4.1E: Exercises

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Exercises $\PageIndex{1}$

1) Find $\frac{dy}{dt}$ at $x=1$ and $y=x^2+3$ if $\frac{dx}{dt}=4.$

2) Find $\frac{dx}{dt}$ at $x=−2$ and $y=2x^2+1$ if $\frac{dy}{dt}=−1.$

3) Find $\frac{dz}{dt}$ at $(x,y)=(1,3)$ and $z^2=x^2+y^2$ if $\frac{dx}{dt}=4$ and $\frac{dy}{dt}=3$.

Answers to odd numbered questions

1. $\dfrac{dy}{dt} = 8$

3. $\dfrac{dz}{dt} = \frac{13}{\sqrt{10}}$

Solution

1) Given $y=x^2+3.$ Differentiating this equation with respect to time, $\dfrac{dy}{dt}= 2x\frac{dx}{dt}$. Now, $\dfrac{dy}{dt}_{x=1}= 2(1)(4 )=8.$

3) Given $z^2=x^2+y^2$. Differentiating this equation with respect to time, $2z\dfrac{dz}{dt}= 2x\frac{dx}{dt}+2y \dfrac{dy}{dt}$.

Since $(x,y)=(1,3)$, $z^2=1^2+3^2=10 \implies z=\sqrt{10}.$ Now $\dfrac{dz}{dt}|_{(x,y)=(1,3)}=\dfrac{(1)(4)+(3)(3)}{\sqrt{10}}=\dfrac{13}{\sqrt{10}}.$

Exercise $\PageIndex{2}$

[T] If two electrical resistors are connected in parallel, the total resistance (measured in ohms, denoted by the Greek capital letter omega, $Ω$) is given by the equation $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}.$ If $R_1$ is increasing at a rate of $0.5Ω/\text{min}$ and $R_2$ decreases at a rate of $1.1Ω/\text{min}$, at what rate does the total resistance change when $R_1=20Ω$ and $R_2=50Ω/\text{min}$?

Exercise $\PageIndex{3}$

A $10$-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of $2$ ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is $5$ ft from the wall?

$alt$

Answer

$2\sqrt{3}$ ft/sec.

Solution

Figure $\PageIndex{3}$: Ladder. (Copyright; Pamini Thangarajah)

As shown, $x$ denotes the distance between the bottom of the ladder and the wall and $y$ denotes the distance between the top of the ladder and the wallat time $t$. Given that $\dfrac{dy}{dt}=-2$ ft/sec. Question: how fast is the bottom moving along the ground when the bottom of the ladder is $5$ ft from the wall, which is $\dfrac{d(x)}{dt}|_{x=5}=?$

From the figure, we can use the Pythagorean theorem to write an equation relating $x$ and $y$:

$x^2+y^2=10^2.$

Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

$2x\dfrac{dx}{dt} +2y\dfrac{dy}{dt}=0.$ Thus

$\dfrac{dx}{dt}= -\dfrac{y}{x}\dfrac{dy}{dt}.$

We need to find $y$ when $x=5$. $y^2=10^2-5^2=(10-5)(10+5)=(5)(15)=(3)(5^2) \implies y=5\sqrt{3}$ ft. Now,

$\dfrac{dx}{dt}|_{x=5}= -\dfrac{5\sqrt{3}}{5} (-2)= 2\sqrt{3}$ ft/sec.

The rate that bottom moving along the ground when the bottom of the ladder is $5$ ft from the wall is $2\sqrt{3}$ ft/sec.

Exercise $\PageIndex{4}$

A $25$-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of $1$ ft/sec, and the bottom of the ladder is initially $20$ ft away from the wall, how fast does the ladder move up the wall $5$ sec after we start pushing?

Exercise $\PageIndex{5}$

Two airplanes are flying in the air at the same height: airplane A is flying east at $250$ mi/h and airplane B is flying north at $300$ mi/h. If they are both heading to the same airport, located $30$ miles east of airplane A and $40$ miles north of airplane B, at what rate is the distance between the airplanes changing?

$alt$

Answer: The distance is decreasing at $390$ mi/h.

Exercise $\PageIndex{6}$

You and a friend are riding your bikes to a restaurant that you think is east; your friend thinks the restaurant is north. You both leave from the same point, with you riding at $16$ mph east and your friend riding $12$ mph north. After you traveled $4$ mi, at what rate is the distance between you changing?

Exercise $\PageIndex{7}$

Two buses are driving along parallel freeways that are $5$ mi apart, one heading east and the other heading west. Assuming that each bus drives a constant $55$ mph, find the rate at which the distance between the buses is changing when they are $13$ mi apart (as the crow flies), heading toward each other.

Answer

The distance between them shrinks at a rate of $\frac{1320}{13}≈101.5$ mph.

Solution

Figure $\PageIndex{7}$: buses. (Copyright; Pamini Thangarajah)

As shown, $x$ denotes the horizontal distance between the two buses and $s$ denotes the distance between the two buses at time $t$. Since each bus drives a constant $55$ mph, $\dfrac{dx}{dt}=-(55)(2)$ mph. Question: The rate at which the distance between the buses changes when they are $13$ mi apart (as the crow flies), heading toward each other. is $\dfrac{d(s)}{dt}|_{s=13}=?$

From the figure, we can use the Pythagorean theorem to write an equation relating $x$ and $s$:

$x^2+5^2=s^2.$

Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

$2x\dfrac{dx}{dt} =2s\dfrac{ds}{dt}.$ Thus

$\dfrac{ds}{dt}= \dfrac{x}{s}\dfrac{dx}{dt}.$

We need to find $x$ when $s=13$. $x^2=13^2-5^2=(13-5)(13+5)=(8)(18)=(4^2)(3^2) \implies x=(4)(3)=12$ mi. Now,

$\dfrac{ds}{dt}|_{s=13}= \dfrac{12}{13} (-55)(2)= \dfrac{-1320}{13}$ mph.

The distance between them shrinks at a rate of $\frac{1320}{13}$ mph.

Exercise $\PageIndex{8}$

1) A $6$-ft-tall person walks away from a $10$-ft lamppost at a constant rate of $3$ ft/sec. What is the rate at which the tip of the shadow moves away from the pole when the person is $10$ ft away from the pole?

$alt$ 10

Answer

$\frac{15}{2}$ ft/sec.

Solution

$clipboard_e7cd8567659cac46793bbe158519d77f1.png$ Figure $\PageIndex{8}$: Shadow. (Copyright; Pamini Thangarajah)

As shown, $x$ denotes the length of the shadow and $y$ denotes the distance between the man and the lamppost at time $t$. Given that $\dfrac{dy}{dt}=3$ ft/sec. Question: The rate at which the tip of the shadow moves away from the pole when the person is $10$ ft away from the pole is $\dfrac{d(x+y)}{dt}|_{y=10}=?$

From the figure, using similar triangles, $\dfrac{x}{x+y}=\dfrac{6}{10}=\dfrac{3}{5} \implies 5x=3x+3y \implies 2x=3y.$ Differentiating this equation with respect to time $2\frac{dx}{dt}=3\frac{dy}{dt} \implies \frac{dx}{dt}=\frac{3}{2}\frac{dy}{dt} =\frac{3}{2}(3)= \frac{9}{2}$ ft/sec.

Note that the rate at which the tip of the shadow moves away from the person when the person is $10$ ft away from the pole is $\frac{dx}{dt}.$

The rate at which the tip of the shadow moves away from the pole when the person is $10$ ft away from the pole is $\dfrac{d(x+y)}{dt}|_{y=10}=3+ \frac{9}{2}=\frac{15}{2}$ ft/sec.

2) Using the previous problem, what is the rate at which the tip of the shadow moves away from the person when the person is $10$ ft from the pole?

Answer

$\frac{9}{2}$ ft/sec.

Solution

Figure $\PageIndex{8}$: Shadow (Copyright; Pamini Thangarajah)

Note that the rate at which the tip of the shadow moves away from the person when the person is $10$ ft away from the pole is $\frac{dx}{dt}= \frac{9}{2}$ ft/sec.

The rate at which the tip of the shadow moves away from the pole when the person is $10$ ft away from the pole is $\dfrac{d(x+y)}{dt}|_{y=10}=3+ \frac{9}{2}=\frac{15}{2}$ ft/sec.

Exercise $\PageIndex{9}$

A $5$-ft-tall person walks toward a wall at a rate of $2$ ft/sec. A spotlight is located on the ground $40$ ft from the wall. How fast does the height of the person’s shadow on the wall change when the person is $10$ ft from the wall?

Answer: The shadow's height decreases at a rate $\frac{4}{9}$ ft/sec.

Solution

Figure $\PageIndex{9}$: Shadow on the wall. (Copyright; Pamini Thangarajah)

As shown, $y$ denotes the length of the shadow on the wall and $x$ denotes the distance between the man and the spotlight at time $t$. Given that $\dfrac{dx}{dt}=2$ ft/sec. Question: The rate at which the height of the person’s shadow on the wall change when the person is $10$ ft is $\dfrac{d(y)}{dt}|_{x=30}=?$

From the figure, using similar triangles, $\dfrac{y}{40}=\dfrac{5}{x}.$ Differentiating this equation with respect to time $\dfrac{1}{40}\frac{dy}{dt}=\dfrac{-5}{x^2}\frac{dx}{dt} \implies \frac{dy}{dt}=\frac{-200}{x^2}\frac{dx}{dt} =\frac{-200}{30^2}(2)= \frac{-4}{9}$ ft/sec.

Hence, the shadow's height decreases at a rate $\frac{4}{9}$ ft/sec.

Exercise $\PageIndex{10}$

Using the previous problem, what is the rate at which the shadow changes when the person is $10$ ft from the wall, if the person is walking away from the wall at a rate of $2$ ft/sec?

Answer

It grows at a rate $\frac{4}{9}$ ft/sec.

Solution

Figure $\PageIndex{10}$: Shadow on the wall. (Copyright; Pamini Thangarajah)

As shown, $y$ denotes the length of the shadow on the wall and $x$ denotes the distance between the man and the spotlight at time $t$. Given that $\dfrac{dx}{dt}=-2$ ft/sec. Question: The rate at which the height of the person’s shadow on the wall change when the person is $10$ ft is $\dfrac{d(y)}{dt}|_{x=30}=?$

Hence, the shadow's height grows at a rate $\frac{4}{9}$ ft/sec.

Exercise $\PageIndex{11}$

1) A helicopter starting on the ground is rising directly into the air at a rate of $25$ ft/sec. You are running on the ground starting directly under the helicopter at a rate of $10$ ft/sec. Find the rate of change of the distance between the helicopter and yourself after $5$ sec.

Answer: The distance is increasing at $\frac{145\sqrt{29}}{29}$ ft/sec.

Solution

Figure $\PageIndex{11}$: Helicopter and a person. (Copyright;Pamini Thangarajah)

As shown, $x$ denotes the position's distance on the person's ground at time $t$. The variable $s$ denotes the distance between the person and the plane at time $t$. $y$ denotes the plane's height above the ground at time $t$.

Given: You are running on the ground starting directly under the helicopter at a rate of $10$ ft/sec,$\dfrac{dx}{dt}=10$ ft/sec and a helicopter starting on the ground is rising directly into the air at a rate of $25$ ft/sec, $\dfrac{dy}{dt}=25$ ft/sec.

Question: what is the rate at which the distance between you and the helicopter changes after $5$ sec, that is $\dfrac{ds}{dt}|_{t=5}=?$.

From the figure, we can use the Pythagorean theorem to write an equation relating $x,y$ and $s$:

$x^2+(y)^2=s^2.$

Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

$x\dfrac{dx}{dt}+ (y)\dfrac{dy}{dt}=s\dfrac{ds}{dt}.$

We need to find $x,y$ and $s$ when $t=5$ sec. Since $\dfrac{dy}{dt}=25$ ft/sec, y=(25)(5)={125}\) ft. Now, Since $\dfrac{dx}{dt}=10$ ft/sec, $x=10(5)=50$ ft. Solving the equation $(50)^2+(125)^2=s^2$ for $s$, we have $s=25 \sqrt{29}$ft at the time of interest. Using these values, we conclude that $ds/dt$ is a solution of the equation \begin{align*}\dfrac{ds}{dt}& = \dfrac{x\frac{dx}{dt}+ (y)\frac{dy}{dt}}{s}\\ &=\dfrac{(50)(10)+(125)(25)}{25 \sqrt{29}}\\ &=\dfrac{20+125}{ \sqrt{29}}\\ &=\dfrac{145}{\sqrt{29}}=\dfrac{145\sqrt{29}}{29} \end{align*}.

Hence, the distance is increasing at $\frac{145\sqrt{29}}{29}$ ft/sec.

2) Using the previous problem, assuming the helicopter is again rising at a rate of $25$ ft/sec and you are running on the ground starting directly under the helicopter at a rate of $10$ ft/sec, what is the rate at which the distance between you and the helicopter is changing when the helicopter has risen to a height of $60$ ft in the air, assuming that, initially, it was $30$ ft above you?

Answer

The distance is increasing at $\frac{135\sqrt{26}}{26}$ ft/sec.

Solution

$clipboard_e4a292b450aadec767fb419e2b630354e.png$

Figure $\PageIndex{11}$: Helicopter and a Person. (Copyright; Pamini Thangarajah)

As shown, $x$ denotes the position's distance on the person's ground at time $t$. The variable $s$ denotes the distance between the person and the plane at time $t$. $y+30$ denotes the plane's height above the ground at time $t$.

Question: what is the rate at which the distance between you and the helicopter changes when the helicopter has risen to a height of $60$ ft in the air, that is $\dfrac{ds}{dt}|_{y=30}=?$.

From the figure, we can use the Pythagorean theorem to write an equation relating $x,y$ and $s$:

$x^2+(y+30)^2=s^2.$

Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

$x\dfrac{dx}{dt}+ (y+30)\dfrac{dy}{dt}=s\dfrac{ds}{dt}.$

We need to find $s$ and $x$ when $y=30$ ft. Since $\dfrac{dy}{dt}=25$ ft/sec, the time taken to travel $y=30$ ft is $t=\dfrac{30}{25}=\dfrac{6}{5}$ sec. Now, Since $\dfrac{dx}{dt}=10$ ft/sec, $x=10 \left( \dfrac{6}{5}\right)=12$ ft. Solving the equation $(12)^2+(60)^2=s^2$ for $s$, we have $s=12 \sqrt{26}$ft at the time of interest. Using these values, we conclude that $ds/dt$ is a solution of the equation \begin{align*}\dfrac{ds}{dt}& = \dfrac{x\frac{dx}{dt}+ (y+30)\frac{dy}{dt}}{s}\\ &=\dfrac{(12)(10)+(60)(25)}{12 \sqrt{26}}\\ &=\dfrac{10+(5)(25)}{ \sqrt{26}}\\ &=\dfrac{135}{\sqrt{26}}=\dfrac{135\sqrt{26}}{26} \end{align*}.

Hence, the distance is increasing at $\frac{135\sqrt{26}}{26}$ ft/sec.

Exercise $\PageIndex{12}$ For the following exercises, draw and label diagrams to help solve the related-rates problems.

1) The side of a cube increases at a rate of $\frac{1}{2}$ m/sec. Find the rate at which the volume of the cube increases when the side of the cube is $4$ m.

2) The volume of a cube decreases at a rate of $10 \text{ m}^3$/sec. Find the rate at which the side of the cube changes when the side of the cube is $2$ m.

Answer: $−\frac{5}{6}$ m/sec

3) The radius of a circle increases at a rate of $2$ m/sec. Find the rate at which the area of the circle increases when the radius is $5$ m.

4) The radius of a sphere decreases at a rate of $3$ m/sec. Find the rate at which the surface area decreases when the radius is $10$ m.

Answer: $240π \,\text{m}^2\text{/sec}$

5) The radius of a sphere increases at a rate of $1$ m/sec. Find the rate at which the volume increases when the radius is $20$ m.

6) The radius of a sphere is increasing at a rate of $9$ cm/sec. Find the radius of the sphere when the volume and the radius of the sphere are increasing at the same numerical rate.

Answer: $\frac{1}{2\sqrt{π}}$ cm

7) The base of a triangle is shrinking at a rate of $1$ cm/min and the height of the triangle is increasing at a rate of $5$ cm/min. Find the rate at which the area of the triangle changes when the height is $22$ cm and the base is $10$ cm.

8) A triangle has two constant sides of length $3$ ft and $5$ ft. The angle between these two sides is increasing at a rate of $0.1$ rad/sec. Find the rate at which the area of the triangle is changing when the angle between the two sides is $π/6.$

Answer: The area is increasing at a rate $\frac{3\sqrt{3}}{8}\,\text{ft}^2\text{/sec}$.

Solution

Figure $\PageIndex{12}$: Triangle. (Copyright; Pamini Thangarajah)

Let $ab=5$ft , $ac=3$ft and $\theta$ be the angle between the sides $ab$ and $ac$ at time $t$. Also let $A$ be the area of the triangle at time $t.$

Given: $\dfrac{d\theta}{dt}=0.1$ rad/sec.

Question: $\dfrac{dA}{dt}\left|_{\theta=\pi/6}\right.$.

Consider the area formula, $ A=\frac{1}{2}(3)(5) \sin(\theta)$. Differentiating this equation with respect to time, we get

$\dfrac{dA}{dt}=\frac{15}{2} \cos(\theta) \dfrac{d\theta}{dt}.$

Hence, $\dfrac{dA}{dt}\left|_{\theta=\pi/6}\right.=\frac{15}{2} \cos(\pi/6) (0.1)=\frac{15}{20}\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{8}\,\text{ft}^2\text{/sec}$.

Thus, the area is increasing at a rate $\frac{3\sqrt{3}}{8}\,\text{ft}^2\text{/sec}$.

9) A triangle has a height that is increasing at a rate of $2$ cm/sec and its area is increasing at a rate of $4 \,\text{cm}^2\text{/sec}$. Find the rate at which the base of the triangle is changing when the height of the triangle is $4$ cm and the area is $20 \,\text{cm}^2$.

Exercise $\PageIndex{13}$

For the following exercises, consider an inverted right cone that is leaking water. (Inverted means the cone's point is facing down, like a funnel.) The dimensions of the conical tank are a height of 16 ft and a radius of 5 ft.

1) How fast does the depth of the water change when the water is $10$ ft high if the cone leaks water at a rate of $10 \,\text{ft}^3\text{/min}$?

2) Find the rate at which the surface area of the water changes when the water is $10$ ft high if the cone leaks water at a rate of $10 \,\text{ft}^3\text{/min}$.

3) If the water level is decreasing at a rate of $3$ in/min when the depth of the water is $8$ ft, determine the rate at which water is leaking out of the cone.

Answers to odd numbered questions

1. The depth of the water decreases at $\frac{128}{125π}$ ft/min.

3. The volume is decreasing at a rate of $\frac{(25π)}{16}ft^3/min.$

Solution

1. Let $h$ denote the height of the water in the funnel, r denotes the radius of the water at its surface, and $V$ denotes the volume of the water at time $t$. We need to determine $\dfrac{dh}{dt}$ when $h=10 ft$. We know that $\dfrac{dV}{dt}=-10ft^3/min$

The volume of water in the cone is

$V=\dfrac{1}{3}\pi r^2h.$

From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, $\frac{r}{h}=\frac{5}{16}$ or $r=\dfrac{5h}{16}.$ Using this fact, the equation for volume can be simplified to

$V=\dfrac{1}{3}\pi \left(\dfrac{5h}{16}\right)^2h=\dfrac{25 \pi}{(3)(4^4)}h^3$.

Applying the chain rule while differentiating both sides of this equation with respect to time $t$, we obtain

$\dfrac{dV}{dt}=\dfrac{25 \pi}{4^4}h^2\dfrac{dh}{dt}.$ Hence $\dfrac{dh}{dt}_{h=10}= \dfrac {4^4}{(25 \pi)(10^2)}\dfrac{dV}{dt}= \dfrac {(4^4)(10)}{(25 \pi)(10^2)}=\dfrac {(4^3)(2)}{(25 \pi)(5)}=\dfrac {128}{125 \pi}$ ft/min.

The depth of the water decreases at $\dfrac {128}{125 \pi}$ ft/min.

3. Let $h$ denote the height of the water in the funnel, r denotes the radius of the water at its surface, and $V$ denote the volume of the water at time $t$. We need to determine $\dfrac{dV}{dt}$ $\dfrac{dh}{dt}$ when $h=8 ft$. We know that $\dfrac{dh}{dt}=-3 in/min=\dfrac{-3}{12}ft/min$, when $h=8 ft$

The volume of water in the cone is

$V=\dfrac{1}{3}\pi r^2h.$

From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, $\frac{r}{h}=\frac{5}{16}$ or $r=\frac{5h}{16}.$ Using this fact, the equation for volume can be simplified to

$V=\dfrac{1}{3}\pi \left(\dfrac{5h}{16}\right)^2h=\dfrac{25 \pi}{(3)(4^4)}h^3$.

Applying the chain rule while differentiating both sides of this equation with respect to time $t$, we obtain

$\dfrac{dV}{dt}=\dfrac{25 \pi}{4^4}h^2\dfrac{dh}{dt}.$ Hence

$\dfrac{dV}{dt}_{h=8}= \left( \dfrac{25 \pi}{4^4} \right) (8^2)\left(\dfrac{-3}{12}\right)= \dfrac {-25 \pi}{16}ft^3/min.$ Hence, the volume is decreasing at a rate of $\dfrac{25\pi}{16}ft^3/min.$

Exercise $\PageIndex{14}$

A vertical cylinder is leaking water at a rate of $1 \,\text{ft}^3\text{/sec}$. If the cylinder has a height of $10$ ft and a radius of $1$ ft, at what rate is the height of the water changing when the height is $6$ ft?

Exercise $\PageIndex{15}$

A cylinder is leaking water but you are unable to determine at what rate. The cylinder has a height of $2$ m and a radius of $2$ m. Find the rate at which the water is leaking out of the cylinder if the rate at which the height is decreasing is $10$ cm/min when the height is $1$ m.

Answer: The water flows out at rate $\frac{2π}{5}\,\text{m}^3\text{/min}$.

Exercise $\PageIndex{16}$

A trough has ends shaped like isosceles triangles, with width $3$ m and height $4$ m, and the trough is $10$ m long. Water is being pumped into the trough at a rate of $5\,\text{m}^3\text{/min}$. At what rate does the height of the water change when the water is $1$ m deep?

$alt$

Answer

$\dfrac{2}{3}$ m/min

Solution

$Screen Shot 2024-10-31 at 10.06.15 PM.png$

Figure $\PageIndex{16}$: A trough

Let $h$ be the height of the water at time $t$ , $x$ be the length of the side at time$t$ as shown in the figure, and let $V$ be the volume of the water at time $t$. Given: $\dfrac{dy}{dt}=5\,\text{m}^3\text{/min}$. Question: (\dfrac{dh}{dt}|_{h=1}=?\).

The volume of the water, $V=(10)\dfrac{xh}{2}.$ We can use the similar triangles to see $\dfrac{h}{4}=\dfrac{x}{3} \implies x=\dfrac{3h}{4}.$

Now $V=5\left(\dfrac{3h}{4}\right) h= \dfrac{15h^2}{4}.$. Differentiating this equation with respect to time, we get $\dfrac{dv}{dt}=\dfrac{15h}{2} \dfrac{dh}{dt}.$ Thus $\dfrac{dh}{dt}|_{h=1}=\dfrac{10}{15(1)}=\dfrac{2}{3}$ m/min

Exercise $\PageIndex{17}$

A tank is shaped like an upside-down square pyramid, with base of $4$ m by $4$ m and a height of $12$ m (see the following figure). How fast does the height increase when the water is $2$ m deep if water is being pumped in at a rate of $\frac{2}{3} \text{ m}^3$/sec?

$alt$

Answer: $\frac{3}{2} $ m/sec.

Exercise $\PageIndex{18}$

For the following exercises, consider a pool shaped like the bottom half of a sphere, that is being filled at a rate of $25 \,\text{ft}^3$/min. The radius of the pool is $10$ ft.

1) Find the rate at which the depth of the water is changing when the water has a depth of $5$ ft.

2) Find the rate at which the depth of the water is changing when the water has a depth of $1$ ft.

3) If the height is increasing at a rate of $1$ in./sec when the depth of the water is $2$ ft, find the rate at which water is being pumped in.

Answers to even numbered questions: 2. $\frac{25}{19π} ft/min$

Exercise $\PageIndex{19}$

Gravel is being unloaded from a truck and falls into a pile shaped like a cone at a rate of $10 \,\text{ft}^3/\text{min}$. The radius of the cone base is three times the height of the cone. Find the rate at which the height of the gravel changes when the pile has a height of $5$ ft.

Answer: $\frac{2}{45π} $ ft/min.

Exercise $\PageIndex{21}$

Using a similar setup from the preceding problem, find the rate at which the gravel is being unloaded if the pile is $5$ ft high and the height is increasing at a rate of $4$ in./min.

Answer: Under Construction

Exercise $\PageIndex{22}$

You are stationary on the ground and are watching a bird fly horizontally at a rate of $10$ m/sec. The bird is located 40 m above your head. How fast does the angle of elevation change when the horizontal distance between you and the bird is $9$ m?

Answer

The angle decreases at $\dfrac{400}{1681} $ rad/sec.

Solution

Figure $\PageIndex{22}$: triangle. (Copyright; Pamini Thangarajah)

As shown, $x$ denotes the horizontal distance between the person's ground and the bird at time $t$. Let $\alpha$ be the elevation angle between the person and the bird at time $t$.

Given: $\dfrac{dx}{dt}=10$ m/sec.

Question: $\dfrac{d\alpha}{dt}\left|_{x=9}\right.$.

Consider , $ \tan(\alpha)=\dfrac{x}{40}$. Differentiating this equation with respect to time, we get

$\sec^2 (\alpha) \dfrac{d\alpha}{dt}=\dfrac{1}{40} \dfrac{dx}{dt}.$

Then $\dfrac{d\alpha}{dt}|_{x=9}=\dfrac{1}{40sec^2(\alpha)} \dfrac{dx}{dt} . $

We need to find $ \sec(\alpha) $ when $x=9.$ We know that $sec^2(\alpha)= 1+\tan^2(\alpha)=1+\left(\dfrac{9}{40}\right)^2=\dfrac{1681}{40^2}.$

Now, $\dfrac{d\alpha}{dt}\left|_{x=9}\right.=\dfrac{1}{40\left(\dfrac{1681}{40^2}\right)} 10= \dfrac{400}{1681} $ rad/sec.

Thus the angle decreases at $\dfrac{400}{1681} $ rad/sec.

Exercise $\PageIndex{23}$

You stand 40 ft from a bottle rocket on the ground and watch as it takes off vertically into the air at a rate of 20 ft/sec. Find the rate at which the angle of elevation changes when the rocket is 30 ft in the air.

Answer: Under Construction

Exercise $\PageIndex{24}$

1) A lighthouse, L, is on an island 4 mi away from the closest point, P, on the beach (see the following image). If the lighthouse light rotates clockwise at a constant rate of 10 revolutions/min, how fast does the beam of light move across the beach 2 mi away from the closest point on the beach?

2) Determine at what rate the beam of light moves across the beach 1 mi away from the closest point on the beach.

$alt$

Answer to odd numbered questions: 1. $100π/min$

Exercise $\PageIndex{25}$

You are walking to a bus stop at a right-angle corner. You move north at a rate of 2 m/sec and are 20 m south of the intersection. The bus travels west at a rate of 10 m/sec away from the intersection – you have missed the bus! What is the rate at which the angle between you and the bus is changing when you are 20 m south of the intersection and the bus is 10 m west of the intersection?

Answer: The angle is changing at a rate of $\frac{21}{25}rad/sec$.

Exercises $\PageIndex{26}$

For the following exercises, refer to the figure of a baseball diamond, which has sides of 90 ft.

$alt$

1) A batter hits a ball toward third base at 75 ft/sec and runs toward first base at a rate of 24 ft/sec. At what rate does the distance between the ball and the batter change when 2 sec have passed?

2) A batter hits a ball toward second base at 80 ft/sec and runs toward first base at a rate of 30 ft/sec. At what rate does the distance between the ball and the batter change when the runner has covered one-third of the distance to first base? (Hint: Recall the law of cosines.)

3) A batter hits the ball and runs toward first base at a speed of 22 ft/sec. At what rate does the distance between the runner and second base change when the runner has run 30 ft?

4) Runners start at first and second base. When the baseball is hit, the runner at first base runs at a speed of 18 ft/sec toward second base and the runner at second base runs at a speed of 20 ft/sec toward third base. How fast is the distance between runners changing 1 sec after the ball is hit?

Answers to even numbered questions

2. The distance is increasing at a rate of $62.50$ ft/sec.

4. The distance is decreasing at a rate of $11.99$ ft/sec

Source

Calculus Applets using GeoGebra by Marc Renault is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

Contributors and Attributions

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

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