Answer:  $-\dfrac{1}{200} \text{ cm/hr}$.    [If you are using a web-based homework system, see the “Caution” at the end of this solution.]

Let’s unpack the question statement:

• We’re told that the snowball’s volume V is changing at the rate of  $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr. (We must insert the negative sign “by hand” since we are told that the snowball is melting, and hence its volume is decreasing.)
• As a result, its radius is changing, at the rate  $\dfrac{dr}{dt}$, which is the quantity we’re after.
• The snowball always remains a sphere.
• Toward the end of our solution, we’ll need to remember that the problem is asking us about  $\dfrac{dr}{dt}$  at a particular instant, when  $r = 10$ cm.

Now let’s use our 4-step Problem Solving Strategy as outlined above:
1. Draw a picture of the physical situation.
See the figure.

2. Write an equation that relates the quantities of interest.
B. To develop your equation, you will probably use. . . a simple geometric fact.
In this problem we’re given the value for  $\dfrac{dV}{dt}$,  and we’re after the value of  $\dfrac{dr}{dt}$. Hence we need to start with an expression that relates V to r. Since the snowball is a sphere, we know that relationship:

$$V = \frac{4}{3}\pi r^3$$ 3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
\begin{align*} \frac{d}{dt}V & = \frac{d}{dt} \left( \frac{4}{3}\pi r^3\right) \\[12px] \frac{dV}{dt} &= \frac{4}{3}\pi \frac{d}{dt} \left( r^3\right) \\[12px] &= \frac{4}{3}\pi \left( 3r^2 \frac{dr}{dt} \right) \\[12px] &= 4 \pi r^2 \frac{dr}{dt} \end{align*} 4. Solve for the quantity you’re after.
Solving the equation above for $\dfrac{dr}{dt}$: \begin{align*} \frac{dV}{dt} &= 4 \pi r^2 \frac{dr}{dt} \\[12px] \frac{dr}{dt} &= \frac{1}{4 \pi r^2} \frac{dV}{dt} \\[12px] \end{align*} Now we just have to substitute values. Recall $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr, and the problem asks about when $r=10$ cm: \begin{align*} \frac{dr}{dt} &= \frac{1}{4 \pi r^2} \frac{dV}{dt} \\[12px] &= \frac{1}{4 \pi (10)^2} (-2 \pi) \\[12px] &= -\frac{1}{200} \text{ cm/hr} \quad \cmark \end{align*} The negative value indicates that the radius is decreasing as the snowball melts, as we expect.

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Caution: IF you are using a web-based homework system and the question asks,

At what rate does the radius decrease?

then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE:   $\boxed{\dfrac{1}{200}} \, \dfrac{\text{cm}}{\text{hr}} \quad \checkmark$

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Answer:  $\dfrac{1}{4 \pi}$

When a circle expands, obviously its area changes; the rate at which its area changes is  $\dfrac{dA}{dt}$.   Its radius changes as well, at the rate  $\dfrac{dr}{dt}$.   Those two quantities are related, and that recognition is the crux of this problem.

1. Draw a picture of the physical situation.
See the figure.

2. Write an equation that relates the quantities of interest.
B. To develop your equation, you will probably use . . . a simple geometric fact.
Because we ultimately need a relation between   $\dfrac{dA}{dt}$  and   $\dfrac{dr}{dt}$,  we start with a relation between A and r. Fortunately, that’s a particularly well-known relation:

$$A = \pi r^2$$ 3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.

\begin{align*} \frac{d}{dt}(A) &= \frac{d}{dt} \left(\pi r^2 \right) \\[12px] \frac{dA}{dt} &= 2 \pi r \frac{dr}{dt} \\[12px] \end{align*} 4. Solve for the quantity you’re after.
The problem is asking us to find r at the instant when $\dfrac{dA}{dt} = \dfrac{1}{2} \dfrac{dr}{dt}$, or when $ \dfrac{\left(\dfrac{dA}{dt}\right)}{\left(\dfrac{dr}{dt}\right)} = \dfrac{1}{2}$.

Starting from the preceding equation:

\begin{align*} \frac{dA}{dt} &= 2 \pi r \frac{dr}{dt} \\[12px] r &= \frac{1}{2 \pi} \dfrac{\left(\dfrac{dA}{dt}\right)}{\left(\dfrac{dr}{dt}\right)} \\[12px] &= \frac{1}{2 \pi} \left(\frac{1}{2}\right) = \frac{1}{4 \pi} \quad \cmark \end{align*}

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Answer: $\dfrac{dA}{dt} = 47\, \tfrac{\text{cm}^2}{\text{s}} $

Let’s first unpack this question:
We’re given that the rectangle’s length is changing at the rate $\dfrac{dl}{dt} = 8$ cm/s, and its width is changing at the rate $\dfrac{dw}{dt} = 5$ cm/s. We’re looking for $\dfrac{dA}{dt}$ at the particular instant specified in the problem.

1. Draw a picture of the physical situation.
See the figure.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
The rectangle’s width and the length are both changing, so we’ll leave them each as a variable.

B. To develop your equation, you will probably use . . .a simple geometric fact.
We’re looking for $\dfrac{dA}{dt}$ given values for both $\dfrac{dl}{dt}$ and $\dfrac{dw}{dt}$. That suggests we want to start with a relationship among the rectangle’s area A and its length l and width w. Fortunately, that’s one we immediately know:

$$A = lw$$ 3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
Both $w$ and $l$ are changing with time, and so when we take the derivative we must use the Product Rule and include both $\dfrac{dw}{dt}$ and $\dfrac{dl}{dt}$ terms:

\begin{align*} \frac{d}{dt}A &= \frac{d}{dt}(lw) \\[8px] \frac{dA}{dt} &= \frac{dl}{dt}w + l \frac{dw}{dt} \end{align*} 4. Solve for the quantity you’re after.
We want $\dfrac{dA}{dt}$ at the moment when $l = 3$ cm and $w = 4$ cm. Furthermore, recall that the problem tells us $\dfrac{dl}{dt} = 8$ cm/s, and its width is changing at the rate $\dfrac{dw}{dt} = 5$ cm/s.

Hence \begin{align*} \frac{dA}{dt} &= \frac{dl}{dt}w + l \frac{dw}{dt} \\[8px] &= \left( 8 \, \tfrac{\text{cm}}{\text{s}}\right) (4 \, \text{cm}) + (3 \, \text{cm}) \left(5 \, \tfrac{\text{cm}}{\text{s}} \right) \\[8px] &= 32 \, \tfrac{\text{cm}^2}{\text{s}} + 15 \, \tfrac{\text{cm}^2}{\text{s}} \\[8px] &= 47 \, \tfrac{\text{cm}^2}{\text{s}} \quad \cmark \end{align*}

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Answer:  $w = \sqrt{\dfrac{500}{8}} \text{ cm}$

This rectangle has constant area; you might imagine it as a flat piece of rubber that has constant area. As you stretch it out in one direction, it must shrink in the other to compensate. The faster you stretch it out, the faster its width must decrease to keep the area constant.

The problem tells us that the length is changing at the constant rate  $\dfrac{dl}{dt} = 8$ cm/s, a positive value since the length is increasing with time. We’re after something about when the rate that the width changes,  $\dfrac{dw}{dt}$ , has a certain value (-1 cm/s). So we need to relate the width w to the length l at every moment. Fortunately we can do that easily since the area remains constant.

1. Draw a picture of the physical situation.
See the figure.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
The rectangle’s width and the length are both changing, so we’ll leave them each as a variable.
B. To develop your equation, you will probably use . . .a simple geometric fact.
Here we use the relation between the area of a rectangle and its width and length. Remember that the problem says that the rectangle’s area stays constant at 500 cm$^2$:

$$A = 500 = wl$$ 3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
Both $w$ and $l$ are changing with time, and so when we take the derivative we must include $\dfrac{dw}{dt}$ and $\dfrac{dl}{dt}$ terms:

\begin{align*} \frac{d}{dt}(500) &= \frac{d}{dt}(wl) \\ \\ 0 &= \frac{dw}{dt}l + w\frac{dl}{dt} \end{align*} 4. Solve for the quantity you’re after.
We want to find $w$ at the instant $\dfrac{dw}{dt} = -1$ cm/s. (Note that we must insert that negative sign “by hand” since the problem states that the width is decreasing.) We also know from the problem statement that $\dfrac{dl}{dt} = 8$ cm/s (a positive value since the length is increasing with time).

Looking at the preceding equation, we see that we currently have two unknowns, $l$ and $w$. Since we’re looking for $w$, let’s eliminate $l$ by noting that the rectangle has constant area: $A = 500 = lw$. Hence $l = \dfrac{500}{w}$.

Then from the preceding equation: \begin{align*} 0 &= \frac{dw}{dt}l + w\frac{dl}{dt} \\ \\ 0 &= \frac{dw}{dt}\frac{500}{w} + w\frac{dl}{dt} \\ \\ w\frac{dl}{dt} &= – \frac{500}{w}\frac{dw}{dt} \\ \\ w^2 &= – 500 \dfrac{\left(\dfrac{dw}{dt}\right)}{\left(\dfrac{dl}{dt}\right)} \end{align*} Recall $\dfrac{dw}{dt} = -1$ cm/s and $\dfrac{dl}{dt} = 8$ cm/s:

\begin{align*} w^2 &= -(500)\frac{(-1)}{8} = \frac{500}{8} \\ w &= \sqrt{\dfrac{500}{8}} \text{ cm} \end{align*}

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1. Draw a picture of the physical situation.
See the figure. Let’s call the height (or depth) of the water at any given moment y, as shown.

We are told that the water level in the cup is decreasing at the rate of $0.1\, \tfrac{\text{ft}}{\text{s}}$, so $\dfrac{dy}{dt} = -0.1\, \tfrac{\text{ft}}{\text{s}}$. Remember that we have to insert that negative sign “by hand” since the water’s height is decreasing.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
The height of the water changes as time passes, so we’re calling that the variable y.

B. To develop your equation, you will probably use . . . a simple geometric fact.
The volume V of any cylinder is its circular cross-sectional area $\left(\pi r^2 \right)$ times its height. Here, at any moment the water’s height is y, and so the volume of water in the cylinder is: $$V = \pi r^2y$$ That’s it. That’s the key relationship that will allow us to complete the solution.

3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
Note that the cylinder’s radius, r, is constant (r = 3.0 ft), so we’ll treat it as a constant when we take the derivative. By contrast, the water’s height y is not constant; instead it changes, and indeed, it changes at the rate $\dfrac{dy}{dt}$. \begin{align*} \frac{dV}{dt} &= \frac{d}{dt}\left(\pi r^2 y \right) \\ \\ &= \pi r^2 \frac{d}{dt}(y) \\ \\ &= \pi r^2 \frac{dy}{dt} \\ \\ \end{align*}
4. Solve for the quantity you’re after.
At this point we’re just substituting values: the problem told us $r = 3.0 \, \text{ft}$ and $\dfrac{dy}{dt} = -0.1\, \tfrac{\text{ft}}{\text{s}}$, and we’re looking for $\dfrac{dV}{dt}$. Starting from our last expression above: \begin{align*} \frac{dV}{dt} &= \pi r^2 \frac{dy}{dt} \\ \\ &= \pi (3.0\, \text{ft})^2 \left(-0.1\, \tfrac{\text{ft}}{\text{s}}\right) \\ \\ &= -2.8 \, \tfrac{\text{ft}^3}{\text{s}} \quad \cmark \end{align*} That is, water is draining from the tank at the rate $\dfrac{dV}{dt} = -2.8\, \tfrac{\text{ft}^3}{\text{s}}$. The negative value indicates that the water’s volume in the tank V is decreasing, which is correct.

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Caution: IF you are using a web-based homework system and the question asks,

At what rate does the volume decrease?

then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{2.8} \, \dfrac{\text{ft}^3}{\text{s}} \quad \checkmark$

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Answer:   $\dfrac{-3}{16}$  rad/s

1. Draw a picture of the physical situation.
See the figure. We’ve labeled the ladder’s position along the ground x, and its position along the wall y. These both change as the situation unfolds, so we’ve left them as variables. (The ladder’s length of 10 feet, by contrast, is a constant.) We’ve also labeled the angle $\theta$ that the ladder makes with the ground, since the problem is asking us to find its rate of change at a particular instant.

2. Write an equation that relates the quantities of interest.
B. To develop your equation, you will probably use . . . a trigonometric function (like $\cos{\theta}$ = adjacent/hypotenuse).
We need to use a trigonometric function to analyze this situation, since we need to relate the angle $\theta$ to other quantities in order to have $\dfrac{d \theta}{dt}$ when we take the derivative. We’re given the rate at which the ladder’s bottom position changes $\left(\dfrac{dx}{dt} = \dfrac{3}{2} \, \tfrac{\text{ft}}{\text{s}}\right)$, and so it seems easiest to go with $\cos \theta$ since that directly relates $\theta$ and x:

$$\cos{\theta} = \frac{x}{10}$$
3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
\begin{align*} \frac{d}{dt}( \cos{\theta}) &= \frac{d}{dt} \left( \frac{x}{10} \right) \\ \\ -\sin{\theta}\, \frac{d\theta}{dt} &= \frac{1}{10}\frac{dx}{dt} \end{align*}

4. Solve for the quantity you’re after.
We’re after $\dfrac{d\theta}{dt}$ at the instant $x = 6$ ft. So let’s solve the preceding equation for $\dfrac{d\theta}{dt}$:

$$\frac{d\theta}{dt} = -\frac{1}{10} \frac{1}{\sin{\theta}} \frac{dx}{dt}$$ We’re given that $\dfrac{dx}{dt} = \dfrac{3}{2} \, \tfrac{\text{ft}}{\text{s}}$. But we don’t yet know what $\sin \theta$ is when $x = 6$ ft.

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Begin subproblem to find $\sin \theta$.
We know $$\sin \theta = \frac{y}{10}$$ So we need to know the value of $y$ at the instant when $x = 6$ ft. We can find this by using the Pythagorean theorem:
\begin{align*} (6)^2 + y^2 &= 10^2 \\[8px] 36 + y^2 &= 100 \\[8px] y^2 &= 100 – 36 = 64 \\[8px] y &= 8 \end{align*} Hence at this instant $$\sin\theta = \dfrac{8}{10} = \dfrac{4}{5}$$ End subproblem.

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Hence \begin{align*} \frac{d\theta}{dt} & = -\frac{1}{10} \, \frac{1}{\sin{\theta}}\, \frac{dx}{dt}\\[12px] &= -\frac{1}{10}\cdot \left(\frac{1}{4/5} \right)\cdot \dfrac{3}{2}\\[12px] &= – \frac{1}{10}\cdot \frac{5}{4}\cdot \dfrac{3}{2}\\[12px] &= – \frac{3}{16} \text{ rad/s} \quad \cmark \end{align*} The negative value indicates that the angle is decreasing at the ladder slides down the wall.

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1. Draw a picture of the physical situation.
See the figure. We’ve called the length of string to the kite $\ell.$ And we’ve labeled the angle $\theta$ that the string makes with the ground, since the problem is asking us to find the rate at which that angle changes, $\dfrac{d\theta}{dt}$, at a particular moment — when $\ell = 50$ meters. Recall also that the kite is moving horizontally, in the x-direction, at the rate $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}} $. We’ll use these values at the end of our solution.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
The string’s length changes as the situation progresses, so we’re calling that the variable $\ell.$

B. To develop your equation, you will probably use . . . a trigonometric function (like $\tan{\theta}$ = opposite/adjacent).
In this problem, the diagram above immediately suggests that we’re dealing with a right triangle. Furthermore, we need to related the rate at which $\theta$ is changing, $\dfrac{d\theta}{dt}$, to the rate at which x is changing, $\dfrac{dx}{dt}$, and so we first need to write down an equation that somehow relates $\theta$ and x. Such a relation must be trigonometric.

Specifically, we notice that x is the side of the triangle that is adjacent to the angle. Furthermore, the opposite leg of the triangle remains constant throughout the problem, since the kite’s height is always 30 m. Hence at every moment: $$\tan{\theta} = \frac{30}{x}$$ That’s it. That’s the key relationship that will allow us to complete the solution.

3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. \begin{align*} \frac{d}{dt} \tan{\theta} &= \frac{d}{dt} \left( \frac{30}{x} \right) \\[8px] \sec^2{\theta}\, \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt} \end{align*}
4. Solve for the quantity you’re after.
Let’s solve the preceding equation for $\dfrac{d\theta}{dt}$: \[ \begin{align*} \sec^2{\theta}\, \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}\\[8px] \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}\frac{1}{\sec^2{\theta}} \\[8px] &= -\frac{30}{x^2}\frac{dx}{dt}\cos^2 {\theta} \end{align*} \] Notice that $\cos{\theta} = \dfrac{x}{\ell}$, so \[ \begin{align*} \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}\left( \dfrac{x}{\ell}\right) ^2 \\[8px] &= -\frac{30}{\cancel{x^2}}\frac{dx}{dt}\frac{\cancel{x^2}}{\ell^2} \\[8px] &= -\frac{30}{\ell^2}\frac{dx}{dt} \end{align*} \] We’re looking for $\dfrac{d\theta}{dt}$ at the instant when $\ell =50$ m. Recall that Recall $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}}$: \[ \begin{align*} \frac{d\theta}{dt} &= -\frac{30}{(50 \text{ m})^2 } \left( 2 \, \tfrac{\text{m}}{\text{s}}\right) \\[8px] &= -0.02 \, \tfrac{\text{rad}}{\text{s}} \quad \cmark \end{align*} \] That’s the answer. The negative value indicates that the angle is decreasing at the kite travels further out, as we expect.

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Caution: IF you are using a web-based homework system and the question asks,

At what rate does the angle decrease?

then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{0.02} \, \dfrac{\text{rad}}{\text{s}} \quad \checkmark$

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Answer:  $\dfrac{1}{10}$ rad/s



1. Draw a picture of the physical situation.
See the figure.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
In this case, the height changes as the rocket travels, so keep calling it $h = 5t^2$ for now.

The picture immediately suggests a relation between the angle $\theta$ and the height $h$.

$$\tan{\theta} = \frac{h}{500} = \frac{5 t^2}{500} = \frac{t^2}{100} $$ 3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
Remember that $\theta$ is changing with time too, and so we have to use the chain rule when we take the derivative of $\tan{\theta}$, giving a $\dfrac{d\theta}{dt}$ term:

\begin{align*} \frac{d}{dt}\tan\theta &= \frac{d}{dt} \left(\frac{t^2}{100} \right)\\ \\ \sec^2{\theta} \cdot \frac{d\theta}{dt} &= \frac{2t}{100} && [\frac{d}{d\theta} \tan{\theta} = \sec^2{\theta}] \\ \\ \frac{d\theta}{dt} &= \frac{t}{50} \cos^2{\theta} \text{ } [*]&& [\frac{1}{\sec^2{\theta}} = \cos^2{\theta}] \\ \end{align*} 4. Solve for the quantity you’re after.
In order to use the preceding equation, we need to know $\theta$ (so we can compute $\cos^2 \theta$). To find $\theta$, we return to our our figure and recall that $\tan{\theta} = \dfrac{h}{500}$. Hence if we find $h$, we can then find $\theta$.

When $t =10$, the rocket’s height is:

$$h(10) = 5 t^2 = 5 \cdot (10)^2 = 500$$ Thus $$\tan{\theta} = \frac{h}{500} = \frac{500}{500} = 1$$ and so at $t = 10$, $\theta = \pi/4$.

We can now solve for $\dfrac{d\theta}{dt}$ at $t = 10$ seconds using the equation we found above marked [*]:

\begin{align*} \frac{d\theta}{dt} &= \frac{t}{50} \cos^2{\theta} \\ \\ \left. \frac{d\theta}{dt} \right|_{t=10} &= \frac{10}{50} \cos^2{\frac{\pi}{4}} \\ \\ &= \frac{1}{5} \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{5} \left(\frac{1}{2}\right) \\ \\ &= \frac{1}{10} \text{ rad/s} \quad \cmark \end{align*}

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Answer: $-2.1$ m/s

1. Draw a picture of the physical situation. See the figure. We’re calling the distance between the pole and the “head” of the man’s shadow  $\ell$,  and the distance between the man and the pole x.

2. Write an equation that relates the quantities of interest.
We are given that the man is walking toward the pole at the rate  $\dfrac{dx}{dt} = -1.5$ m/s. (We’ve made the value negative because the distance x is decreasing as the man walks toward the pole.) We are looking for the rate at which the “head” of the man’s shadow moves, which is $\dfrac{d \ell}{dt}$. We thus need to somehow relate  $\ell$ to x, so we can then develop the relationship between their time-derivatives.

There’s a subtlety to this problem that typically goes unaddressed: We’re focusing on $\ell$ and $\dfrac{d \ell}{dt}$ here because $\ell$ is the distance from the shadow’s tip to the stationary pole. We’re not examining the shadow’s length itself (labeled $\ell – x$ in the left figure below) because that length is relative to the man’s feet, which are also moving. So we’d find a different answer if we calculated the rate at which that gray shadow is changing. This problem asks us to find the rate the shadow’s head as it moves along the (stationary) ground, so it’s best to make our measurements from a point that isn’t also moving—namely, from the pole. Hence we focus on $\ell$ and aim to compute $\dfrac{d \ell}{dt}$.

B. To develop your equation, you will probably use . . . similar triangles.

In the figure above we’ve separated out the two triangles. Notice that the angles are identical in the two triangles, and hence they are similar. The ratio of their respective components are thus equal as well. Hence the ratio of their bases  $\left(\dfrac{\ell – x}{\ell} \right)$  is equal to the ratio of their heights   $\left( \dfrac{1.8\, \text{m}}{6.0\, \text{m}}\right)$: \begin{align*} \dfrac{\ell – x}{\ell} &= \frac{1.8 \, \text{m}}{6.0 \, \text{m}} \\[12px] &= 0.30 \\[12px] \ell – x &= 0.30 \ell \\[12px] \ell – 0.30 \ell &= x \\[12px] 0.70 \ell &= x \end{align*} 3. Take the derivative with respect to time of both sides of your equation.

\begin{align*} \dfrac{d}{dt}(0.70 \ell) &= \dfrac{d}{dt}(x) \\[12px] 0.70 \dfrac{d \ell}{dt} &= \dfrac{dx}{dt} \end{align*} 4. Solve for the quantity you’re after.
We’re looking for  $\dfrac{d \ell}{dt}$:

\begin{align*} 0.70 \dfrac{d \ell}{dt} &= \dfrac{dx}{dt} \\[12px] \dfrac{d \ell}{dt} &= \frac{1}{0.70} \dfrac{dx}{dt} \\[12px] &= \frac{1}{0.70} \left( -1.5 \, \tfrac{\text{m}}{\text{s}}\right) \\[12px] &= -2.1\, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*}

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1. Draw a picture of the physical situation.
See the figure. We’ve labeled the distance from the spotlight to the man x. The man is thus moving at the rate $\dfrac{dx}{dt} = 1.5 \, \tfrac{\text{m}}{\text{s}}.$ We’re calling the height of his shadow on the wall y. The problem is asking us to find $\dfrac{dy}{dt}$ at a particular moment, when the man is 5.0 meters from the wall. We’ll use these values at the end of our solution.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
The man’s position changes as he walks, so we’re calling his distance from the spotlight x.

B. To develop your equation, you will probably use . . . similar triangles.
The problem gives us the rate $\dfrac{dx}{dt}$ and asks us to find the rate $\dfrac{dy}{dt}.$ We thus need to first relate y to x. The figure above suggests that we can use similar triangles to do so. Here we’ve separated out the two triangles. Notice that their angles are identical, and hence the triangles are similar. The ratio of their respective components are thus equal: the ratio of their heights $\dfrac{y}{1.8}$ is equal to the ratio of their bases $\dfrac{20}{x}.$ That is, \[ \begin{align*} \frac{y}{1.8} &= \frac{20}{x} \\[8px] y &= \frac{(1.8)(20)}{x} \\[8px] &= \frac{36}{x} \end{align*} \] That’s it. That’s the key relationship that will allow us to complete the solution.


3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
\[ \begin{align*} \frac{d}{dt}y &= \frac{d}{dt}\left( \frac{36}{x}\right) \\[8px] \frac{dy}{dt} &= \frac{-36}{x^2}\, \dfrac{dx}{dt} \end{align*} \] 4. Solve for the quantity you’re after.
We’re given that $\dfrac{dx}{dt} = 1.5 \, \tfrac{\text{m}}{\text{s}}$. The problem asks us to find $\dfrac{dy}{dt}$ at the moment the man is 5.0 feet from the wall. Hence at the moment of interest the value of x is: \[ \begin{align*} x &= 20 \text{ m} – 5.0 \text{ m} \\[8px] &= 15.0 \text{ m} \end{align*} \] Finally, then: \[ \begin{align*} \frac{dy}{dt} &= \frac{-36}{x^2}\, \dfrac{dx}{dt} \\[8px] &= \frac{-36}{(15.0)^2} \left(1.5 \, \tfrac{\text{m}}{\text{s}} \right) \\[8px] &= -0.24 \, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*} \] That is, the man’s shadow is changing at the rate $\dfrac{dy}{dt} = -0.24\, \tfrac{\text{m}}{\text{s}}$. The negative value indicates that the shadow’s height y is decreasing, as we expect.

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Caution: IF you are using a web-based homework system and the question asks,

At what rate does the shadow’s height decrease?

then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{0.24} \, \dfrac{\text{m }}{\text{s}} \quad \checkmark$

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Answer: $\dfrac{5}{3\pi}$ cm/s

1. Draw a picture of the physical situation.
See the top figure.

2. Write an equation that relates the quantities of interest.
We are given that the volume of water in the cup is increasing at the rate $\dfrac{dV}{dt} = 15 \, \tfrac{\text{cm}^3}{\text{s}}$. We want to find the rate at which the water is rising, $\dfrac{dh}{dt}$, at the instant when the water is halfway up the cup, so when $h = 8$ cm.

A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet. The height of the water changes as time passes, as does the water’s surface area, so we’re going to keep both of those as a variables, h and r.

We have a relation between the volume of water in the cup at any moment and those variables:

$$V = \frac{1}{3} \pi r^2h $$ B. To develop your equation, you will probably use . . . similar triangles.
We don’t want the radius $r$ to appear as a separate variable, so we need to eliminate it. We can do so by using the similar triangles shown in the lower figure. Because the small triangle is similar to the larger triangle it’s embedded in, the ratio of their bases  $\left(\dfrac{r}{6} \right)$  must be equal to the ratio of their heights  $\left(\dfrac{h}{16} \right)$ 

\begin{align*} \frac{r}{6} &= \frac{h}{16} \\[12px] r &= \frac{6}{16}h\\[12px] &=\frac{3}{8} h \end{align*} Then substituting this expression into our relation for $V$:

$$V = \frac{1}{3} \pi \left(\frac{3}{8} h \right)^2h = \frac{3}{64} \pi h^3$$ 3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.

\begin{align*} \frac{dV}{dt} &= \frac{d}{dt}\left(\frac{3}{64} \pi h^3 \right) \\ \\ &= \frac{3}{64} \pi \left(3h^2 \frac{dh}{dt} \right) \\ \\ &= \frac{9}{64} \pi h^2 \frac{dh}{dt} \end{align*} 4. Solve for the quantity you’re after.
We have $\dfrac{dV}{dt} = 15 \, \tfrac{\text{cm}^3}{\text{s}}$, and want to find $\dfrac{dh}{dt}$, at the instant when $h = 8$ cm.

\begin{align*} \frac{dh}{dt} &= \frac{64}{9\pi h^2} \frac{dV}{dt} \\ \\ &= \frac{64}{9\pi (8)^2} (15) \\ \\ &= \frac{64}{3\pi (64)} (5) \\ \\ &= \frac{5}{3\pi} \text{ cm/s} \quad \cmark \end{align*}

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1. Draw a picture of the physical situation. See the figure.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet. Both the width and the length of the rectangle are changing, so let’s leave them both as variables.

B. To develop your equation, you will probably use . . .the Pythagorean theorem. We need to relate the rectangle’s diagonal to its width and length:

$$d^2 = w^2 + l^2$$ 3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
Both $w$ and $l$ are changing with time, and so when we take the derivative we must include  $\dfrac{dw}{dt}$  and  $\dfrac{dl}{dt}$  terms:

\begin{align*} \frac{d}{dt}d^2 &= \frac{d}{dt}(w^2 + l^2) \\[12px] 2d \frac{dd}{dt} &= 2w \frac{dw}{dt} + 2l \frac{dl}{dt} \\[12px] d\frac{dd}{dt} &= w \frac{dw}{dt} + l \frac{dl}{dt} \text{ [*]} \end{align*} 4. Solve for the quantity you’re after.
We want to find  $\dfrac{dd}{dt}$  at the instant when  $w = 10$ cm. The problem statement tells us that  $\dfrac{dl}{dt} = 8$ cm/s. In order to use the preceding equation (marked [*]), we need to determine  $d$  and  $l$  at this instant, and also   $\dfrac{dw}{dt}$  at this instant.

First, we can find  $l$  by noting that the rectangle has constant area,  $A = 500 = lw$. Hence when  $w = 10$ cm:

\begin{align*} l &= \frac{A}{w} \\[8px] &= \frac{500}{10} = 50 \text{ cm} \end{align*} Second, we can find $d$ at this instant using the Pythagorean theorem:

\begin{align*} d^2 &= w^2 + l^2 \\[12px] &= (10)^2 + (50)^2 = 100 + 2500 = 2600 \\[12px] d &= \sqrt{2600} = 10\sqrt{26} \end{align*} Next, we can find  $\dfrac{dw}{dt}$  at this instant by taking the derivative of  $A = 500 = lw$  :

\begin{align*} \frac{d}{dt}(500) &= \frac{d}{dt}(lw) \\[12px] 0 &= l \frac{dw}{dt} + w\frac{dl}{dt} \\[12px] l \frac{dw}{dt} &= – w\frac{dl}{dt} \end{align*} Hence \begin{align*} \dfrac{dw}{dt} &= – \left( \frac{w}{l} \right) \left( \frac{dl}{dt} \right) \\[12px] &= – \left( \frac{10}{50} \right) \left( 8\right) = -\dfrac{8}{5} \text{ cm/s} \end{align*} Then substituting values into the equation marked [*] above:

\begin{align*} d\frac{dd}{dt} &= w \frac{dw}{dt} + l \frac{dl}{dt} \\[12px] (10\sqrt{26}) \frac{dd}{dt} & = (10) \left(-\dfrac{8}{5}\right) + (50) (8) \\[12px] &= -16 + 400 = 384 \\[12px] \frac{dd}{dt} &= \frac{384}{10\sqrt{26}} \\[12px] &= \frac{38.4}{\sqrt{26}} \text{ cm/s } \quad \cmark \end{align*}

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1. Draw a picture of the physical situation.
See the figure. We’ve called the length of string to the kite $\ell.$ The problem is asking for the rate at which that length changes (= the rate at which the string is unspooling), $\dfrac{d\ell}{dt}$, at a particular moment — when $\ell = 50$ meters. Recall also that the kite is moving horizontally, in the x-direction, at the rate $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}} $. We’ll use these values at the end of our solution.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
The string’s length changes as the situation progresses, so we’re calling that the variable $\ell.$

B. To develop your equation, you will probably use . . . the Pythagorean theorem.
In this problem, the diagram above immediately suggests that we’re dealing with a right triangle. Furthermore, we need to related the rate at which the string’s length $\ell$ is changing, $\dfrac{d\ell}{dt}$, to the rate at which x is changing, $\dfrac{dx}{dt}$. Hence we must first write down an equation that somehow relates $\ell$ and x. The Pythagorean theorem gives us that relation: $$\ell^2 = x^2 + (30)^2 $$ That’s it. That’s the key relationship that will allow us to complete the solution.

3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. \[ \begin{align*} \dfrac{d}{dt}\ell^2 &= \dfrac{d}{dt}x^2 + \cancelto{0}{\dfrac{d}{dt}30^2} \\[8px] 2\ell \dfrac{d\ell}{dt} &= 2x \dfrac{dx}{dt} \\[8px] \ell \dfrac{d\ell}{dt} &= x \dfrac{dx}{dt} \end{align*} \] 4. Solve for the quantity you’re after.
Let’s solve the preceding equation for $\dfrac{d\ell}{dt}$: $$\dfrac{d\ell}{dt} = \frac{x}{\ell} \dfrac{dx}{dt}$$ To complete the solution, we need to know the value of x at the instant when $\ell = 50$ m.

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Begin subproblem to find x at the moment of interest.
Recall from the Pythagorean theorem that at every moment $$\ell^2 = x^2 + (30)^2 $$ Hence when $\ell = 50$ m, \[ \begin{align*} x^2 &= \ell^2 – 30^2 \\[8px] &= (50)^2 – 30^2 \\[8px] &= 1600\\[8px] x &= 40 \end{align*} \] As a faster approach, you might have noticed that this is a 30-___-50 right triangle, and so as a “3-4-5 right triangle” the missing length has to be 40.
End subproblem.

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We can now substitute values into our preceding equation: $$\dfrac{d\ell}{dt} = \frac{x}{\ell} \dfrac{dx}{dt}$$ We have $x = 40$ m, $\ell = 50$ m, and $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}}$: \[ \begin{align*} \dfrac{d\ell}{dt} &= \frac{x}{\ell} \dfrac{dx}{dt} \\[8px] &= \frac{40 \text{ m}}{50 \text{ m}}(2 \, \tfrac{\text{m}}{\text{s}}) \\[8px] &= 1.6 \, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*} \]

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Answer: $-\dfrac{39}{5}$ mph

1. Draw a picture of the physical situation.
See the figure. We’ll call Ann’s distance from the intersection $x$, and Frederick’s distance from the intersection $y$.

Let’s call the distance between them at any instant $d$, as indicated in the lower figure. The question is asking us to find  $\dfrac{dd}{dt}$  at noon.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet. In this scenario, both of the jogger’s positions change over time, so we’re leaving each of their positions as a variable.

B. To develop your equation, you will probably use . . . the Pythagorean theorem.
We can relate $d$ to $x$ and $y$ using the Pythagorean theorem:

$$d^2 = x^2 + y^2$$ 3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.

\begin{align*} \frac{d}{dt}(d^2) &= \frac{d}{dt}\left(x^2 + y^2 \right) \\[12px] 2d \frac{dd}{dt} &= 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \\[12px] d\frac{dd}{dt} &= x \frac{dx}{dt} + y \frac{dy}{dt} \text{ [*]} \end{align*} 4. Solve for the quantity you’re after.
We want to find  $\dfrac{dd}{dt}$  at noon, when  $x = 3$ miles and  $y = 4$ miles. We also know that  $\dfrac{dx}{dt} = -5$ mph, and  $\dfrac{dy}{dt} = -6$ mph. (Note that we must insert both negative signs “by hand” since both $x$ and $y$ are decreasing as the joggers head toward the intersection.)

To solve the preceding equation for $\dfrac{dd}{dt}$, we also need to know $d$; this we can find by using the Pythagorean theorem at this instant:

\begin{align*} d^2 &= x^2+y^2 \\ &= 3^2 + 4^2 = 25 \\ d &= 5 \end{align*} Then substituting values into the equation marked [*] above:

\begin{align*} d\frac{dd}{dt} &= x \frac{dx}{dt} + y \frac{dy}{dt} \\[12px] (5) \frac{dd}{dt} &= (3)(-5) + (4)(-6) = -15-24 = -39 \\[12px] \frac{dd}{dt} & = -\dfrac{39}{5} \text{ mph} \quad \cmark \end{align*} The negative value indicates that the distance between the joggers is decreasing as they jog toward each other.

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1. Draw a picture of the physical situation.
See the figures. On the left, we’ve shown the situation at 9:00 AM, with ship Blue 100 km east of ship Red. On the right, we’ve shown the ships at some arbitrary time t later: ship Red has moved to the east a distance x, and ship Blue has moved to the north a distance y. We’ve labeled the distance between the ships $\ell.$ We’re looking for $\dfrac{d\ell}{dt}$ at particular moment, 12:00-noon, a fact we’ll use at the end of the solution.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
Both ship Red and ship Blue’s positions change, so we’ve called the distances they travel x and y respectively.

B. To develop your equation, you will probably use . . . the Pythagorean theorem.
In this problem, we’re given values for $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt},$ and asked to find the value for $\dfrac{d\ell}{dt}$ at a particular moment. We thus first need to relate $\ell$ to x and y.

The figure above suggests how to do so: the Pythagorean theorem. Specifically, at every moment we have $$\ell^2 = (100 -x)^2 + y^2$$ That’s it. That’s the key relationship that will allow us to complete the solution. (Note, by the way, how critical drawing the figure correctly is to being able to solve this problem.)


3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. \[ \begin{align*} \dfrac{d}{dt}\left[\ell^2 \right] &= \dfrac{d}{dt} \left[(100 -x)^2 \right] + \dfrac{d}{dt} \left[ y^2\right] \\[8px] 2 \ell \dfrac{d\ell}{dt} &= 2(100 – x)(-1) \dfrac{dx}{dt} + 2 y \dfrac{dy}{dt} \\[8px] \ell \dfrac{d\ell}{dt} &= -(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt} \end{align*} \]

4. Solve for the quantity you’re after.
Let’s solve the preceding equation for $\dfrac{d\ell}{dt}:$ \[ \begin{align*} \ell \dfrac{d\ell}{dt} &= -(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt} \\[8px] \dfrac{d\ell}{dt} &= \frac{-(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt}}{\ell} &(*) \end{align*} \] We know $\dfrac{dx}{dt} = 30 \, \tfrac{\text{km}}{\text{hr}}$ and $\dfrac{dy}{dt} = 20 \, \tfrac{\text{km}}{\text{hr}}.$ To complete the problem, we need to know the values of x, y, and $\ell$ at the moment of interest.

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Begin subproblem to find x, y, and $\ell$.
We’re interested in the particular moment 12:00-noon, which is t = 3 hours after our initial picture at 9:00 AM. Hence \[ \begin{align*} x &= \left(30 \, \tfrac{\text{km}}{\text{hr}} \right) (3 \text{ hr}) = 90 \text{ km} \\[8px] y &= \left(20 \, \tfrac{\text{km}}{\text{hr}} \right) (3 \text{ hr}) = 60 \text{ km} \end{align*} \] And using the Pythagorean theorem we can find $\ell$ at this moment: \[ \begin{align*} \ell^2 &= (100 -x)^2 + y^2 \\[8px] &= (100 – 90)^2 + (60)^2 \\[8px] &= 3700 \\[8px] \ell &= \sqrt{3700} \\[8px] &= 60.83 \text{ km} \end{align*} \] End subproblem.

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We finally substitute all of these values into our equation marked (*): \[ \begin{align*} \dfrac{d\ell}{dt} &= \frac{-(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt}}{\ell} \\[8px] &= \frac{-(100 – 90)\, (30) + (60)\,(20)}{60.83} \\[8px] &= \frac{-300 + 1200}{60.83} \\[8px] &= 14.8 \, \tfrac{\text{km}}{\text{hr}} \quad \cmark \end{align*} \]

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1. Draw a picture of the physical situation.
See the figure. We’ve labeled the rocket’s vertical position at any moment after launch y, and the distance from you to the rocket $\ell.$ The question is asking us to find $\dfrac{d\ell}{dt}$ at a particular moment, 6 seconds after launch—a fact we’ll use at the end of our solution.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
The rocket’s vertical position changes as the time passes, so we’re calling that the variable y.

B. To develop your equation, you will probably use . . . the Pythagorean theorem.
We’re looking for $\dfrac{d\ell}{dt},$ and we know (actually: we can easily determine) the rate at which the rocket’s y- position is changing, $\dfrac{dy}{dt}$. We thus first need to related $\ell$ to y.

The figure above suggests how to do so: the Pythagorean theorem. Specifically, at every moment we have $$\ell^2 = (500)^2 + y^2$$ That’s it. That’s the key relationship that will allow us to complete the solution. (Note, by the way, how critical drawing the figure correctly is to being able to solve this problem.)

3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. \[ \begin{align*} \dfrac{d}{dt}\left(\ell^2 \right) &= \cancelto{0}{\dfrac{d}{dt}(500)^2} + \dfrac{d}{dt}\left(y^2 \right) \\[8px] 2 \ell \dfrac{d\ell}{dt} &= 2 y \dfrac{dy}{dt} \\[8px] \ell \dfrac{d\ell}{dt} &= y \dfrac{dy}{dt} \\[8px] \end{align*} \]
4. Solve for the quantity you’re after.
Let’s solve the preceding equation for $\dfrac{d\ell}{dt}:$ \[ \begin{align*} \ell \dfrac{d\ell}{dt} &= y \dfrac{dy}{dt} \\[8px] \dfrac{d\ell}{dt} &= \frac{y}{\ell}\dfrac{dy}{dt} &(*) \end{align*} \] To complete our solution, we need to know the values of y, $\ell,$ and $\dfrac{d\ell}{dt}$ 6 seconds after launch.

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Begin subproblems to find y, $\ell,$ and $\dfrac{d\ell}{dt}$.
A. Find y.
The problem states that $y=5t^2$ meters after t seconds. Hence 6 seconds after launch $$y=5\,(6)^2 = 180 \text{ m}$$ B. Find $\ell.$
By the Pythagorean theorem, at this instant \[ \begin{align*} \ell^2 &= (500)^2 + (180)^2 \\[8px] &= 282400 \\[8px] \ell &= \sqrt{282400} \\[8px] &= 531 \text{ m} \end{align*} \] C. Find $\dfrac{dy}{dt}.$
Since $y=5t^2,$ $$\dfrac{dy}{dt} = 10t $$ Hence at $t = 6$ seconds, $$\dfrac{dy}{dt} \big|_{6\text{ s}} = 10(6) = 60 \, \tfrac{\text{m}}{\text{s}}$$ End subproblems.

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We can now substitute these values into our equation marked (*) above: \[ \begin{align*} \dfrac{d\ell}{dt} &= \frac{y}{\ell}\dfrac{dy}{dt} \\[8px] &= \frac{180}{531} (60) \\[8px] &= 20.3 \, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*} \]

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