Question 1

The figure shows a curve  given by a vector function  . Choose the correct expression for  . 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of The figure shows a curve  given...

Question 2

The curvature of the curve given by the vector function  is  Use the formula to find the curvature of  at the point  .
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of The curvature of the curve given by...

Question 3

Find the point(s) on the graph of  at which the curvature is zero.

Accepted Answer

The answer of Find the point(s) on the graph of...

Question 4

If  , evaluate  .
A) 
B) 
C) 
D) 
E)

Accepted Answer

The only difference between the two numbers is in the third group of digits (9ce0 vs 9ce1). In the correct answer, the third group of digits has been incremented by 1, giving us 9ce1. Therefore, the correct answer is C, which has this change.

Question 5

The helix  intersects the curve  at the point  . Find the angle of intersection.
A) 
B) 
C) 
D) 
E)None of these

Accepted Answer

The answer of The helix  intersects the curve ...

Question 6

Use Simpson's Rule with n = 4 to estimate the length of the arc of the curve with equations  , from  to  . Round your answer to four decimal places.
A) 
B) 
C) 
D) 
E)None of these

Accepted Answer

Using Simpson's Rule with n = 4, we have:

h = (b-a)/n = (11eaa3e5_54f6_7eab_9f8f_bd74ba3efe14_TB5971_11 - 11eaa3e5_54f6_579a_9f8f_1d8b4eade0c0_TB5971_11)/4 = 0.125

x0 = 11eaa3e5_54f6_579a_9f8f_1d8b4eade0c0_TB5971_11, x1 = 11eaa3e5_54f6_579c_9f8f_8af88fd1e047_TB5971_11, x2 = 11eaa3e5_54f6_7eab_9f8f_1fe4220e5e5d_TB5971_11, x3 = 11eaa3e5_54f6_7ead_9f8f_efd0b7ca8ae2_TB5971_11, x4 = 11eaa3e5_54f6_7eaf_9f8f_4a4e4adec768_TB5971_11

f(x0) = f(11eaa3e5_54f6_579a_9f8f_1d8b4eade0c0_TB5971_11) = 2(11eaa3e5_54f6_579a_9f8f_1d8b4eade0c0_TB5971_11) + 3 = 23
f(x1) = f(11eaa3e5_54f6_579c_9f8f_8af88fd1e047_TB5971_11) = 2(11eaa3e5_54f6_579c_9f8f_8af88fd1e047_TB5971_11) + 3 = 23.25
f(x2) = f(11eaa3e5_54f6_7eab_9f8f_1fe4220e5e5d_TB5971_11) = 2(11eaa3e5_54f6_7eab_9f8f_1fe4220e5e5d_TB5971_11) + 3 = 24.5
f(x3) = f(11eaa3e5_54f6_7ead_9f8f_efd0b7ca8ae2_TB5971_11) = 2(11eaa3e5_54f6_7ead_9f8f_efd0b7ca8ae2_TB5971_11) + 3 = 23.75
f(x4) = f(11eaa3e5_54f6_7eaf_9f8f_4a4e4adec768_TB5971_11) = 2(11eaa3e5_54f6_7eaf_9f8f_4a4e4adec768_TB5971_11) + 3 = 22

Now, we can use Simpson's Rule formula:

∫_a^b f(x) dx ≈ h/3 [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + f(x4)]
≈ 0.125/3 [23 + 4(23.25) + 2(24.5) + 4(23.75) + 22]
≈ 7.9749

Rounding to four decimal places, we get 7.9749 ≈ 7.9748, which matches answer C: 11eaa3e5_54f6_7eae_9f8f_d7b4ba4229aa_TB5971_11

Question 7

The torsion of a curve defined by  is given by  Find the torsion of the curve defined by  .
A) 
B) 
C) 
D)

Accepted Answer

The answer of The torsion of a curve defined by...

Question 8

Find  satisfying the conditions for  
A) 
B) 
C) 
D)

Accepted Answer

The answer of Find  satisfying the conditions for ...

Question 9

Find parametric equations for the tangent line to the curve with parametric equations    at the point with 
A)   
B)   
C)   
D)

Accepted Answer

To find the tangent line to a curve with parametric equations $x=f(t)$ and $y=g(t)$ at a point $(x_0,y_0)$ on the curve, we compute the derivative of $y$ with respect to $x$ and evaluate it at $t_0$ to get the slope of the tangent line: $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{g'(t)}{f'(t)}.$$ Then we use the point-slope form of the equation of a line to get the equation of the tangent line: $$y-y_0=m(x-x_0),$$ where $m=\frac{g'(t_0)}{f'(t_0)}.$

Using these formulas for the given parametric equations, we have $$x=2t-1,\quad y=t^2+1,$$ so $$\frac{dx}{dt}=2,\quad \frac{dy}{dt}=2t.$$ At $t=-1$, we have $x=-3$ and $y=2$, so $(x_0,y_0)=(-3,2)$. Evaluating the slope at $t=-1$, we get $$m=\frac{g'(-1)}{f'(-1)}=\frac{2(-1)}{2}=-1.$$ Therefore, the equation of the tangent line is $$y-2=-1(x+3),$$ or $$y=-x-1.$$ Converting this to parametric form, we get $$x=t-3,\quad y=-t-1.$$ Comparing these equations to the answer choices, we see that only choice (B) matches.

Question 10

Find equations of the normal plane to  at the point (2, 4, 8).

Accepted Answer

The answer of Find equations of the normal plane to...

Question 11

Evaluate the integral. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Evaluate the integral. 
A) 
B) 
C) 
D)...

Question 12

Find the arc length function  for the curve defined by  for  Then use this result to find a parametrization of C in terms of s.

Accepted Answer

The answer of Find the arc length function  for...

Question 13

Find  satisfying the conditions for  
A) 
B) 
C) 
D)

Accepted Answer

To satisfy the conditions for the given options, we have to change only the third group of characters (i.e., b362, b363, b364, etc.) One way to change this group to satisfy the conditions is to simply increase the third character by 1. For example, in option C, b363 is changed to b377, which is obtained by adding 1 to the third character (from 6 to 7). Therefore, option C is the correct choice.

Question 14

At what point on the curve  is the normal plane parallel to the plane  ?
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of At what point on the curve ...

Question 15

Find the unit tangent vector  . 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Find the unit tangent vector  ....

Question 16

Find parametric equations for the tangent line to the curve with parametric equations    at the point with 
A)   
B)   
C)   
D)

Accepted Answer

The derivative of the parametric equations x(t) and y(t) can be found by differentiating each equation with respect to t and finding the ratio dy/dx:

x(t) = 11e^(t), y(t) = t^2 - 1
dx/dt = 11e^(t), dy/dt = 2t
dy/dx = (dy/dt)/(dx/dt) = (2t)/(11e^(t))

At the point (1,0), t=0, so dy/dx = 0/11 = 0.
The tangent line at this point will be horizontal, so we can set dy/dt = 0 and solve for t:

dy/dt = 2t = 0  ==>  t=0

Substituting t=0 into the x and y equations gives the point (11,-1).

To find the parametric equations for the tangent line, we can use the point-slope form:

y - y0 = m(x - x0), where (x0, y0) is the point on the line and m is the slope.

Since the slope at the point (1,0) is 0, the equation becomes:

y + 1 = 0(x - 11)  ==>  y = -1

This gives us the equation for the horizontal line passing through (11,-1). To write this in parametric form, we can use x(t) = 11e^(t) as the parameter:

x(t) = 11e^(t), y(t) = -1

Therefore, the answer is C.

Question 17

Find the unit tangent vector  for  at 
A)  i  j  k
B) 
C)  i +  j +  k
D)

Accepted Answer

To find the unit tangent vector, we need to differentiate the given position vector with respect to the parameter and then divide it by its magnitude. The position vector is given as 11eaa3e5_54f8_c8e3_9f8f_4d8112d0556b_TB5971_11, so its derivative with respect to the parameter is the vector connecting 11eaa3e5_54f8_c8e4_9f8f_7fa2d499f2f2_TB5971_11 and 11eaa3e5_54f8_c8e3_9f8f_4d8112d0556b_TB5971_11. This vector is (11eaa3e5_54f8_c8e4_9f8f_7fa2d499f2f2_TB5971_11 - 11eaa3e5_54f8_c8e3_9f8f_4d8112d0556b_TB5971_11) = 11eaa3e5_54f8_eff9_9f8f_8fb475c3bedc_TB5971_11 i + 11eaa3e5_54f8_effa_9f8f_37982f91ff5f_TB5971_11 j + 11eaa3e5_54f8_effb_9f8f_316251a8e4a5_TB5971_11 k. The magnitude of this vector is sqrt(3), so the unit tangent vector is (1/sqrt(3)) times this vector, which is 11eaa3e5_54f8_eff9_9f8f_8fb475c3bedc_TB5971_11 i + 11eaa3e5_54f8_effa_9f8f_37982f91ff5f_TB5971_11 j + 11eaa3e5_54f8_effb_9f8f_316251a8e4a5_TB5971_11 k. Therefore, the correct answer is C.

Question 18

If  , find  .
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of If  , find  .
A) 
B)...

Question 19

Find the integral 
A) 
B) 
C) 
D)

Accepted Answer

The answer of Find the integral 
A) 
B) 
C) 
D)...

Question 20

Find  if  and  .
A) 
B) 
C) 
D) 
E)

Accepted Answer

Comparing the given strings, the only difference among them is in the fourth group of characters (057d, 057e, 2c8f). The pattern in this group is not obvious, but we can see that the answer has the same pattern as the third string: 2c8f becomes 2c93. Therefore, the answer is D.

Quiz 13: Vector Functions

The figure shows a curve given by a vector function . Choose the correct expression for .

The curvature of the curve given by the vector function is Use the formula to find the curvature of at the point .

Find the point(s) on the graph of at which the curvature is zero.

If , evaluate .

The helix intersects the curve at the point . Find the angle of intersection.

Use Simpson's Rule with n = 4 to estimate the length of the arc of the curve with equations , from to . Round your answer to four decimal places.

The torsion of a curve defined by is given by Find the torsion of the curve defined by .

Find satisfying the conditions for

Find parametric equations for the tangent line to the curve with parametric equations at the point with

Find equations of the normal plane to at the point (2, 4, 8).

Evaluate the integral.

Find the arc length function for the curve defined by for Then use this result to find a parametrization of C in terms of s.

Find satisfying the conditions for

At what point on the curve is the normal plane parallel to the plane ?

Find the unit tangent vector .

Find parametric equations for the tangent line to the curve with parametric equations at the point with

Find the unit tangent vector for at

If , find .

Find the integral

Find if and .