Calculus/Differentiation/Applications of Derivatives/Solutions

Relative Extrema

Find the relative maximum(s) and minimum(s), if any, of the following functions.

1.

f(x)={\frac {x}{x+1}}\,

f'(x)={\frac {1}{(x+1)^{2}}}

There are no roots of the derivative. The derivative fails to exist when x=-1 , but the function also fails to exists at that point, so it is not an extremum. Thus, the function has no relative extrema.

f'(x)={\frac {1}{(x+1)^{2}}}

There are no roots of the derivative. The derivative fails to exist when x=-1 , but the function also fails to exists at that point, so it is not an extremum. Thus, the function has no relative extrema.

2.

f(x)=(x-1)^{2/3}\,

f'(x)={\frac {2}{3}}(x-1)^{-1/3}={\frac {2}{3{\sqrt[{3}]{x-1}}}}

There are no roots of the derivative. The derivative fails to exist at

x=1

.

f(1)=0

. The point $(1,0)$ is a minimum since

f(x)

is nonnegative because of the even numerator in the exponent. The function has no relative maximum.

f'(x)={\frac {2}{3}}(x-1)^{-1/3}={\frac {2}{3{\sqrt[{3}]{x-1}}}}

There are no roots of the derivative. The derivative fails to exist at

x=1

.

f(1)=0

. The point $(1,0)$ is a minimum since

f(x)

is nonnegative because of the even numerator in the exponent. The function has no relative maximum.

3.

f(x)=x^{2}+{\frac {2}{x}}\,

f'(x)=2x-{\frac {2}{x^{2}}}

$2x-{\frac {2}{x^{2}}}=0\implies 2x^{3}-2=0\implies x=1$
$f^{\prime \prime }(x)=2+{\frac {4}{x^{3}}}$
$f^{\prime \prime }(1)=6$
Since the second derivative is positive, $x=1$ corresponds to a relative minimum.

The derivative fails to exist when

x=0

, but so does the function. There is no relative maximum.

f'(x)=2x-{\frac {2}{x^{2}}}

$2x-{\frac {2}{x^{2}}}=0\implies 2x^{3}-2=0\implies x=1$
$f^{\prime \prime }(x)=2+{\frac {4}{x^{3}}}$
$f^{\prime \prime }(1)=6$
Since the second derivative is positive, $x=1$ corresponds to a relative minimum.

The derivative fails to exist when

x=0

, but so does the function. There is no relative maximum.

4.

f(s)={\frac {s}{1+s^{2}}}\,

f'(x)={\frac {(1+s^{2})-2s^{2}}{(1+s^{2})^{2}}}={\frac {1-s^{2}}{(1+s^{2})^{2}}}

${\frac {1-s^{2}}{(1+s^{2})^{2}}}=0\implies s=\pm 1$
$f^{\prime \prime }(x)={\frac {-2s(1+s^{2})^{2}-2(1+s^{2})(2s)(1-s^{2})}{(1+s^{2})^{4}}}$
$f^{\prime \prime }(-1)={\frac {-2(-1)(1+(-1)^{2})^{2}-2(1+(-1)^{2})(2(-1))(1-(-1)^{2})}{(1+(-1)^{2})^{4}}}={\frac {8}{16}}={\frac {1}{2}}$
Since the second derivative of $f(x)$ at $x=-1$ is positive, $x=-1$ corresponds to a relative mimimum.
$f^{\prime \prime }(1)={\frac {-2(1+1^{2})^{2}-2(1+1^{2})(2)(1-1^{2})}{(1+1^{2})^{4}}}={\frac {-8}{16}}=-{\frac {1}{2}}$

Since the second derivative of

f(x)

at

x=1

is negative, $x=1$ corresponds to a relative maximum.

f'(x)={\frac {(1+s^{2})-2s^{2}}{(1+s^{2})^{2}}}={\frac {1-s^{2}}{(1+s^{2})^{2}}}

${\frac {1-s^{2}}{(1+s^{2})^{2}}}=0\implies s=\pm 1$
$f^{\prime \prime }(x)={\frac {-2s(1+s^{2})^{2}-2(1+s^{2})(2s)(1-s^{2})}{(1+s^{2})^{4}}}$
$f^{\prime \prime }(-1)={\frac {-2(-1)(1+(-1)^{2})^{2}-2(1+(-1)^{2})(2(-1))(1-(-1)^{2})}{(1+(-1)^{2})^{4}}}={\frac {8}{16}}={\frac {1}{2}}$
Since the second derivative of $f(x)$ at $x=-1$ is positive, $x=-1$ corresponds to a relative mimimum.
$f^{\prime \prime }(1)={\frac {-2(1+1^{2})^{2}-2(1+1^{2})(2)(1-1^{2})}{(1+1^{2})^{4}}}={\frac {-8}{16}}=-{\frac {1}{2}}$

Since the second derivative of

f(x)

at

x=1

is negative, $x=1$ corresponds to a relative maximum.

5.

f(x)=x^{2}-4x+9\,

f'(x)=2x-4

$2x-4=0\implies x=2$
$f^{\prime \prime }(x)=2$

Since the second derivative is positive, $x=2$ corresponds to a relative minimum. There is no relative maximum.

f'(x)=2x-4

$2x-4=0\implies x=2$
$f^{\prime \prime }(x)=2$

Since the second derivative is positive, $x=2$ corresponds to a relative minimum. There is no relative maximum.

6.

f(x)={\frac {x^{2}+x+1}{x^{2}-x+1}}\,

{\begin{aligned}f'(x)&={\frac {(x^{2}-x+1)(2x+1)-(2x-1)(x^{2}+x+1)}{(x^{2}-x+1)^{2}}}\\&={\frac {2x^{3}-2x^{2}+2x+x^{2}-x+1-2x^{3}-2x^{2}-2x+x^{2}+x+1}{(x^{2}-x+1)^{2}}}\\&={\frac {-2x^{2}+2}{(x^{2}-x+1)^{2}}}\end{aligned}}

${\frac {-2x^{2}+2}{(x^{2}-x+1)^{2}}}=0\implies x=\pm 1$
$f^{\prime \prime }(x)={\frac {(-4x)(x^{2}-x+1)^{2}-(2x-1)(-2x^{2}+2)}{(x^{2}-x+1)^{4}}}$
$f^{\prime \prime }(-1)={\frac {(-4(-1))((-1)^{2}-(-1)+1)^{2}-(2(-1)-1)(-2(-1)^{2}+2)}{((-1)^{2}-(-1)+1)^{4}}}={\frac {36}{81}}={\frac {4}{9}}$
Since $f^{\prime \prime }(-1)$ is positive, $x=-1$ corresponds to a relative minimum.
$f^{\prime \prime }(1)={\frac {-4}{1}}=-4$

Since

f^{\prime \prime }(1)

is negative, $x=1$ corresponds to a relative maximum.

{\begin{aligned}f'(x)&={\frac {(x^{2}-x+1)(2x+1)-(2x-1)(x^{2}+x+1)}{(x^{2}-x+1)^{2}}}\\&={\frac {2x^{3}-2x^{2}+2x+x^{2}-x+1-2x^{3}-2x^{2}-2x+x^{2}+x+1}{(x^{2}-x+1)^{2}}}\\&={\frac {-2x^{2}+2}{(x^{2}-x+1)^{2}}}\end{aligned}}

${\frac {-2x^{2}+2}{(x^{2}-x+1)^{2}}}=0\implies x=\pm 1$
$f^{\prime \prime }(x)={\frac {(-4x)(x^{2}-x+1)^{2}-(2x-1)(-2x^{2}+2)}{(x^{2}-x+1)^{4}}}$
$f^{\prime \prime }(-1)={\frac {(-4(-1))((-1)^{2}-(-1)+1)^{2}-(2(-1)-1)(-2(-1)^{2}+2)}{((-1)^{2}-(-1)+1)^{4}}}={\frac {36}{81}}={\frac {4}{9}}$
Since $f^{\prime \prime }(-1)$ is positive, $x=-1$ corresponds to a relative minimum.
$f^{\prime \prime }(1)={\frac {-4}{1}}=-4$

Since

f^{\prime \prime }(1)

is negative, $x=1$ corresponds to a relative maximum.

Range of Function

7. Show that the expression

x+1/x

cannot take on any value strictly between 2 and -2.

f(x)=x+{\frac {1}{x}}

$f'(x)=1-{\frac {1}{x^{2}}}$
$1-{\frac {1}{x^{2}}}=0\implies x=\pm 1$
$f^{\prime \prime }(x)={\frac {2}{x^{3}}}$
$f^{\prime \prime }(-1)=-2$
Since $f^{\prime \prime }(-1)$ is negative, $x=-1$ corresponds to a relative maximum.
$f(-1)=-2$
$\lim \limits _{x\to -\infty }f(x)=-\infty$
For $x<-1$ , $f'(x)$ is positive, which means that the function is increasing. Coming from very negative $x$ -values, $f$ increases from a very negative value to reach a relative maximum of $-2$ at $x=-1$ .
For $-1<x<1$ , $f'(x)$ is negative, which means that the function is decreasing.
$\lim _{x\to 0^{-}}f(x)=-\infty$
$\lim _{x\to 0^{+}}f(x)=+\infty$
$f^{\prime \prime }(1)=2$
Since $f^{\prime \prime }(1)$ is positive, $x=1$ corresponds to a relative minimum.
$f(1)=2$
Between $[-1,0)$ the function decreases from $-2$ to $-\infty$ , then jumps to $+\infty$ and decreases until it reaches a relative minimum of $2$ at $x=1$ .
For $x>1$ , $f'(x)$ is positive, so the function increases from a minimum of $2$ .

The above analysis shows that there is a gap in the function's range between

-2

and

2

.

f(x)=x+{\frac {1}{x}}

$f'(x)=1-{\frac {1}{x^{2}}}$
$1-{\frac {1}{x^{2}}}=0\implies x=\pm 1$
$f^{\prime \prime }(x)={\frac {2}{x^{3}}}$
$f^{\prime \prime }(-1)=-2$
Since $f^{\prime \prime }(-1)$ is negative, $x=-1$ corresponds to a relative maximum.
$f(-1)=-2$
$\lim \limits _{x\to -\infty }f(x)=-\infty$
For $x<-1$ , $f'(x)$ is positive, which means that the function is increasing. Coming from very negative $x$ -values, $f$ increases from a very negative value to reach a relative maximum of $-2$ at $x=-1$ .
For $-1<x<1$ , $f'(x)$ is negative, which means that the function is decreasing.
$\lim _{x\to 0^{-}}f(x)=-\infty$
$\lim _{x\to 0^{+}}f(x)=+\infty$
$f^{\prime \prime }(1)=2$
Since $f^{\prime \prime }(1)$ is positive, $x=1$ corresponds to a relative minimum.
$f(1)=2$
Between $[-1,0)$ the function decreases from $-2$ to $-\infty$ , then jumps to $+\infty$ and decreases until it reaches a relative minimum of $2$ at $x=1$ .
For $x>1$ , $f'(x)$ is positive, so the function increases from a minimum of $2$ .

The above analysis shows that there is a gap in the function's range between

-2

and

2

.

Absolute Extrema

Determine the absolute maximum and minimum of the following functions on the given domain

8.

f(x)={\frac {1}{3}}x^{3}-{\frac {1}{2}}x^{2}+1

on

[0,3]

f

is differentiable on

[0,3]

, so the extreme value theorem guarantees the existence of an absolute maximum and minimum on

[0,3]

. Find and check the critical points:

$f'(x)=x^{2}-x$
$x^{2}-x=0\implies x=0,1$
$f(0)=1$
$f(1)={\frac {5}{6}}$
Check the endpoint:
$f(3)={\frac {11}{2}}$

Maximum at $(3,{\frac {11}{2}})$ ; minimum at $(1,{\frac {5}{6}})$

f

is differentiable on

[0,3]

, so the extreme value theorem guarantees the existence of an absolute maximum and minimum on

[0,3]

. Find and check the critical points:

$f'(x)=x^{2}-x$
$x^{2}-x=0\implies x=0,1$
$f(0)=1$
$f(1)={\frac {5}{6}}$
Check the endpoint:
$f(3)={\frac {11}{2}}$

Maximum at $(3,{\frac {11}{2}})$ ; minimum at $(1,{\frac {5}{6}})$

9.

f(x)=({\frac {4}{3}}x^{2}-1)x

on

[-{\frac {1}{2}},2]

f'(x)=({\frac {4}{3}}x^{2}-1)+x({\frac {8}{3}}x)=4x^{2}-1

$4x^{2}-1=0\implies x=\pm {\frac {1}{2}}$
$f(-{\frac {1}{2}})=({\frac {1}{3}}-1)(-{\frac {1}{2}})={\frac {1}{3}}$
$f({\frac {1}{2}})=({\frac {1}{3}}-1)({\frac {1}{2}})=-{\frac {1}{3}}$
$f(2)=({\frac {16}{3}}-1)(2)={\frac {26}{3}}$

Maximum at $(2,{\frac {26}{3}})$ ; minimum at $({\frac {1}{2}},-{\frac {1}{3}})$

f'(x)=({\frac {4}{3}}x^{2}-1)+x({\frac {8}{3}}x)=4x^{2}-1

$4x^{2}-1=0\implies x=\pm {\frac {1}{2}}$
$f(-{\frac {1}{2}})=({\frac {1}{3}}-1)(-{\frac {1}{2}})={\frac {1}{3}}$
$f({\frac {1}{2}})=({\frac {1}{3}}-1)({\frac {1}{2}})=-{\frac {1}{3}}$
$f(2)=({\frac {16}{3}}-1)(2)={\frac {26}{3}}$

Maximum at $(2,{\frac {26}{3}})$ ; minimum at $({\frac {1}{2}},-{\frac {1}{3}})$

Determine Intervals of Change

Find the intervals where the following functions are increasing or decreasing

10.

f(x)=10-6x-2x^{2}

f'(x)=-6-4x

$-6-4x=0\implies x=-{\frac {3}{2}}$
$f'(x)$ is the equation of a line with negative slope, so $f'(x)$ is positive for $x<-{\frac {3}{2}}$ and negative for $x>{\frac {3}{2}}$ .

This means that the function is increasing on $(-\infty ,-{\frac {3}{2}})$ and decreasing on $({\frac {3}{2}},+\infty )$ .

f'(x)=-6-4x

$-6-4x=0\implies x=-{\frac {3}{2}}$
$f'(x)$ is the equation of a line with negative slope, so $f'(x)$ is positive for $x<-{\frac {3}{2}}$ and negative for $x>{\frac {3}{2}}$ .

This means that the function is increasing on $(-\infty ,-{\frac {3}{2}})$ and decreasing on $({\frac {3}{2}},+\infty )$ .

11.

f(x)=2x^{3}-12x^{2}+18x+15

f'(x)=6x^{2}-24x+18

${\begin{aligned}6x^{2}-24x+18=0&\implies x-4x+3=0\\&\implies (x-1)(x-3)=0\\&\implies x=1,3\end{aligned}}$
$f'(x)$ is the equation of a bowl-shaped parabola that crosses the $x$ -axis at $1$ and $3$ , so $f'(x)$ is negative for $1<x<3$ and positive elsewhere.

This means that the function is decreasing on $(1,3)$ and increasing elsewhere.

f'(x)=6x^{2}-24x+18

${\begin{aligned}6x^{2}-24x+18=0&\implies x-4x+3=0\\&\implies (x-1)(x-3)=0\\&\implies x=1,3\end{aligned}}$
$f'(x)$ is the equation of a bowl-shaped parabola that crosses the $x$ -axis at $1$ and $3$ , so $f'(x)$ is negative for $1<x<3$ and positive elsewhere.

This means that the function is decreasing on $(1,3)$ and increasing elsewhere.

12.

f(x)=5+36x+3x^{2}-2x^{3}

f'(x)=36+6x-6x^{2}

${\begin{aligned}36+6x-6x^{2}=0&\implies 6+x-x^{2}=0\\&\implies (x+2)(-x+3)=0\\&\implies x=-2,3\end{aligned}}$
$f'(x)$ is the equation of a hill-shaped parabola that crosses the $x$ -axis at $-2$ and $3$ , so $f'(x)$ is positive for $-2<x<3$ and negative elsewhere.

This means that the function is increasing on $(-2,3)$ and decreasing elsewhere.

f'(x)=36+6x-6x^{2}

${\begin{aligned}36+6x-6x^{2}=0&\implies 6+x-x^{2}=0\\&\implies (x+2)(-x+3)=0\\&\implies x=-2,3\end{aligned}}$
$f'(x)$ is the equation of a hill-shaped parabola that crosses the $x$ -axis at $-2$ and $3$ , so $f'(x)$ is positive for $-2<x<3$ and negative elsewhere.

This means that the function is increasing on $(-2,3)$ and decreasing elsewhere.

13.

f(x)=8+36x+3x^{2}-2x^{3}

If you did the previous exercise then no calculation is required since this function has the same derivative as that function and thus is increasing and decreasing on the same intervals; i.e., the function is increasing on $(-2,3)$ and decreasing elsewhere.

14.

f(x)=5x^{3}-15x^{2}-120x+3

f'(x)=15x^{2}-30x-120

${\begin{aligned}15x^{2}-30x-120=0&\implies x^{2}-2x-8=0\\&\implies (x+2)(x-4)=0\\&\implies x=-2,4\end{aligned}}$
$f'$ is negative on $(-2,4)$ and positive elsewhere.

So

f

is decreasing on $(-2,4)$ and increasing elsewhere.

f'(x)=15x^{2}-30x-120

${\begin{aligned}15x^{2}-30x-120=0&\implies x^{2}-2x-8=0\\&\implies (x+2)(x-4)=0\\&\implies x=-2,4\end{aligned}}$
$f'$ is negative on $(-2,4)$ and positive elsewhere.

So

f

is decreasing on $(-2,4)$ and increasing elsewhere.

15.

f(x)=x^{3}-6x^{2}-36x+2

f'(x)=3x^{2}-12x-36

${\begin{aligned}3x^{2}-12x-36=0&\implies x^{2}-4x-12=0\\&\implies (x+2)(x-6)=0\\&\implies x=-2,6\end{aligned}}$

f

is decreasing on $(-2,6)$ and increasing elsewhere.

f'(x)=3x^{2}-12x-36

${\begin{aligned}3x^{2}-12x-36=0&\implies x^{2}-4x-12=0\\&\implies (x+2)(x-6)=0\\&\implies x=-2,6\end{aligned}}$

f

is decreasing on $(-2,6)$ and increasing elsewhere.

Determine Intervals of Concavity

Find the intervals where the following functions are concave up or concave down

16.

f(x)=10-6x-2x^{2}

f'(x)=-6-4x

$f''(x)=-4$

The function is concave down everywhere.

f'(x)=-6-4x

$f''(x)=-4$

The function is concave down everywhere.

17.

f(x)=2x^{3}-12x^{2}+18x+15

f'(x)=6x^{2}-24x+18

$f^{\prime \prime }(x)=12x-24$
$12x-24=0\implies x=2$
When $x<2$ , $f^{\prime \prime }(x)$ is negative, and when $x>2$ , $f^{\prime \prime }(x)$ is positive.

This means that the function is concave down on $(-\infty ,2)$ and concave up on $(2,+\infty )$ .

f'(x)=6x^{2}-24x+18

$f^{\prime \prime }(x)=12x-24$
$12x-24=0\implies x=2$
When $x<2$ , $f^{\prime \prime }(x)$ is negative, and when $x>2$ , $f^{\prime \prime }(x)$ is positive.

This means that the function is concave down on $(-\infty ,2)$ and concave up on $(2,+\infty )$ .

18.

f(x)=5+36x+3x^{2}-2x^{3}

f'(x)=36+6x-6x^{2}

$f^{\prime \prime }(x)=6-12x$
$6-12x=0\implies x={\frac {1}{2}}$
$f^{\prime \prime }(x)$ is positive when $x<{\frac {1}{2}}$ and negative when $x>{\frac {1}{2}}$ .

This means that the function is concave up on $(-\infty ,{\frac {1}{2}})$ and concave down on $({\frac {1}{2}},+\infty )$ .

f'(x)=36+6x-6x^{2}

$f^{\prime \prime }(x)=6-12x$
$6-12x=0\implies x={\frac {1}{2}}$
$f^{\prime \prime }(x)$ is positive when $x<{\frac {1}{2}}$ and negative when $x>{\frac {1}{2}}$ .

This means that the function is concave up on $(-\infty ,{\frac {1}{2}})$ and concave down on $({\frac {1}{2}},+\infty )$ .

19.

f(x)=8+36x+3x^{2}-2x^{3}

If you did the previous exercise then no calculation is required since this function has the same second derivative as that function and thus is concave up and concave down on the same intervals; i.e., the function is concave up on $(-\infty ,{\frac {1}{2}})$ and concave down on $({\frac {1}{2}},+\infty )$ .

20.

f(x)=5x^{3}-15x^{2}-120x+3

f'(x)=15x^{2}-30x-120

$f^{\prime \prime }(x)=30x-30$
$30x-30=0\implies x=1$
$f^{\prime \prime }(x)$ is positive when $x>1$ and negative when $x<1$ .

This means that the function is concave down on $(-\infty ,1)$ and concave up on $(1,+\infty )$ .

f'(x)=15x^{2}-30x-120

$f^{\prime \prime }(x)=30x-30$
$30x-30=0\implies x=1$
$f^{\prime \prime }(x)$ is positive when $x>1$ and negative when $x<1$ .

This means that the function is concave down on $(-\infty ,1)$ and concave up on $(1,+\infty )$ .

21.

f(x)=x^{3}-6x^{2}-36x+2

f'(x)=3x^{2}-12x-36

$f^{\prime \prime }(x)=6x-12$
$6x-12=0\implies x=2$
$f^{\prime \prime }(x)$ is positive when $x>2$ and negative when $x<2$ .

This means that the function is concave down on $(-\infty ,2)$ and concave up on $(2,+\infty )$ .

f'(x)=3x^{2}-12x-36

$f^{\prime \prime }(x)=6x-12$
$6x-12=0\implies x=2$
$f^{\prime \prime }(x)$ is positive when $x>2$ and negative when $x<2$ .

This means that the function is concave down on $(-\infty ,2)$ and concave up on $(2,+\infty )$ .

Word Problems

22. You peer around a corner. A velociraptor 64 meters away spots you. You run away at a speed of 6 meters per second. The raptor chases, running towards the corner you just left at a speed of

4t

meters per second (time

t

measured in seconds after spotting). After you have run 4 seconds the raptor is 32 meters from the corner. At this time, how fast is death approaching your soon to be mangled flesh? That is, what is the rate of change in the distance between you and the raptor?

Velocity is the rate in change of position with respect to time. The raptor's velocity relative to you is given by

${\frac {dx}{dt}}=v(t)=4t-6$
After 4 seconds, the rate of change in position with respect to time is

v(4)=16-6=\mathbf {10{\frac {m}{s}}}

Velocity is the rate in change of position with respect to time. The raptor's velocity relative to you is given by

${\frac {dx}{dt}}=v(t)=4t-6$
After 4 seconds, the rate of change in position with respect to time is

v(4)=16-6=\mathbf {10{\frac {m}{s}}}

23. Two bicycles leave an intersection at the same time. One heads north going 12 mph and the other heads east going 5 mph. How fast are the bikes getting away from each other after one hour?

Set up a coordinate system with the origin at the intersection and the

+y

-axis pointing north. We assume that the position of the bike heading north is a function of the position of the bike heading east.

$y={\frac {12}{5}}x$
The distance between the bikes is given by
$s={\sqrt {x^{2}+y^{2}}}={\sqrt {x^{2}+({\frac {12}{5}}x)^{2}}}={\sqrt {{\frac {144+25}{25}}x^{2}}}={\sqrt {{\frac {169}{25}}x^{2}}}={\frac {13}{5}}x$
Let $t$ represent the elapsed time in hours. We want ${\frac {ds}{dt}}$ when $t=1$ . Apply the chain rule to $s$ :
${\frac {ds}{dt}}={\frac {ds}{dx}}{\frac {dx}{dt}}={\frac {13}{5}}\cdot 5=13$

Thus, the bikes are moving away from one another at 13 mph.

Set up a coordinate system with the origin at the intersection and the

+y

-axis pointing north. We assume that the position of the bike heading north is a function of the position of the bike heading east.

$y={\frac {12}{5}}x$
The distance between the bikes is given by
$s={\sqrt {x^{2}+y^{2}}}={\sqrt {x^{2}+({\frac {12}{5}}x)^{2}}}={\sqrt {{\frac {144+25}{25}}x^{2}}}={\sqrt {{\frac {169}{25}}x^{2}}}={\frac {13}{5}}x$
Let $t$ represent the elapsed time in hours. We want ${\frac {ds}{dt}}$ when $t=1$ . Apply the chain rule to $s$ :
${\frac {ds}{dt}}={\frac {ds}{dx}}{\frac {dx}{dt}}={\frac {13}{5}}\cdot 5=13$

Thus, the bikes are moving away from one another at 13 mph.

24. You're making a can of volume 200 m

^{3}

with a gold side and silver top/bottom. Say gold costs 10 dollars per m

^{2}

and silver costs 1 dollar per m

^{2}

. What's the minimum cost of such a can?

The volume of the can as a function of the radius,

r

, and the height,

h

, is

$V=\pi r^{2}h$
We are constricted to have a can with a volume of $200m^{3}$ , so we use this fact to relate the radius and the height:
$\pi r^{2}h=200\implies h={\frac {200}{\pi r^{2}}}$
The surface area of the side is
$A_{side}=2\pi rh=2\pi r{\frac {200}{\pi r^{2}}}={\frac {400}{r}}$
and the cost of the side is
$C_{side}={\frac {4000}{r}}$
The surface area of the top and bottom (which is also the cost) is
$A_{tb}=C_{tb}=2\pi r^{2}$
The total cost is given by
$C={\frac {4000}{r}}+2\pi r^{2}$
We want to minimize $C$ , so take the derivative:
$C'=-{\frac {4000}{r^{2}}}+4\pi r$
Find the critical points:
$-{\frac {4000}{r^{2}}}+4\pi r=0\implies 4\pi r={\frac {4000}{r^{2}}}\implies 4\pi r^{3}=4000\implies r={\sqrt[{3}]{\frac {1000}{\pi }}}={\frac {10}{\sqrt[{3}]{\pi }}}$
Check the second derivative to see if this point corresponds to a maximum or minimum:
$C^{\prime \prime }={\frac {8000}{r^{3}}}+4\pi$
$C^{\prime \prime }({\frac {10}{\sqrt[{3}]{\pi }}})={\frac {8000}{1000/\pi }}+4\pi =8\pi +4\pi =12\pi$
Since the second derivative is positive, the critical point corresponds to a minimum. Thus, the minimum cost is

C({\frac {10}{\sqrt[{3}]{\pi }}})={\frac {4000}{10/{\sqrt[{3}]{\pi }}}}+2\pi {\frac {10}{\sqrt[{3}]{\pi }}}\approx \mathbf {\$878.76}

The volume of the can as a function of the radius,

r

, and the height,

h

, is

$V=\pi r^{2}h$
We are constricted to have a can with a volume of $200m^{3}$ , so we use this fact to relate the radius and the height:
$\pi r^{2}h=200\implies h={\frac {200}{\pi r^{2}}}$
The surface area of the side is
$A_{side}=2\pi rh=2\pi r{\frac {200}{\pi r^{2}}}={\frac {400}{r}}$
and the cost of the side is
$C_{side}={\frac {4000}{r}}$
The surface area of the top and bottom (which is also the cost) is
$A_{tb}=C_{tb}=2\pi r^{2}$
The total cost is given by
$C={\frac {4000}{r}}+2\pi r^{2}$
We want to minimize $C$ , so take the derivative:
$C'=-{\frac {4000}{r^{2}}}+4\pi r$
Find the critical points:
$-{\frac {4000}{r^{2}}}+4\pi r=0\implies 4\pi r={\frac {4000}{r^{2}}}\implies 4\pi r^{3}=4000\implies r={\sqrt[{3}]{\frac {1000}{\pi }}}={\frac {10}{\sqrt[{3}]{\pi }}}$
Check the second derivative to see if this point corresponds to a maximum or minimum:
$C^{\prime \prime }={\frac {8000}{r^{3}}}+4\pi$
$C^{\prime \prime }({\frac {10}{\sqrt[{3}]{\pi }}})={\frac {8000}{1000/\pi }}+4\pi =8\pi +4\pi =12\pi$
Since the second derivative is positive, the critical point corresponds to a minimum. Thus, the minimum cost is

C({\frac {10}{\sqrt[{3}]{\pi }}})={\frac {4000}{10/{\sqrt[{3}]{\pi }}}}+2\pi {\frac {10}{\sqrt[{3}]{\pi }}}\approx \mathbf {\$878.76}

25. A farmer is investing in

216{\rm {\,cm}}

of fencing so that he can create an outdoor pen to display three different animals to sell. To make it cost effective, he used one of the walls of the outdoor barn as one of the sides of the fenced in area, which is able to enclose the entire area. He wants the internal areas for the animals to roam in to be congruent (i.e. he wants to segment the total area into three equal areas). What is the maximum internal area that the animals can roam in, given these conditions?

One important factor to consider when maximizing the area given the length of material given is that one of the sides is not made of the same material, but one that already exists (i.e. the barn wall is used in place of the fencing). Additionally, the roaming areas are congruent, so

A=3lw

for some length and width, and the perimeter is therefore

3w+4l=216

. Try to draw the situation as presented and the area and perimeter arguments may make more sense.

Notice that $3w+4l=216\Leftrightarrow 3w=216-4l$ , so $A=3lw=l(3w)\Rightarrow A(l)=l(216-4l)=216l-4l^{2}$ . Therefore, to maximize the total area (which will also maximize the internal roaming areas for the animal since each area is congruent), we take the derivative of the area function with respect to length.

A^{\prime }(l)=216-8l

A^{\prime }(l)=0\Rightarrow l=27

Notice that since $A^{\prime \prime }(l)=-8<0$ for all $l$ , it implies that $A(27)$ is a local maximum.

A(27)=27(216-4\cdot 27)=2916

Therefore, the maximum internal roaming area for one of the animals is

2916/3=\mathbf {972} {\rm {\,cm^{2}}}

.

One important factor to consider when maximizing the area given the length of material given is that one of the sides is not made of the same material, but one that already exists (i.e. the barn wall is used in place of the fencing). Additionally, the roaming areas are congruent, so

A=3lw

for some length and width, and the perimeter is therefore

3w+4l=216

. Try to draw the situation as presented and the area and perimeter arguments may make more sense.

Notice that $3w+4l=216\Leftrightarrow 3w=216-4l$ , so $A=3lw=l(3w)\Rightarrow A(l)=l(216-4l)=216l-4l^{2}$ . Therefore, to maximize the total area (which will also maximize the internal roaming areas for the animal since each area is congruent), we take the derivative of the area function with respect to length.

A^{\prime }(l)=216-8l

A^{\prime }(l)=0\Rightarrow l=27

Notice that since $A^{\prime \prime }(l)=-8<0$ for all $l$ , it implies that $A(27)$ is a local maximum.

A(27)=27(216-4\cdot 27)=2916

Therefore, the maximum internal roaming area for one of the animals is

2916/3=\mathbf {972} {\rm {\,cm^{2}}}

.

26. What is the maximum area of a rectangle inscribed (fitted so that the corners of the rectangle are on the circumference) inside a circle of radius

r

?

A rectangle's diagonals meet at the same point. Since it is inscribed inside a circle, the midpoint of the circle is the same as the intersection of the diagonals of the rectangle. Therefore, the radius and a point

(x,y)

on the circle and rectangle are related by the relation

x^{2}+y^{2}=r^{2}

.

Since it must be the case that $x^{2}+y^{2}=r^{2}$ , it also means that the set of points $P=\{(x,y),(-x,y),(x,-y),(-x,-y)\}$ also satisfy the relation. The distance between two horizontally or vertically related points is thus $2x$ and $2y$ respectively, so the area of the rectangle is given by $A=2x\cdot 2y=4xy$ .

Therefore, in terms of $x$ , the area of the rectangle is $A(x)=4x{\sqrt {r^{2}-x^{2}}}$ where $0\leq x\leq r$ . Taking the derivative to find the maximum,

A^{\prime }(x)=4{\sqrt {r^{2}-x^{2}}}-{\frac {4x^{2}}{\sqrt {r^{2}-x^{2}}}}

Notice that there exists a critical point on $x=\pm r$ (since the derivative does not exist on those points). We only worry about $x=r$ since $x=-r$ is outside of the points of interest.

Set the derivative to zero to find the other critical points.

{\begin{aligned}A^{\prime }(x)=0&\Rightarrow 4{\sqrt {r^{2}-x^{2}}}-{\frac {4x^{2}}{\sqrt {r^{2}-x^{2}}}}=0\\&\Rightarrow 4\left(r^{2}-x^{2}\right)-4x^{2}=0\\&\Rightarrow 4r^{2}-8x^{2}=0\\&\Rightarrow x^{2}={\frac {r^{2}}{2}}{\text{ or }}x=\pm {\frac {r}{\sqrt {2}}}\end{aligned}}

Since $x=-{\frac {r}{\sqrt {2}}}$ falls outside our points of interest, the only critical points we care about is at $x={\frac {r}{\sqrt {2}}}$ . Therefore, we test the three critical points:

{\begin{aligned}A(0)&=4\cdot (0)\cdot {\sqrt {r^{2}-(0)^{2}}}=0\\A\left({\frac {r}{\sqrt {2}}}\right)&=4\cdot \left({\frac {r}{\sqrt {2}}}\right)\cdot {\sqrt {r^{2}-\left({\frac {r}{\sqrt {2}}}\right)^{2}}}\\&=4\cdot {\frac {r}{\sqrt {2}}}\cdot {\sqrt {\frac {r^{2}}{2}}}\\&=4\cdot {\frac {r^{2}}{2}}=2r^{2}\\A(r)&=4\cdot (r)\cdot {\sqrt {r^{2}-(r)^{2}}}=0\end{aligned}}

The largest area from the critical points tested is

{\textbf {2r}}^{\textbf {2}}

.

A rectangle's diagonals meet at the same point. Since it is inscribed inside a circle, the midpoint of the circle is the same as the intersection of the diagonals of the rectangle. Therefore, the radius and a point

(x,y)

on the circle and rectangle are related by the relation

x^{2}+y^{2}=r^{2}

.

Since it must be the case that $x^{2}+y^{2}=r^{2}$ , it also means that the set of points $P=\{(x,y),(-x,y),(x,-y),(-x,-y)\}$ also satisfy the relation. The distance between two horizontally or vertically related points is thus $2x$ and $2y$ respectively, so the area of the rectangle is given by $A=2x\cdot 2y=4xy$ .

Therefore, in terms of $x$ , the area of the rectangle is $A(x)=4x{\sqrt {r^{2}-x^{2}}}$ where $0\leq x\leq r$ . Taking the derivative to find the maximum,

A^{\prime }(x)=4{\sqrt {r^{2}-x^{2}}}-{\frac {4x^{2}}{\sqrt {r^{2}-x^{2}}}}

Notice that there exists a critical point on $x=\pm r$ (since the derivative does not exist on those points). We only worry about $x=r$ since $x=-r$ is outside of the points of interest.

Set the derivative to zero to find the other critical points.

{\begin{aligned}A^{\prime }(x)=0&\Rightarrow 4{\sqrt {r^{2}-x^{2}}}-{\frac {4x^{2}}{\sqrt {r^{2}-x^{2}}}}=0\\&\Rightarrow 4\left(r^{2}-x^{2}\right)-4x^{2}=0\\&\Rightarrow 4r^{2}-8x^{2}=0\\&\Rightarrow x^{2}={\frac {r^{2}}{2}}{\text{ or }}x=\pm {\frac {r}{\sqrt {2}}}\end{aligned}}

Since $x=-{\frac {r}{\sqrt {2}}}$ falls outside our points of interest, the only critical points we care about is at $x={\frac {r}{\sqrt {2}}}$ . Therefore, we test the three critical points:

{\begin{aligned}A(0)&=4\cdot (0)\cdot {\sqrt {r^{2}-(0)^{2}}}=0\\A\left({\frac {r}{\sqrt {2}}}\right)&=4\cdot \left({\frac {r}{\sqrt {2}}}\right)\cdot {\sqrt {r^{2}-\left({\frac {r}{\sqrt {2}}}\right)^{2}}}\\&=4\cdot {\frac {r}{\sqrt {2}}}\cdot {\sqrt {\frac {r^{2}}{2}}}\\&=4\cdot {\frac {r^{2}}{2}}=2r^{2}\\A(r)&=4\cdot (r)\cdot {\sqrt {r^{2}-(r)^{2}}}=0\end{aligned}}

The largest area from the critical points tested is

{\textbf {2r}}^{\textbf {2}}

.

27. A cylinder is to be fitted inside a glass spherical display case with a radius of

R=3{\rm {\,in}}

. (The sphere will form around the cylinder.) What is the largest volume that a cylinder will have inside such a display case?

Notice in the figure that the corners of the cylinder can connect to the center of the sphere

O

. Additionally, the center of the sphere is also the center of the cylinder, so given a height

h

and radius

r

, the relation of the corners of the cylinder to the sphere are thus given by

r^{2}+\left({\frac {h}{2}}\right)^{2}=R^{2}

.

Given the volume of a cylinder is $V=\pi r^{2}h$ , and $r^{2}=R^{2}-\left({\frac {h}{2}}\right)^{2}=R^{2}-{\frac {h^{2}}{4}}$ , the volume of the cylinder with respect to the height is

V(h)=\pi \left(R^{2}-{\frac {h^{2}}{4}}\right)h=\pi hR^{2}-{\frac {h^{3}\pi }{4}}

V^{\prime }(h)=\pi R^{2}-{\frac {3h^{2}\pi }{4}}

Setting the derivative to zero yields the critical points.

{\begin{aligned}V^{\prime }(h)=0&\Rightarrow \pi R^{2}-{\frac {3h^{2}\pi }{4}}=0\\&\Rightarrow \pi \left(R^{2}-{\frac {3h^{2}}{4}}\right)=0\\&\Rightarrow R^{2}-{\frac {3h^{2}}{4}}=0\Rightarrow R^{2}={\frac {3h^{2}}{4}}\\&\Rightarrow {\frac {4}{3}}R^{2}=h^{2}\\&\Rightarrow h=\pm {\sqrt {{\frac {4}{3}}R^{2}}}\end{aligned}}

.

Keep in mind that the volume must be positive, so $h>0$ . As such, we only need to look at the critical point $h={\sqrt {{\frac {4}{3}}R^{2}}}$ . Taking the second derivative

V^{\prime \prime }(h)=-{\frac {6\pi }{4}}h=-{\frac {3\pi }{2}}h

Since the critical point $h={\sqrt {{\frac {4}{3}}R^{2}}}$ is positive, and $V^{\prime \prime }(h)$ is linear, $V^{\prime \prime }(0)=0$ , and $V^{\prime \prime }(h)$ is trending downward,

V^{\prime \prime }\left({\sqrt {{\frac {4}{3}}R^{2}}}\right)<0

, so by the second derivative test,

V\left({\sqrt {{\frac {4}{3}}R^{2}}}\right)

is a local maximum.

Recall the radius of the sphere is $R=3$ , so $h={\sqrt {{\frac {4}{3}}\cdot 3^{2}}}={\sqrt {12}}$ .

{\begin{aligned}V({\sqrt {12}})&=\pi \left(3^{2}-{\frac {\left({\sqrt {12}}\right)^{2}}{4}}\right){\sqrt {12}}\\&=\pi \left(9-3\right){\sqrt {4\cdot 3}}\\&=\pi \left(6\right)2{\sqrt {3}}=12\pi {\sqrt {3}}\end{aligned}}

Therefore, the maximum volume of the cylinder inscribed inside a sphere is

{\textbf {12}}{\boldsymbol {\pi }}{\sqrt {\textbf {3}}}\,{\textbf {in}}^{\textbf {3}}

.

Notice in the figure that the corners of the cylinder can connect to the center of the sphere

O

. Additionally, the center of the sphere is also the center of the cylinder, so given a height

h

and radius

r

, the relation of the corners of the cylinder to the sphere are thus given by

r^{2}+\left({\frac {h}{2}}\right)^{2}=R^{2}

.

Given the volume of a cylinder is $V=\pi r^{2}h$ , and $r^{2}=R^{2}-\left({\frac {h}{2}}\right)^{2}=R^{2}-{\frac {h^{2}}{4}}$ , the volume of the cylinder with respect to the height is

V(h)=\pi \left(R^{2}-{\frac {h^{2}}{4}}\right)h=\pi hR^{2}-{\frac {h^{3}\pi }{4}}

V^{\prime }(h)=\pi R^{2}-{\frac {3h^{2}\pi }{4}}

Setting the derivative to zero yields the critical points.

{\begin{aligned}V^{\prime }(h)=0&\Rightarrow \pi R^{2}-{\frac {3h^{2}\pi }{4}}=0\\&\Rightarrow \pi \left(R^{2}-{\frac {3h^{2}}{4}}\right)=0\\&\Rightarrow R^{2}-{\frac {3h^{2}}{4}}=0\Rightarrow R^{2}={\frac {3h^{2}}{4}}\\&\Rightarrow {\frac {4}{3}}R^{2}=h^{2}\\&\Rightarrow h=\pm {\sqrt {{\frac {4}{3}}R^{2}}}\end{aligned}}

.

Keep in mind that the volume must be positive, so $h>0$ . As such, we only need to look at the critical point $h={\sqrt {{\frac {4}{3}}R^{2}}}$ . Taking the second derivative

V^{\prime \prime }(h)=-{\frac {6\pi }{4}}h=-{\frac {3\pi }{2}}h

Since the critical point $h={\sqrt {{\frac {4}{3}}R^{2}}}$ is positive, and $V^{\prime \prime }(h)$ is linear, $V^{\prime \prime }(0)=0$ , and $V^{\prime \prime }(h)$ is trending downward,

V^{\prime \prime }\left({\sqrt {{\frac {4}{3}}R^{2}}}\right)<0

, so by the second derivative test,

V\left({\sqrt {{\frac {4}{3}}R^{2}}}\right)

is a local maximum.

Recall the radius of the sphere is $R=3$ , so $h={\sqrt {{\frac {4}{3}}\cdot 3^{2}}}={\sqrt {12}}$ .

{\begin{aligned}V({\sqrt {12}})&=\pi \left(3^{2}-{\frac {\left({\sqrt {12}}\right)^{2}}{4}}\right){\sqrt {12}}\\&=\pi \left(9-3\right){\sqrt {4\cdot 3}}\\&=\pi \left(6\right)2{\sqrt {3}}=12\pi {\sqrt {3}}\end{aligned}}

Therefore, the maximum volume of the cylinder inscribed inside a sphere is

{\textbf {12}}{\boldsymbol {\pi }}{\sqrt {\textbf {3}}}\,{\textbf {in}}^{\textbf {3}}

.

28. A

6\,{\rm {ft}}

tall man is walking away from a light that is

15

-feet above the ground. The man is walking away from the light at

6

feet per second. How fast (speed not velocity) is the shadow, cast by the man, changing its length with respect to time?

The light and the man are related by a triangle relation as shown. The man, ${\overline {PB}}$ , is moving away from the light at a rate of $6$ feet per second, so the length of ${\overline {AB}}$ must be changing. Therefore, set that length to be $x$ . Assuming the length of the shadow and the distance from the post is $L$ , i.e. $m{\overline {AQ}}=L$ , the shadow has $m{\overline {BQ}}=L-x$ .

Notice the similar triangles involved: $\triangle OCP\sim \triangle PBQ$ because $\angle OPC\cong \angle PQB$ (show this yourself!). As such, we may relate lengths as shown:

{\begin{aligned}{\frac {m{\overline {BQ}}}{m{\overline {PB}}}}&={\frac {m{\overline {CP}}}{m{\overline {OC}}}}\\{\frac {L-x}{6}}&={\frac {x}{15-6}}\\(L-x)(15-6)=9L-9x&=6x\\L&={\frac {15}{9}}x\end{aligned}}

The length of the shadow as a function of the person's distance from the post is

S(x)=L-x={\frac {15}{9}}x-x={\frac {2}{3}}x

. Therefore, the rate of change of the length of the shadow is

{\frac {dS}{dt}}={\frac {2}{3}}{\frac {dx}{dt}}={\frac {2}{3}}\cdot 6\,{\rm {ft/s}}={\textbf {4}}\,{\textbf {ft/s}}

.

The light and the man are related by a triangle relation as shown. The man, ${\overline {PB}}$ , is moving away from the light at a rate of $6$ feet per second, so the length of ${\overline {AB}}$ must be changing. Therefore, set that length to be $x$ . Assuming the length of the shadow and the distance from the post is $L$ , i.e. $m{\overline {AQ}}=L$ , the shadow has $m{\overline {BQ}}=L-x$ .

Notice the similar triangles involved: $\triangle OCP\sim \triangle PBQ$ because $\angle OPC\cong \angle PQB$ (show this yourself!). As such, we may relate lengths as shown:

{\begin{aligned}{\frac {m{\overline {BQ}}}{m{\overline {PB}}}}&={\frac {m{\overline {CP}}}{m{\overline {OC}}}}\\{\frac {L-x}{6}}&={\frac {x}{15-6}}\\(L-x)(15-6)=9L-9x&=6x\\L&={\frac {15}{9}}x\end{aligned}}

The length of the shadow as a function of the person's distance from the post is

S(x)=L-x={\frac {15}{9}}x-x={\frac {2}{3}}x

. Therefore, the rate of change of the length of the shadow is

{\frac {dS}{dt}}={\frac {2}{3}}{\frac {dx}{dt}}={\frac {2}{3}}\cdot 6\,{\rm {ft/s}}={\textbf {4}}\,{\textbf {ft/s}}

.

29. A canoe is being pulled toward a dock (normal to the water) using a taut rope. The canoe is normal to the water while it is being pulled. The rope is hauled in at a constant $5\,{\rm {ft/s}}$ . The dock is $5\,{\rm {ft}}$ above the water. Answer items (a) through (b).

(a) How fast is the boat approaching the dock when

13\,{\rm {ft}}

of rope are out?

Since the boat is

x

feet from the dock, of height

y=5\,{\rm {ft}}

, and the two measures are perpendicular, the pythagorean theorem applies. The length of the rope,

r=13\,{\rm {ft}}

, is the hypotenuse between

x

and

y

, so

x^{2}+5^{2}=13^{2}\Rightarrow x=12\,{\rm {ft}}

.

Implicitly taking the derivative of the relation ( $x^{2}+y^{2}=r^{2}$ ) tells us that $2x\cdot {\frac {dx}{dt}}=2r\cdot {\frac {dr}{dt}}$ (notice that the height of the dock is constant). Isolating for ${\frac {dx}{dt}}={\dot {x}}$ :

{\begin{aligned}{\dot {x}}&={\frac {r}{x}}{\dot {r}}\\&={\frac {13}{5}}\cdot 5\,{\rm {ft/s}}\\&={\textbf {13}}\,{\textbf {ft/s}}\end{aligned}}

Since the boat is

x

feet from the dock, of height

y=5\,{\rm {ft}}

, and the two measures are perpendicular, the pythagorean theorem applies. The length of the rope,

r=13\,{\rm {ft}}

, is the hypotenuse between

x

and

y

, so

x^{2}+5^{2}=13^{2}\Rightarrow x=12\,{\rm {ft}}

.

Implicitly taking the derivative of the relation ( $x^{2}+y^{2}=r^{2}$ ) tells us that $2x\cdot {\frac {dx}{dt}}=2r\cdot {\frac {dr}{dt}}$ (notice that the height of the dock is constant). Isolating for ${\frac {dx}{dt}}={\dot {x}}$ :

{\begin{aligned}{\dot {x}}&={\frac {r}{x}}{\dot {r}}\\&={\frac {13}{5}}\cdot 5\,{\rm {ft/s}}\\&={\textbf {13}}\,{\textbf {ft/s}}\end{aligned}}

(b) Hence, what is the rate of change of the angle between the rope and the dock?

In this situation, the angle between the dock and the rope is

\sin(\theta )={\frac {x}{r}}={\frac {12}{13}}

. Note that with the given quantities,

\cos(\theta )={\frac {5}{13}}

. Implicitly taking the derivative tells us that

{\begin{aligned}{\dot {\theta }}\cos(\theta )&={\frac {r{\dot {x}}-x{\dot {r}}}{r^{2}}}\\{\dot {\theta }}&={\frac {r{\dot {x}}-x{\dot {r}}}{r^{2}\cos(\theta )}}\\{\dot {\theta }}&={\frac {13\cdot 13-12\cdot 13}{(13)^{2}\cdot {\frac {5}{13}}}}\\&={\frac {1}{5}}={\textbf {0.2}}\,{\textbf {rad/s}}\end{aligned}}

One additional note to keep in mind is that the angle is a unit-less quantity quantitatively, but in context, it was useful to define the units.

In this situation, the angle between the dock and the rope is

\sin(\theta )={\frac {x}{r}}={\frac {12}{13}}

. Note that with the given quantities,

\cos(\theta )={\frac {5}{13}}

. Implicitly taking the derivative tells us that

{\begin{aligned}{\dot {\theta }}\cos(\theta )&={\frac {r{\dot {x}}-x{\dot {r}}}{r^{2}}}\\{\dot {\theta }}&={\frac {r{\dot {x}}-x{\dot {r}}}{r^{2}\cos(\theta )}}\\{\dot {\theta }}&={\frac {13\cdot 13-12\cdot 13}{(13)^{2}\cdot {\frac {5}{13}}}}\\&={\frac {1}{5}}={\textbf {0.2}}\,{\textbf {rad/s}}\end{aligned}}

One additional note to keep in mind is that the angle is a unit-less quantity quantitatively, but in context, it was useful to define the units.

30. A very enthusiastic parent is video taping a runner in your class during a

100\,{\rm {m}}

race. The parent has the runner center frame and is recording

2\,{\rm {m}}

from the straight-line track. The runner in your class is running at a constant

4\,{\rm {m/s}}

. What is the rate of change of the shooting angle if the runner passes the parent half a second after the parent's direct shot (after the point in which the runner's motion and the parent's line of sight are perpendicular)?

In this situation, the angle between the runner and the direct line from the track is

\tan(\theta )={\frac {x}{2}}

. Implicitly taking the derivative tells us that

{\begin{aligned}{\dot {\theta }}\sec ^{2}(\theta )&={\frac {\dot {x}}{2}}\\{\dot {\theta }}&={\frac {\dot {x}}{2\sec ^{2}(\theta )}}\\&={\frac {\dot {x}}{2}}\cos ^{2}(\theta )\end{aligned}}

Notice that $\theta =\tan ^{-1}\left({\frac {x}{2}}\right)$ . Since $x(t)=4t$ after the direct line shot, $x(0.5)=4\cdot 0.5=2$ . Therefore, because $\tan ^{-1}\left(1\right)={\frac {\pi }{4}}$ ,

{\begin{aligned}{\dot {\theta }}&={\frac {4}{2}}\cdot \cos ^{2}\left({\frac {\pi }{4}}\right)\\&=2\cdot {\frac {1}{2}}={\textbf {1}}\;{\textbf {rad/s}}\end{aligned}}

One additional note to keep in mind is that the angle is a unit-less quantity quantitatively, but in context, it was useful to define the units.

In this situation, the angle between the runner and the direct line from the track is

\tan(\theta )={\frac {x}{2}}

. Implicitly taking the derivative tells us that

{\begin{aligned}{\dot {\theta }}\sec ^{2}(\theta )&={\frac {\dot {x}}{2}}\\{\dot {\theta }}&={\frac {\dot {x}}{2\sec ^{2}(\theta )}}\\&={\frac {\dot {x}}{2}}\cos ^{2}(\theta )\end{aligned}}

Notice that $\theta =\tan ^{-1}\left({\frac {x}{2}}\right)$ . Since $x(t)=4t$ after the direct line shot, $x(0.5)=4\cdot 0.5=2$ . Therefore, because $\tan ^{-1}\left(1\right)={\frac {\pi }{4}}$ ,

{\begin{aligned}{\dot {\theta }}&={\frac {4}{2}}\cdot \cos ^{2}\left({\frac {\pi }{4}}\right)\\&=2\cdot {\frac {1}{2}}={\textbf {1}}\;{\textbf {rad/s}}\end{aligned}}

One additional note to keep in mind is that the angle is a unit-less quantity quantitatively, but in context, it was useful to define the units.

Approximation Problems

By assumption, for these problems, assume $\pi \approx 3.14$ and $e\approx 2.718$ unless stated otherwise. One may use a calculator or design a computer program, but one must indicate the method and reasoning behind every step where necessary.

35. Approximate

\sin(3)

using whatever method. If you use Newton's or Euler's method, do it in a maximum of THREE (3) iterations.

This is an example solution and does not mean that other methods do not exist. We will use Euler's method. To use this, express the derivative

f(x,y)={\frac {dy}{dx}}

in terms of

y(x)

.

{\frac {d}{dx}}\left[\sin(x)\right]=\cos(x)

We know $\cos(x)={\sqrt {1-\sin ^{2}(x)}}$ . Since $y=\sin(x)$ ,

f(x,y)={\sqrt {1-y^{2}}}

.

We will use step size $\Delta x=\pi -3\approx 0.14$ since it will be the easiest to calculate without the use of a calculator. Let $y_{0}=y(\pi )$ .

y_{1}=0+0.14f\left(\pi ,0\right)

y_{1}=0.14\cdot 1=0.14

With this,

\sin(3)\approx \mathbf {0.14}

.

This is an example solution and does not mean that other methods do not exist. We will use Euler's method. To use this, express the derivative

f(x,y)={\frac {dy}{dx}}

in terms of

y(x)

.

{\frac {d}{dx}}\left[\sin(x)\right]=\cos(x)

We know $\cos(x)={\sqrt {1-\sin ^{2}(x)}}$ . Since $y=\sin(x)$ ,

f(x,y)={\sqrt {1-y^{2}}}

.

We will use step size $\Delta x=\pi -3\approx 0.14$ since it will be the easiest to calculate without the use of a calculator. Let $y_{0}=y(\pi )$ .

y_{1}=0+0.14f\left(\pi ,0\right)

y_{1}=0.14\cdot 1=0.14

With this,

\sin(3)\approx \mathbf {0.14}

.

36. Approximate

{\sqrt {2}}

using whatever method. If you use Newton's or Euler's method, do it in a maximum of THREE (3) iterations.

This is an example solution and does not mean that other methods do not exist. We will use the Newton-Rhapson method. To use this, we need to find an equation that allows

{\sqrt {2}}

to be the root of the function. One obvious example is

f(x)=x^{2}-2

. From there,

f^{\prime }(x)=2x

. Let

x_{0}=1

x_{1}=1-{\frac {-1}{2}}=1.5

x_{2}=1.5-{\frac {0.25}{3}}\approx 1.4167

With this,

{\sqrt {2}}\approx \mathbf {1.4167}

This is an example solution and does not mean that other methods do not exist. We will use the Newton-Rhapson method. To use this, we need to find an equation that allows

{\sqrt {2}}

to be the root of the function. One obvious example is

f(x)=x^{2}-2

. From there,

f^{\prime }(x)=2x

. Let

x_{0}=1

x_{1}=1-{\frac {-1}{2}}=1.5

x_{2}=1.5-{\frac {0.25}{3}}\approx 1.4167

With this,

{\sqrt {2}}\approx \mathbf {1.4167}

37. Approximate

\ln(3)

using whatever method. If you use Newton's or Euler's method, do it in a maximum of THREE (3) iterations.

This is an example solution and does not mean that other methods do not exist. We will use the local-point linear approximation. We know

\ln(e)=1

. Let

f(x)=\ln(x)

;

f^{\prime }(e)={\frac {1}{e}}

. The tangent line equation at the point

(e,1)

is

y={\frac {1}{e}}\left(x-e\right)+1={\frac {x}{e}}

. Because

3-e

is a small difference, we may assume that

y\approx f(x)

. Therefore

f(3)\approx {\frac {3}{e}}\approx 1.10

. With this,

\ln(3)\approx \mathbf {1.10} {\text{ OR }}\mathbf {1.09}

using linearization.

This is an example solution and does not mean that other methods do not exist. We will use the local-point linear approximation. We know

\ln(e)=1

. Let

f(x)=\ln(x)

;

f^{\prime }(e)={\frac {1}{e}}

. The tangent line equation at the point

(e,1)

is

y={\frac {1}{e}}\left(x-e\right)+1={\frac {x}{e}}

. Because

3-e

is a small difference, we may assume that

y\approx f(x)

. Therefore

f(3)\approx {\frac {3}{e}}\approx 1.10

. With this,

\ln(3)\approx \mathbf {1.10} {\text{ OR }}\mathbf {1.09}

using linearization.

Advanced Understanding

45. Consider the differentiable function $f(x)$ for all $x>-2$ and continuous function $g(x)$ below, where $g(x)$ is linear for all $x\geq 1$ and differentiable for all $x\in (-2,1)\cup (1,\infty )$ , and $f(x)$ and $g(x)$ are continuous for all $x\geq -2$ .

a. Approximate

f^{\prime }(1.6)

.

To approximate, choose two nearby points.

f^{\prime }(1.6)\approx {\frac {1-3}{2-1.6}}=\mathbf {-5}

To approximate, choose two nearby points.

f^{\prime }(1.6)\approx {\frac {1-3}{2-1.6}}=\mathbf {-5}

b. Using your answer from (a), find

\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}

.

\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}\to {\frac {0}{0}}

. Additionally,

f{\text{ and }}g

are differentiable at

x=1.6

since

f

is differentiable for all

x\geq -2

and

g

is linear for all

x\geq 1

. Therefore, L'hôpital's rule applies.

{\begin{aligned}\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&=\lim _{x\to 1.6}{\frac {{\frac {d}{dx}}\left[f(x)-3\right]}{{\frac {d}{dx}}\left[3-g(x)\right]}}\\&=\lim _{x\to 1.6}{\frac {f^{\prime }(x)}{-g^{\prime }(x)}}\end{aligned}}

Since $g(x)$ is linear for all $x\geq 1$ ,

-g^{\prime }(1.6)=-{\frac {3+4}{1.6-1}}=-{\frac {7}{\frac {3}{5}}}

Using the answer from (a),

{\begin{aligned}\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&={\frac {f^{\prime }(1.6)}{-g^{\prime }(1.6)}}\\&={\frac {-5}{-{\frac {7}{\frac {3}{5}}}}}\\&={\frac {15}{35}}\\\therefore \lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&=\mathbf {\frac {3}{7}} \end{aligned}}

\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}\to {\frac {0}{0}}

. Additionally,

f{\text{ and }}g

are differentiable at

x=1.6

since

f

is differentiable for all

x\geq -2

and

g

is linear for all

x\geq 1

. Therefore, L'hôpital's rule applies.

{\begin{aligned}\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&=\lim _{x\to 1.6}{\frac {{\frac {d}{dx}}\left[f(x)-3\right]}{{\frac {d}{dx}}\left[3-g(x)\right]}}\\&=\lim _{x\to 1.6}{\frac {f^{\prime }(x)}{-g^{\prime }(x)}}\end{aligned}}

Since $g(x)$ is linear for all $x\geq 1$ ,

-g^{\prime }(1.6)=-{\frac {3+4}{1.6-1}}=-{\frac {7}{\frac {3}{5}}}

Using the answer from (a),

{\begin{aligned}\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&={\frac {f^{\prime }(1.6)}{-g^{\prime }(1.6)}}\\&={\frac {-5}{-{\frac {7}{\frac {3}{5}}}}}\\&={\frac {15}{35}}\\\therefore \lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&=\mathbf {\frac {3}{7}} \end{aligned}}

c. Assume

f^{\prime }(1.6)=f^{\prime }(2)

. Find an approximation of the first positive root of

f(x)

shown on the graph. Use only ONE (1) iteration.

There exists a

c\in \mathbb {R}

that satisfies

f(c)=0

. Using only one iteration of the Newton-Rhapson method, choose

x_{0}=2

.

f(2)=1

and

f^{\prime }(2)=-5

by assumption.

{\begin{aligned}x_{1}=c&=x_{0}-{\frac {f(x_{0})}{f^{\prime }(x_{0})}}\\&=2-{\frac {1}{-5}}=2.2\\\end{aligned}}

c=\mathbf {2.2}

is the first positive root of the function

f(x)

.

There exists a

c\in \mathbb {R}

that satisfies

f(c)=0

. Using only one iteration of the Newton-Rhapson method, choose

x_{0}=2

.

f(2)=1

and

f^{\prime }(2)=-5

by assumption.

{\begin{aligned}x_{1}=c&=x_{0}-{\frac {f(x_{0})}{f^{\prime }(x_{0})}}\\&=2-{\frac {1}{-5}}=2.2\\\end{aligned}}

c=\mathbf {2.2}

is the first positive root of the function

f(x)

.

d. A computer program found that there exists only one local maximum and minimum on the function

f(x)

and found no local maximum or minimum for

g(x)

. Based on this finding, what flaw exists in the program and how can it be fixed?

The program is taking the derivatives of functions and is considering all inputs where the derivative evaluated at that point is zero, but it failed to consider the case where the derivative does not exist. In cases where the derivative does not exist, the program must still check to see that the sign of the derivative changes before and after the point. Based on the sign changes, it can determine whether there is a local minimum or maximum. Thus, to fix the program, add one more case that considers when the derivative does not exist.

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