac { \pi } { 2 } + t + 5 \right)
E) g(t)=π245sin(t+5)g ( t ) = \frac { \pi } { 2 } - 45 - \sin ( t + 5 )
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53
Model the curve with a cosine function.  <strong>Model the curve with a cosine function.    Note that the period of the curve is  P = 14  , its range is  [ 0,120 ]  the graph of the cosine function is shifted upward 60 units and shifted to the right 7 units. </strong> A)  f ( x ) = 120 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) + 7  B)  f ( x ) = 120 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) - 7  C)  f ( x ) = 60 \cos \left( \frac { \pi ( x + 7 ) } { 7 } \right) + 60  D)  f ( x ) = 7 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) + 7  E)  f ( x ) = 60 \cos \left( \frac { \pi ( x - 7 ) } { 7 } \right) + 60
Note that the period of the curve is P=14P = 14 , its range is [0,120][ 0,120 ] the graph of the cosine function is shifted upward 60 units and shifted to the right 7 units.

A) f(x)=120cos(π(x60)60)+7f ( x ) = 120 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) + 7
B) f(x)=120cos(π(x60)60)7f ( x ) = 120 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) - 7
C) f(x)=60cos(π(x+7)7)+60f ( x ) = 60 \cos \left( \frac { \pi ( x + 7 ) } { 7 } \right) + 60
D) f(x)=7cos(π(x60)60)+7f ( x ) = 7 \cos \left( \frac { \pi ( x - 60 ) } { 60 } \right) + 7
E) f(x)=60cos(π(x7)7)+60f ( x ) = 60 \cos \left( \frac { \pi ( x - 7 ) } { 7 } \right) + 60
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54
Use the conversion formula cosx=sin(π2x)\cos x = \sin \left( \frac { \pi } { 2 } - x \right) to replace the expression f(t)=5.2cos(6πt)+10f ( t ) = 5.2 \cos ( 6 \pi t ) + 10
By a sine function.

A) f(t)=5.2sin(π26πt)+10f ( t ) = 5.2 \sin \left( \frac { \pi } { 2 } - 6 \pi t \right) + 10
B) f(t)=5.2sin(π6πt2)+10f ( t ) = 5.2 \sin \left( \frac { \pi - 6 \pi t } { 2 } \right) + 10
C) f(t)=5.2sin(π26t)+10f ( t ) = 5.2 \sin \left( \frac { \pi } { 2 } - 6 t \right) + 10
D) f(t)=6sin(π25.2πt)+10f ( t ) = 6 \sin \left( \frac { \pi } { 2 } - 5.2 \pi t \right) + 10
E) f(t)=10sin(π26πt)+5.2f ( t ) = 10 \sin \left( \frac { \pi } { 2 } - 6 \pi t \right) + 5.2
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55
Model the curve with a sine function.  <strong>Model the curve with a sine function.    Note that the period of the curve is  P = 32  and its range is  [ - 40,0 ]  , the graph of the sine function is shifted to the right 7 units. </strong> A)  f ( x ) = 20 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) + 20  B)  f ( x ) = - 20 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) + 20  C)  f ( x ) = 40 \sin \left( \frac { \pi ( x - 7 ) } { 16 } \right) - 40  D)  f ( x ) = 40 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) - 40  E)  f ( x ) = 20 \sin \left( \frac { \pi ( x - 7 ) } { 16 } \right) - 20
Note that the period of the curve is P=32P = 32 and its range is [40,0][ - 40,0 ] , the graph of the sine function is shifted to the right 7 units.

A) f(x)=20sin(π(x+7)16)+20f ( x ) = 20 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) + 20
B) f(x)=20sin(π(x+7)16)+20f ( x ) = - 20 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) + 20
C) f(x)=40sin(π(x7)16)40f ( x ) = 40 \sin \left( \frac { \pi ( x - 7 ) } { 16 } \right) - 40
D) f(x)=40sin(π(x+7)16)40f ( x ) = 40 \sin \left( \frac { \pi ( x + 7 ) } { 16 } \right) - 40
E) f(x)=20sin(π(x7)16)20f ( x ) = 20 \sin \left( \frac { \pi ( x - 7 ) } { 16 } \right) - 20
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56
Model the curve with a sine function.
 <strong>Model the curve with a sine function.   Note that the period of the curve is  P = \frac { 1 } { 5 }  and its range is  [ - 2.2,2.2 ]  and the graph of the sine function is shifted to the left 0.55 units. </strong> A)  f ( x ) = 2.2 \sin ( 10 \pi ( x + 0.55 ) )  B)  f ( x ) = 2.2 \sin ( 10 \pi ( x - 0.55 ) )  C)  f ( x ) = 2.2 \sin ( 10 \pi x + 0.55 )  D)  f ( x ) = 2.2 \sin ( 10 \pi ( 2 x + 0.55 ) )  E)  f ( x ) = 4.4 \sin ( 5 \pi ( x + 0.55 ) )
Note that the period of the curve is P=15P = \frac { 1 } { 5 } and its range is [2.2,2.2][ - 2.2,2.2 ] and the graph of the sine function is shifted to the left 0.55 units.

A) f(x)=2.2sin(10π(x+0.55))f ( x ) = 2.2 \sin ( 10 \pi ( x + 0.55 ) )
B) f(x)=2.2sin(10π(x0.55))f ( x ) = 2.2 \sin ( 10 \pi ( x - 0.55 ) )
C) f(x)=2.2sin(10πx+0.55)f ( x ) = 2.2 \sin ( 10 \pi x + 0.55 )
D) f(x)=2.2sin(10π(2x+0.55))f ( x ) = 2.2 \sin ( 10 \pi ( 2 x + 0.55 ) )
E) f(x)=4.4sin(5π(x+0.55))f ( x ) = 4.4 \sin ( 5 \pi ( x + 0.55 ) )
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57
Calculate the derivative.
Calculate the derivative. ​
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58
The depth of water d(t)d ( t ) at my favorite surfing spot varies from 8 to 20 feet, depending on the time. Last Sunday high tide occurred at 5:00 A.M. and the next high tide occurred at 6:30 P.M. Use a sine function to model the depth of water as a function of time t in hours since midnight on Sunday morning.

A) d(t)=10sin(2π(t1.625)13.5)+4d ( t ) = 10 \sin \left( \frac { 2 \pi ( t - 1.625 ) } { 13.5 } \right) + 4
B) d(t)=6sin(2π(t+1.625)11.5)14d ( t ) = 6 \sin \left( \frac { 2 \pi ( t + 1.625 ) } { 11.5 } \right) - 14
C) d(t)=6sin(2π(t+1.625)13.5)+14d ( t ) = - 6 \sin \left( \frac { - 2 \pi ( t + 1.625 ) } { 13.5 } \right) + 14
D) d(t)=6sin(2π(t1.625)13.5)+14d ( t ) = 6 \sin \left( \frac { 2 \pi ( t - 1.625 ) } { 13.5 } \right) + 14
E) d(t)=14sin(2π(t1.625)13.5)+6d ( t ) = 14 \sin \left( \frac { 2 \pi ( t - 1.625 ) } { 13.5 } \right) + 6
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59
Model the curve with a cosine function.  <strong>Model the curve with a cosine function.     Note that the period of the curve is  P = \frac { 1 } { 5 }  , its range is  [ - 3.3,3.3 ]  and the graph of the cosine function is shifted to the right 0.35 units. </strong> A)  f ( x ) = 6.6 \cos ( 20 \pi ( 2 x - 0.35 ) )  B)  f ( x ) = 3.3 \cos ( 10 ( x - 0.35 ) )  C)  f ( x ) = 6.6 \cos ( 20 \pi ( 2 x + 0.35 ) )  D)  f ( x ) = 3.3 \cos ( 10 \pi ( x - 0.35 ) )  E)  f ( x ) = 3.3 \cos ( 10 \pi ( x + 0.35 ) )
Note that the period of the curve is P=15P = \frac { 1 } { 5 } , its range is [3.3,3.3][ - 3.3,3.3 ] and the graph of the cosine function is shifted to the right 0.35 units.

A) f(x)=6.6cos(20π(2x0.35))f ( x ) = 6.6 \cos ( 20 \pi ( 2 x - 0.35 ) )
B) f(x)=3.3cos(10(x0.35))f ( x ) = 3.3 \cos ( 10 ( x - 0.35 ) )
C) f(x)=6.6cos(20π(2x+0.35))f ( x ) = 6.6 \cos ( 20 \pi ( 2 x + 0.35 ) )
D) f(x)=3.3cos(10π(x0.35))f ( x ) = 3.3 \cos ( 10 \pi ( x - 0.35 ) )
E) f(x)=3.3cos(10π(x+0.35))f ( x ) = 3.3 \cos ( 10 \pi ( x + 0.35 ) )
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60
Model the curve with a cosine function.  <strong>Model the curve with a cosine function.    Note that the period of the curve is  P = \frac { 1 } { 3 }  and its range is  [ - 1,1 ]  . </strong> A)  f ( x ) = \cos ( 6 x )  B)  f ( x ) = \cos ( 6 \pi x )  C)  f ( x ) = \cos \left( \frac { x } { 6 } \right)  D)  f ( x ) = \cos \left( \frac { \pi x } { 6 } \right)  E)  f ( x ) = 6 \cos ( \pi x )
Note that the period of the curve is P=13P = \frac { 1 } { 3 } and its range is [1,1][ - 1,1 ] .

A) f(x)=cos(6x)f ( x ) = \cos ( 6 x )
B) f(x)=cos(6πx)f ( x ) = \cos ( 6 \pi x )
C) f(x)=cos(x6)f ( x ) = \cos \left( \frac { x } { 6 } \right)
D) f(x)=cos(πx6)f ( x ) = \cos \left( \frac { \pi x } { 6 } \right)
E) f(x)=6cos(πx)f ( x ) = 6 \cos ( \pi x )
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61
Starting with the identity Starting with the identity   and then dividing both sides of the equation by a suitable trigonometric function, derive the trigonometric identity. ​  and then dividing both sides of the equation by a suitable trigonometric function, derive the trigonometric identity.
Starting with the identity   and then dividing both sides of the equation by a suitable trigonometric function, derive the trigonometric identity. ​
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62
The uninflated cost of Dugout brand snow shovels currently varies from a high of $30 on January 1 (t=0)( t = 0 ) to a low of $6 on July 1 (t=0.5)( t = 0.5 ) . Assuming this trend were to continue indefinitely, calculate the uninflated cost u(t)u ( t ) of Dugout snow shovels as a function of time t in years. (Use a sine function.)

A) u(t)=15sin(2π(t0.75))+3u ( t ) = 15 \sin ( 2 \pi ( t - 0.75 ) ) + 3
B) u(t)=12sin(2π(t0.75))+18u ( t ) = 12 \sin ( 2 \pi ( t - 0.75 ) ) + 18
C) u(t)=18sin(2π(t0.75))+12u ( t ) = 18 \sin ( 2 \pi ( t - 0.75 ) ) + 12
D) u(t)=12sin(2π(t0.75))18u ( t ) = - 12 \sin ( 2 \pi ( t - 0.75 ) ) - 18
E) u(t)=12sin(2π(t+0.75))18u ( t ) = 12 \sin ( 2 \pi ( t + 0.75 ) ) - 18
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63
The depth of water The depth of water   at my favorite surfing spot varies from 5 to 15 feet, depending on the time. Last Sunday high tide occurred at 5:00 A.M. and the next high tide occurred at 6:30 P.M. Use a sine function to model to the depth of water as a function of time t in hours since midnight in Sunday morning. at my favorite surfing spot varies from 5 to 15 feet, depending on the time. Last Sunday high tide occurred at 5:00 A.M. and the next high tide occurred at 6:30 P.M. Use a sine function to model to the depth of water as a function of time t in hours since midnight in Sunday morning.
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64
Model the curve with a sine function.
Model the curve with a sine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the sine function is shifted to the left 0.9 units. Write the model function as a function of x and π.
Note that the period of the curve is Model the curve with a sine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the sine function is shifted to the left 0.9 units. Write the model function as a function of x and π. , its range is Model the curve with a sine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the sine function is shifted to the left 0.9 units. Write the model function as a function of x and π. and the graph of the sine function is shifted to the left 0.9 units. Write the model function as a function of x and π.
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65
Model the curve with a cosine function.
Model the curve with a cosine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the cosine function is shifted upward 55 units and shifted to the right 14 units. Write the model function as a function of x and π.
Note that the period of the curve is Model the curve with a cosine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the cosine function is shifted upward 55 units and shifted to the right 14 units. Write the model function as a function of x and π. , its range is Model the curve with a cosine function. ​   ​ Note that the period of the curve is   , its range is   and the graph of the cosine function is shifted upward 55 units and shifted to the right 14 units. Write the model function as a function of x and π. and the graph of the cosine function is shifted upward 55 units and shifted to the right 14 units. Write the model function as a function of x and π.
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66
Model the curve with a sine function.

Model the curve with a sine function. ​ ​   ​ Note that the period of the curve is   and its range is   . Write the model function as a function of x and π.
Note that the period of the curve is Model the curve with a sine function. ​ ​   ​ Note that the period of the curve is   and its range is   . Write the model function as a function of x and π. and its range is Model the curve with a sine function. ​ ​   ​ Note that the period of the curve is   and its range is   . Write the model function as a function of x and π. . Write the model function as a function of x and π.
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67
Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function
Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function ​   ​ where ​t is time in quarters (   represents the end of the first quarter of 1995) and   is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales. ​ Maximum sales __________ billions of dollars ​ Minimum sales __________ billions of dollars
where ​t is time in quarters ( Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function ​   ​ where ​t is time in quarters (   represents the end of the first quarter of 1995) and   is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales. ​ Maximum sales __________ billions of dollars ​ Minimum sales __________ billions of dollars represents the end of the first quarter of 1995) and Sales of computers are subject to seasonal fluctuations. Computer City's sales of computers in 1995 and 1996 can be approximated by the function ​   ​ where ​t is time in quarters (   represents the end of the first quarter of 1995) and   is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales. ​ Maximum sales __________ billions of dollars ​ Minimum sales __________ billions of dollars is computer sales (quarterly revenue) in billions of dollars. Estimate Computer City's maximum and minimum quarterly revenue from computer sales.

Maximum sales __________ billions of dollars

Minimum sales __________ billions of dollars
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