Question 1

Solve the problem.
-The velocity of a body of mass $m$ falling from rest under the action of gravity is given by the equation $\mathrm { v } = \sqrt { \frac { \mathrm { mg } } { \mathrm { k } } } \tanh \left( \sqrt { \frac { \mathrm { gk } } { \mathrm { m } } \mathrm { t } } \right)$, where $\mathrm { k }$ is a constant that depends on the body's aerodynamic properties and the density of the air, $g$ is the gravitational constant, and $t$ is the number of seconds into the fall. Find the limiting velocity, $\lim _ { \mathrm { t } \rightarrow \infty }$, of a 210 lb. skydiver $( \mathrm { mg } = 210 )$ when $\mathrm { k } = 0.006$.
A) 59.16 ft/sec
B) 187.08 ft/sec
C) There is no limiting speed.
D) 0.01 ft/sec

Accepted Answer

To find the limiting velocity, we take the limit as $t$ approaches infinity. As $t$ gets very large, the argument of the hyperbolic tangent function approaches infinity, so the function itself approaches 1. Therefore, we have $$\lim_{t\to \infty} \sqrt{\frac{\mathrm{gk}}{\mathrm{m}}t}= \infty$$ and $$\lim_{t\to\infty} \tanh\left(\sqrt{\frac{\mathrm{gk}}{\mathrm{m}}t}\right) = 1.$$ Substituting these limits into the equation for velocity, we get $$\lim_{t\to\infty} \mathrm{v} = \sqrt{\frac{\mathrm{mg}}{\mathrm{k}}} \lim_{t\to\infty} \tanh\left(\sqrt{\frac{\mathrm{gk}}{\mathrm{m}}t}\right) = \sqrt{\frac{(210\;\mathrm{lb})(32.2\;\mathrm{ft/s}^2)}{(0.006)}}(1) \approx 187.08\;\mathrm{ft/s}.$$ Therefore, the answer is B.

Question 2

Solve the problem.
-A region in the first quadrant is bounded above by the curve $y = \tanh x$, below by the $x$-axis, on the left by the $y$-axis, and on the right by the line $x = \ln 5$. Find the volume of the solid generated by revolving the region about the $x$-axis.
A) $2 \pi$
B) $- \frac { 12 } { 13 }$
C) $\pi \left( \ln 5 - \frac { 12 } { 13 } \right)$
D) 0

Accepted Answer

The curve $y = \tanh x$ approaches the $x$-axis and $y$-axis as $x\to -\infty$ and $y\to 0^+$. When $x \ge \ln 5$, it is tangent to the vertical line $x=\ln 5$. Since the curve is symmetric with respect to the $y$-axis, we can focus on the region in the first octant, as shown below.

[asy]
size(180);
import contour;
import graph;
import solids;

real ymin = 0;
real ymax = 0.9999;

add(contour(ymin^2/(2*ymin^2-1) <= x^2-y^2 && x <= log(5), (0.1,0.1), (log(5),ymax), 99), rgb(0.8,0.8,0.8));
draw((0,ymin)--(0,ymax),black+linewidth(1.25),EndArrow(5));
draw((0,0)--(log(5),0),black+linewidth(1.25),EndArrow(5));
draw(arc((0,0),ymax,0,90),black+linewidth(1.25));
draw((0,0)--(log(5),tanh(log(5))),red+linewidth(1.25));
draw((0,ymin)--(log(5),ymin),dashed);

label("$x$",(log(5),0),(2,0));
label("$y$",(0,ymax),(0,2));
label("$0$",(0.1,0.1),NE);
label("$\ln 5$",(log(5),0),S);
label("$y=\tanh x$",(log(5),tanh(log(5))),E);
label("$y=\frac { 1 } { \sqrt { 2 } }$",(0.2,0.2),W);
label("$y=-\frac { 1 } { \sqrt { 2 } }$",(0.2,-0.2),W);
[/asy]

The region is bounded by the curves
$$x^2-y^2 = y\tanh x \quad \text{and} \quad x = \ln 5.$$ Solving for $y$ in $x^2-y^2=y\tanh x$, we get
$$y = -\frac12\tanh x \pm \frac12\sqrt{\tanh^2x+4x^2}.$$ Since $0\le y\le \tanh x$, we take the positive sign and restrict attention to the region lying below the red line $y=-\frac12\tanh x+\frac12\sqrt{\tanh^2x+4x^2}$.

Rotating the region about the $x$-axis generates a solid with cross-sectional area $\pi y^2$ at height $y$. Thus, the desired volume is
\begin{align*}
\int_0^{\ln 5} \pi\left(-\frac 12\tanh x+\frac 12\sqrt{\tanh^2x+4x^2}\right)^2\,dx &= \int_0^{\ln 5} \pi\left(\frac 14-\frac 12\tanh x\sqrt{\tanh^2x+4x^2}+\frac 14(\tanh^2x+4x^2)\right)\,dx \
&= \pi\int_0^{\ln 5} \left(\frac 54x^2-\frac 12\tanh x\sqrt{\tanh^2x+4x^2}+\frac 14\right)\,dx \
&= \pi\left[\frac 35 x^3+\frac 14(\tanh x)\sqrt{\tanh^2x+4x^2}-\frac 12\ln\left|\frac {2(2x-\sqrt{\tanh^2x+4x^2})}{\tanh x+\sqrt{\tanh^2x+4x^2}}\right|\right]_0^{\ln 5} \
&= \boxed{\pi\left(\ln 5-\frac{12}{13}\right)}.
\end{align*}
The last step above follows from the substitution $u = e^{2x}$.
Hence the answer is $\boxed{\textbf{(C)}}$.

Question 3

Find the slowest growing and the fastest growing functions as x→∞ .
-$$\begin{array} { l } 
y = 7 x ^ { 10 } \
y = e ^ { x } \
y = e ^ { x - 2 } \
y = x e ^ { x }
\end{array}$$
A) Slowest: $y = 7 x ^ { 10 }$
Fastest: $y = e ^ { x }$ and $y = e ^ { x - 2 }$ grow at the same rate.
B) Slowest: $y = x e ^ { x }$
Fastest: $\mathrm { y } = \mathrm { e } ^ { \mathrm { x } }$
C) Slowest: $y = e ^ { x - 2 }$
Fastest: $y = x e ^ { x }$
D) Slowest: $\mathrm { y } = 7 \mathrm { x } ^ { 10 }$
Fastest: $y = x e ^ { x }$

Accepted Answer

The slowest growing function is $y = 7x^{10}$ because polynomial functions grow slower than exponential functions. The fastest growing function is $y = xe^x$ because it combines both linear and exponential growth, outpacing pure exponential growth as $x \rightarrow \infty$.

Question 4

Provide an appropriate response.
-Suppose you are looking for an item in an ordered list one million items long. Which would be better, a
sequential search or a binary search? Why?

Accepted Answer

The answer of Provide an appropriate response.
-Suppose you are looking...

Question 5

Solve the problem.
-A region in the first quadrant is bounded above by the curve $y = \cosh x$, below by the curve $y = \sinh x$, on the left by the $y$-axis, and on the right by the line $x = 12$. Find the volume of the solid generated by revolving the region about the $x$-axis.
A) 0
B) $\frac { \pi } { 2 } \left( e ^ { - 24 } + 1 \right)$
C) $12 \pi$
D) $2 \pi$

Accepted Answer

We start by trying to find the points of intersection between the two curves. Setting them equal, we get $\cosh x=\sinh x$, or $\tanh x=1$, which yields $x=\operatorname{arctanh} 1=\infty$. Since that is not a valid solution, we must have that the two curves do not intersect.

The two curves both intersect the origin, so that is the point of our solid. To find the other boundaries of the solid, we need to find the $x$-coordinate of the point on $\cosh x$ that is closest to $x=12$, and the $x$-coordinate of the point on $\sinh x$ that is farthest from $x=0$. The first point can be found using $\cosh x \ge e^x/2$ for all $x$, and so we want the value of $x$ that satisfies $$\frac{e^{12}}{2}=\cosh x=\frac{e^x+e^{-x}}{2}.$$Solving this yields $e^x=15$, or $x=\ln 15$. The second point can be found similarly by observing that $\sinh x \le e^x/2$ for all $x$, and so we want the value of $x$ that satisfies $$\frac{e^x}{2}=\sinh x=\frac{e^x-e^{-x}}{2}.$$Solving this yields $x=\ln 3$.

Thus, the solid is a region of revolution, and its volume can be computed with the formula $$\int_a^b \pi y^2\,dx$$where $y$ is the distance from the curve to the $x$-axis. The two curves don't intersect, so we can choose the limits of $a=0$ and $b=12$ without having to split up the integral.

Thus the volume of the solid is \begin{align*}
\int_0^{12}\pi\left(\cosh x\right)^2\, dx-\int_0^{12}\pi(\sinh x)^2\,dx &=\pi\int_0^{12}\left(\cosh^2 x-\sinh^2 x\right)\,dx \
&=\pi\int_0^{12}\cosh(2x)\,dx \
&=\frac{\pi}{2}\left[\sinh(24)\right] \
&=\boxed{12\pi}.
\end{align*}

Question 6

Solve the problem.
-The velocity of a body of mass $m$ falling from rest under the action of gravity is given by the equation $\mathrm { v } = \sqrt { \frac { \mathrm { mg } } { \mathrm { k } } } \tanh \left( \sqrt { \frac { \mathrm { gk } } { \mathrm { m } } } \mathrm { t } \right)$, where $\mathrm { k }$ is a constant that depends on the body's aerodynamic properties and the density of the air, $g$ is the gravitational constant, and $t$ is the number of seconds into the fall. Find the limiting velocity, $\lim _ { \mathrm { t } \rightarrow } \mathrm { v }$, of a $420 \mathrm { lb }$. skydiver ( $\mathrm { mg } = 420$ ) when $\mathrm { k } = 0.006$.
A) 264.58 ft/sec
B) 0.01 ft/sec
C) There is no limiting speed.
D) 83.67 ft/sec

Accepted Answer

The limiting velocity is found by taking the limit of the velocity equation as $t$ approaches infinity. The hyperbolic tangent function, $\tanh(x)$, approaches 1 as $x$ approaches infinity. Thus, the limiting velocity is $\sqrt{\frac{mg}{k}}$. Given $mg = 420$ and $k = 0.006$, the limiting velocity is $\sqrt{\frac{420}{0.006}} = 264.58$ ft/sec.

Question 7

Solve the problem.
-Consider the area of the region in the first quadrant enclosed by the curve $y = \frac { 1 } { 8 } \cosh 8 x$, the coordinate axes, and the line $x = 6$. This area is the same as the area of a rectangle of a length s, where $s$ is the length of the curve from $x = 0$ to $x = 6$. What is the height of the rectangle?
A) $\frac { 1 } { 8 }$
B) $\sinh 48$
C) $\frac { 1 } { 64 } \sinh 48$
D) 8

Accepted Answer

The answer of Solve the problem.
-Consider the area of the...

Question 8

Find the slowest growing and the fastest growing functions as x→∞ .
-$$\begin{array} { l } 
y = 2 x ^ { 2 } + 9 x \
y = e ^ { x } \
y = e ^ { x } / 6 \
y = \log _ { 4 } x
\end{array}$$
A) Slowest: $y = \log _ { 4 } x$
Fastest: $\mathrm { y } = \mathrm { e } ^ { \mathrm { X } }$
B) Slowest: $2 x ^ { 2 } + 9 x$
Fastest: $\mathrm { y } = \mathrm { e } ^ { \mathrm { x } }$
C) Slowest: $y = e ^ { x } / 6$
Fastest: $2 x ^ { 2 } + 9 x$
D) Slowest: $y = \log _ { 4 } x$
Fastest: $\mathrm { y } = \mathrm { e } ^ { \mathrm { x } }$ and $\mathrm { y } = \mathrm { e } ^ { \mathrm { x } } / 6$ grow at the same rate.

Accepted Answer

The slowest growing function is $y = \log_4 x$ because logarithmic growth is slower than polynomial, which in turn is slower than exponential growth. The fastest growing functions are $y = e^x$ and $y = e^x / 6$, both of which are exponential and grow at the same rate since dividing by a constant does not change the growth rate.

Question 9

Find the slowest growing and the fastest growing functions as x→∞ .
-$$\begin{array} { l } 
y = \ln 2 x \
y = 5 \ln x \
y = \frac { 1 } { x } \
y = \sqrt { x }
\end{array}$$
A) Slowest: $y = \frac { 1 } { x }$
Fastest: $y = \sqrt { x }$
B) Slowest: $y = \frac { 1 } { x }$
Fastest: $y = 5 \ln x$
C) Slowest: $y = \sqrt { x }$
Fastest: $y = \ln 2 x$ and $y = 5 \ln x$ grow at the same rate.
D) Slowest: $y = \ln 2 x$ and $y = 5 \ln x$ grow at the same rate. Fastest: $y = \sqrt { x }$

Accepted Answer

As x approaches infinity, the slowest growing function is always any function of the form $\frac { 1 } { x^k }$ where k is a positive constant. This means that $y = \frac { 1 } { x }$ is the slowest. On the other hand, the fastest-growing function is always any function of the form $x^k$ where k is a positive constant. This means that $y = \sqrt { x }$ is the fastest.

Question 10

Find the slowest growing and the fastest growing functions as x→∞ .
-$$\begin{array} { l } 
y = x ^ { 2 } + 7 x \
y = x ^ { 2 } \
y = \sqrt { x ^ { 4 } + x ^ { 2 } } \
y = 2 x ^ { 2 }
\end{array}$$
A) Slowest: $y = \sqrt { x ^ { 4 } + x ^ { 2 } }$
Fastest: $y = 2 x ^ { 2 }$
B) Slowest: $y = x ^ { 2 }$ and $y = 2 x ^ { 2 }$ grow at the same rate.
Fastest: $y = \sqrt { x ^ { 4 } + x ^ { 2 } }$
C) Slowest: $y = \sqrt { x ^ { 4 } + x ^ { 2 } }$
Fastest: $y = x ^ { 2 } + 7 x$
D) They all grow at the same rate.

Accepted Answer

All functions are asymptotically equivalent to $y = x^2$ as $x \to \infty$, so they all grow at the same rate.

Question 11

Find the slowest growing and the fastest growing functions as x→∞ .
-$\begin{array} { l } 
y = x + 7 \
y = e ^ { x } \
y = x ^ { 3 } + \cos ^ { 2 } x \
y = 4 ^ { x }
\end{array}$
A) Slowest: $y = x + 7$
Fastest: $\mathrm { y } = \mathrm { e } ^ { \mathrm { X } }$

B) Slowest: $y = e ^ { x }$
Fastest: $y = x ^ { 3 } + \cos ^ { 2 } x$

C) Slowest: $y = x + 7$
Fastest: $y = x ^ { 3 } + \cos ^ { 2 } x$

D) Slowest: $y = x + 7$
Fastest: $y = 4 ^ { x }$

Accepted Answer

The slowest growing function is $y = x + 7$ because it is linear, and among the given options, linear growth is slower than polynomial (cubic in this case) and exponential growth. The fastest growing function is $y = 4^x$ because exponential functions grow faster than both linear and polynomial functions, and $4^x$ grows faster than $e^x$ for large values of $x$.

Question 12

Provide an appropriate response.
-Show that $\lim _ { x \rightarrow } \frac { \ln ( x + 1 ) } { \ln x } = \lim _ { x \rightarrow } \frac { \ln ( x + 9982 ) } { \ln x }$. Explain why this is the case.

Accepted Answer

The answer of Provide an appropriate response.
-Show that \(\lim _...

Question 13

Find the slowest growing and the fastest growing functions as x→∞ .
-$$\begin{array} { l } 
y = e ^ { x } \
y = e ^ { x / 7 } \
y = x ^ { x } \
y = 4 ^ { x }
\end{array}$$
A) Slowest: $y = e ^ { x / 7 }$
Fastest: $\mathrm { y } = \mathrm { x } ^ { \mathrm { x } }$
B) Slowest: $\mathrm { y } = \mathrm { e } ^ { \mathrm { x } / 7 }$ and $\mathrm { y } = \mathrm { e } ^ { \mathrm { x } }$ grow at the same rate.
Fastest: $y = 4 ^ { x }$
C) Slowest: $y = x ^ { x }$
Fastest: $y = 4 ^ { x }$
D) Slowest: $y = e ^ { x / 7 }$ and $y = e ^ { x }$ grow at the same rate.
Fastest: $\mathrm { y } = \mathrm { x } ^ { \mathrm { x } }$

Accepted Answer

The slowest growing function is $y = e^{x/7}$ because it has the smallest base and exponent growth rate. The fastest growing function is $y = x^x$ because the exponent grows with $x$, leading to a much faster growth rate than the others as $x \rightarrow \infty$.

Question 14

Provide an appropriate response.
-

Accepted Answer

The answer of Provide an appropriate response.
-...

Question 15

Provide an appropriate response.
-A polynomial $f ( x )$ has a degree smaller than or equal to another polynomial $g ( x )$. Does $f = O ( g )$ and does $\mathrm { g } = \mathrm { O } ( \mathrm { f } )$ ?

Accepted Answer

The answer of Provide an appropriate response.
-A polynomial \(f (...

Question 16

Solve the problem.
-Find the length of the segment of the curve $y = \frac { 1 } { 2 } \cosh 2 x$ from $x = 0$ to $x = \ln \sqrt { 5 }$.
A) $\frac { 13 } { 10 }$
B) $\frac { 6 } { 5 }$
C) 5
D) $\frac { 1 } { 4 } \left( \sqrt { 5 } - \frac { 1 } { \sqrt { 5 } } \right)$

Accepted Answer

The answer of Solve the problem.
-Find the length of the...

Quiz 8: Integrals and Transcendental Functions

Solve the problem. -A region in the first quadrant is bounded above by the curve $y = \tanh x$ , below by the $x$ -axis, on the left by the $y$ -axis, and on the right by the line $x = \ln 5$ . Find the volume of the solid generated by revolving the region about the $x$ -axis.

Find the slowest growing and the fastest growing functions as x→∞ . - $\begin{array} { l } y = 7 x ^ { 10 } \\y = e ^ { x } \\y = e ^ { x - 2 } \\y = x e ^ { x }\end{array}$

Provide an appropriate response. -Suppose you are looking for an item in an ordered list one million items long. Which would be better, a sequential search or a binary search? Why?

Solve the problem. -A region in the first quadrant is bounded above by the curve $y = \cosh x$ , below by the curve $y = \sinh x$ , on the left by the $y$ -axis, and on the right by the line $x = 12$ . Find the volume of the solid generated by revolving the region about the $x$ -axis.

Find the slowest growing and the fastest growing functions as x→∞ . - $\begin{array} { l } y = 2 x ^ { 2 } + 9 x \\y = e ^ { x } \\y = e ^ { x } / 6 \\y = \log _ { 4 } x\end{array}$

Find the slowest growing and the fastest growing functions as x→∞ . - $\begin{array} { l } y = \ln 2 x \\y = 5 \ln x \\y = \frac { 1 } { x } \\y = \sqrt { x }\end{array}$

Find the slowest growing and the fastest growing functions as x→∞ . - $\begin{array} { l } y = x ^ { 2 } + 7 x \\y = x ^ { 2 } \\y = \sqrt { x ^ { 4 } + x ^ { 2 } } \\y = 2 x ^ { 2 }\end{array}$

Find the slowest growing and the fastest growing functions as x→∞ . - $\begin{array} { l } y = x + 7 \\y = e ^ { x } \\y = x ^ { 3 } + \cos ^ { 2 } x \\y = 4 ^ { x }\end{array}$

Provide an appropriate response. -Show that $\lim _ { x \rightarrow } \frac { \ln ( x + 1 ) } { \ln x } = \lim _ { x \rightarrow } \frac { \ln ( x + 9982 ) } { \ln x }$ . Explain why this is the case.

Find the slowest growing and the fastest growing functions as x→∞ . - $\begin{array} { l } y = e ^ { x } \\y = e ^ { x / 7 } \\y = x ^ { x } \\y = 4 ^ { x }\end{array}$

Provide an appropriate response. -

Provide an appropriate response. -A polynomial $f ( x )$ has a degree smaller than or equal to another polynomial $g ( x )$ . Does $f = O ( g )$ and does $\mathrm { g } = \mathrm { O } ( \mathrm { f } )$ ?

Solve the problem. -Find the length of the segment of the curve $y = \frac { 1 } { 2 } \cosh 2 x$ from $x = 0$ to $x = \ln \sqrt { 5 }$ .

Quiz 8: Integrals and Transcendental Functions

Find the slowest growing and the fastest growing functions as x→∞ . - y=7x10y=exy=ex−2y=xex\begin{array} { l } y = 7 x ^ { 10 } \\y = e ^ { x } \\y = e ^ { x - 2 } \\y = x e ^ { x }\end{array}y=7x10y=exy=ex−2y=xex​

Provide an appropriate response. -Suppose you are looking for an item in an ordered list one million items long. Which would be better, a sequential search or a binary search? Why?

Find the slowest growing and the fastest growing functions as x→∞ . - y=2x2+9xy=exy=ex/6y=log⁡4x\begin{array} { l } y = 2 x ^ { 2 } + 9 x \\y = e ^ { x } \\y = e ^ { x } / 6 \\y = \log _ { 4 } x\end{array}y=2x2+9xy=exy=ex/6y=log4​x​

Find the slowest growing and the fastest growing functions as x→∞ . - y=ln⁡2xy=5ln⁡xy=1xy=x\begin{array} { l } y = \ln 2 x \\y = 5 \ln x \\y = \frac { 1 } { x } \\y = \sqrt { x }\end{array}y=ln2xy=5lnxy=x1​y=x​​

Find the slowest growing and the fastest growing functions as x→∞ . - y=x2+7xy=x2y=x4+x2y=2x2\begin{array} { l } y = x ^ { 2 } + 7 x \\y = x ^ { 2 } \\y = \sqrt { x ^ { 4 } + x ^ { 2 } } \\y = 2 x ^ { 2 }\end{array}y=x2+7xy=x2y=x4+x2​y=2x2​

Find the slowest growing and the fastest growing functions as x→∞ . - y=x+7y=exy=x3+cos⁡2xy=4x\begin{array} { l } y = x + 7 \\y = e ^ { x } \\y = x ^ { 3 } + \cos ^ { 2 } x \\y = 4 ^ { x }\end{array}y=x+7y=exy=x3+cos2xy=4x​

Find the slowest growing and the fastest growing functions as x→∞ . - y=exy=ex/7y=xxy=4x\begin{array} { l } y = e ^ { x } \\y = e ^ { x / 7 } \\y = x ^ { x } \\y = 4 ^ { x }\end{array}y=exy=ex/7y=xxy=4x​