Free AP Calculus AB Contextual Applications of Differentiation

AP Calculus AB · Contextual Applications of Differentiation · free lesson

Related Rates

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Rates that are tied together

In a related rates problem two or more quantities change with time, and an equation ties them together at every instant. Because the quantities are tied, so are their rates: if you know how fast one changes, the equation tells you how fast the other changes. The tool is the chain rule. When $x$ is a function of the time $t$ , a term such as $x^{2}$ is a composition, so

\begin{aligned} \frac{d}{dt}\big[x^2\big] &= 2x\,\frac{dx}{dt} \end{aligned}

Every changing quantity picks up its own rate this way: $\frac{d}{dt}\big[y^3\big] = 3y^2\,\frac{dy}{dt}$ , and a product such as $xy$ needs the product rule, $\frac{d}{dt}\big[xy\big] = \frac{dx}{dt}\,y + x\,\frac{dy}{dt}$ . This is implicit differentiation, with $t$ in place of $x$ .

A ladder sliding down a wall

A ladder $13$ feet long leans against a vertical wall. Its foot is pulled away from the wall at a steady $2$ feet per second. Let $x$ be the distance from the wall to the foot of the ladder and $y$ the height of its top, both in feet. Watch it slide. The green arrow, the rate $\frac{dx}{dt}$ of the foot, stays the same length the whole time; the red arrow, the rate $\frac{dy}{dt}$ of the top, grows: the top falls faster and faster.

The equation that ties $x$ and $y$ together is the Pythagorean theorem, $x^{2} + y^{2} = 169$ . It holds at every instant, so both sides have the same derivative with respect to $t$ . Differentiate each term with the chain rule; the constant $169$ has derivative 0:

\begin{aligned} \frac{d}{dt}\big[x^2 + y^2\big] &= \frac{d}{dt}\big[{169}\big] \\ {2x}\,\frac{dx}{dt} + {2y}\,\frac{dy}{dt} &= 0 \end{aligned}

Only now put in the values at the instant you want. When the foot is $x = 5$ feet from the wall, the top is at $y = \sqrt{ {169} - {5}^2} = {12}$ feet, and $\frac{dx}{dt} = {2}$ :

\begin{aligned} 2({5})({2}) + 2({12})\,\frac{dy}{dt} &= 0 \\ \frac{dy}{dt} &= -\frac{5}{6} \end{aligned}

The answer is negative because $y$ is decreasing: at that instant the top of the ladder is sliding down the wall at $\frac{5}{6}$ feet per second. Later, when $x = 12$ and $y = 5$ , the same working gives $\frac{dy}{dt} = {-\frac{24}{5}}$ feet per second: the foot still moves at $2$ feet per second, but the top now falls much faster. The graph shows $\frac{dy}{dt}$ for each position $x$ of the foot. It falls without bound as $x$ nears $13$ and the top reaches the floor.

Rule.

Solving a related rates problem, step by step.

Draw a picture. Name with letters the quantities that change; write the fixed ones as numbers.
Write the rate you are given and the rate you want as derivatives with respect to $t$ . A quantity that is decreasing has a negative rate.
Write an equation that ties the changing quantities together and holds at every instant. If it holds a letter you have no rate for, remove that letter first with a fixed relation (similar triangles, a given shape).
Differentiate both sides with respect to $t$ : each changing quantity brings its own rate (chain rule), and a product needs the product rule.
Only then put in the values at the instant, and solve for the rate you want.
Give the answer with units, and say whether the quantity is increasing or decreasing.

Warning.

Do not put the instant's values in before differentiating. With $x = 5$ and $y = 12$ the equation becomes $5^{2} + 12^{2} = 169$ , a true statement about numbers; its derivative is $0 = 0$ and says nothing about the rates. The values go in after step 4.

A cone filling: remove a variable first

Water runs into a tank shaped like a cone, point down, at a steady $2$ cubic feet per minute. The tank is $12$ feet deep and the radius of its top is $4$ feet. The volume of water is $V = \frac{1}{3}\pi r^2 h$ , where $h$ is the depth of the water and $r$ the radius of its surface. Both $r$ and $h$ change, and we are given no rate for $r$ . But the water always forms a cone of the same shape as the tank, so by similar triangles

\begin{aligned} \frac{r}{h} &= \frac{ {4} }{ {12} } = {\frac{1}{3}} \end{aligned}

that is, $r = {\frac{1}{3}}h$ at every instant. Put this in first, so that $V$ depends on $h$ alone: $V = \frac{1}{3}\pi({\frac{1}{3}}h)^2 h = \pi \cdot {\frac{1}{27}h^{3}}$ . Now differentiate with respect to $t$ :

\begin{aligned} \frac{dV}{dt} &= \pi \cdot {\frac{1}{9}h^{2}}\,\frac{dh}{dt} \end{aligned}

With $\frac{dV}{dt} = {2}$ , the depth rises at $\frac{dh}{dt} = \frac{ {2} }{\pi \cdot {\frac{1}{9}h^{2}}} = \frac{ {18} }{\pi h^2}$ feet per minute. The same inflow raises the level quickly when the water is shallow and its surface small, and slowly when it is deep: at $h = 3$ feet, $\frac{dh}{dt} = \frac{ {2} }{\pi}$ feet per minute; at $h = 9$ feet, only $\frac{ {2} }{ {9}\pi}$ feet per minute. Watch the tank fill at the steady rate.

A shadow: a rate that does not change

A person $6$ feet tall walks away from a street lamp $15$ feet high at $5$ feet per second. Let $x$ be the distance from the foot of the lamppost to the person and $y$ the length of the person's shadow. The light ray from the lamp over the person's head to the tip of the shadow makes two similar right triangles, one with the lamppost and one with the person, so

\begin{aligned} \frac{ {15} }{x + y} &= \frac{ {6} }{y} \end{aligned}

Clearing the fractions, ${15}y = {6}x + {6}y$ , so $y = {\frac{2}{3}}x$ . Differentiating, $\frac{dy}{dt} = {\frac{2}{3}}\,\frac{dx}{dt} = {\frac{2}{3}}({5}) = {\frac{10}{3}}$ feet per second, at every instant: the shadow grows at a constant rate however far the person has gone. The tip of the shadow is at $x + y$ from the post, so it moves at $\frac{dx}{dt} + \frac{dy}{dt} = {\frac{25}{3}}$ feet per second, faster than the person walks.

Note.

On the AP exam the usual slips are: putting in the instant's numbers before differentiating; dropping one of the rates when two quantities change at once; leaving the rate of a decreasing quantity positive; using a diameter where the formula wants the radius; and, for an angle, forgetting the chain factor ( $\frac{d}{dt}\tan\theta = \sec^2\theta\,\frac{d\theta}{dt}$ ). A free-response answer shows the equation that ties the quantities, its derivative with respect to $t$ , the values at the instant, and the answer with units and its meaning (what changes, and whether it increases or decreases).

Worked examples

Each example follows the same steps: name the changing quantities, write the equation that ties them at every instant, differentiate both sides with respect to $t$ , and only then put in the values at the instant. Every number in the working is found by the engine.

Example.

1. A balloon being filled. Air is pumped into a spherical balloon at 200 cubic centimeters per second. How fast is the radius of the balloon increasing at the instant the radius is 5 centimeters?

Let $r$ be the radius and $V$ the volume, both functions of $t$ . We know $\frac{dV}{dt} = {200}$ and want $\frac{dr}{dt}$ when $r = 5$ . The equation that ties them is the volume of a sphere, $V = \frac{4}{3}\pi r^3$ . Differentiate with respect to $t$ ; the chain rule brings in $\frac{dr}{dt}$ :

\begin{aligned} \frac{dV}{dt} &= 4\pi r^2\,\frac{dr}{dt} \end{aligned}

Now put in the instant's values: $4r^2 = {100}$ when $r = 5$ , so

\begin{aligned} 200 &= {100}\pi\,\frac{dr}{dt} \\ \frac{dr}{dt} &= \frac{ {200} }{ {100}\pi} = \frac{ {2} }{\pi} \end{aligned}

The radius is increasing at $\frac{ {2} }{\pi} \approx {0.637}$ centimeters per second at that instant. As the balloon grows, its surface $4\pi r^2$ grows, so the same flow of air makes the radius grow more slowly.

Example.

2. Two cars and the distance between them. Car $A$ is 3 miles east of an intersection, driving west toward it at 30 miles per hour. Car $B$ is 4 miles north of the intersection, driving north away from it at 45 miles per hour. At that instant, is the distance between the cars increasing or decreasing, and at what rate?

Let $x$ be the distance of car $A$ from the intersection, $y$ that of car $B$ , and $z$ the distance between the cars. Car $A$ is getting closer, so $x$ is decreasing: $\frac{dx}{dt} = {-30}$ , with a minus sign. Car $B$ moves away, so $\frac{dy}{dt} = {45}$ .

The roads meet at a right angle, so $x^{2} + y^{2} = z^{2}$ at every instant. Differentiate with respect to $t$ :

\begin{aligned} 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} &= 2z\,\frac{dz}{dt} \end{aligned}

At this instant $x = 3$ and $y = 4$ , so $z = \sqrt{ {3}^2 + {4}^2} = {5}$ . Put in the values and the two rates:

\begin{aligned} 2({3})({-30}) + 2({4})({45}) &= 2({5})\,\frac{dz}{dt} \\ \frac{dz}{dt} &= \frac{ {-180} + {360} }{ {10} } = {18} \end{aligned}

The rate is positive, so the distance between the cars is increasing, at $18$ miles per hour. Car $A$ 's approach is not enough to make up for car $B$ moving away faster. Leaving the minus sign off $\frac{dx}{dt}$ would give $\frac{ {180} + {360} }{ {10} } = {54}$ , a wrong answer.

Example.

3. A conical tank draining. A tank in the shape of a cone, point down, is 10 feet deep, and its top is a circle 10 feet across. Water leaks out at 3 cubic feet per minute. How fast is the depth of the water changing when the water is 4 feet deep?

Let $h$ be the depth of the water and $r$ the radius of its surface. The top is 10 feet across, so its radius is $5$ feet: the formula for a cone wants the radius, not the diameter. Water leaks out, so $\frac{dV}{dt} = {-3}$ .

The water is a cone of the same shape as the tank, so by similar triangles $\frac{r}{h} = \frac{ {5} }{ {10} }$ , that is $r = {\frac{1}{2}}h$ . Put this into $V = \frac{1}{3}\pi r^2 h$ so that only $h$ is left, then differentiate with respect to $t$ :

\begin{aligned} V &= \frac{1}{3}\pi({\frac{1}{2}}h)^2 h = \pi \cdot {\frac{1}{12}h^{3}} \\ \frac{dV}{dt} &= \pi \cdot {\frac{1}{4}h^{2}}\,\frac{dh}{dt} \end{aligned}

At the instant $h = 4$ , the factor $\frac{1}{4}h^{2}$ is $4$ :

\begin{aligned} -3 &= {4}\pi\,\frac{dh}{dt} \\ \frac{dh}{dt} &= \frac{ {-3} }{ {4}\pi} \end{aligned}

The depth is decreasing at $\frac{ {3} }{ {4}\pi} \approx {0.239}$ feet per minute when the water is $4$ feet deep. Taking the diameter 10 as the radius would have made $r = h$ and given a rate four times too small.

Example.

4. An angle of elevation. A hot-air balloon rises straight up from the ground at 4 meters per second. An observer stands 60 meters from the point where it left the ground. How fast is the angle of elevation from the observer to the balloon changing when the balloon is 80 meters up?

Let $y$ be the height of the balloon and $\theta$ the angle of elevation, in radians. The distance 60 is fixed. In the right triangle, $\tan\theta = \frac{y}{ {60} }$ . Differentiate both sides with respect to $t$ ; the left side needs the chain rule, because $\theta$ changes with $t$ :

\begin{aligned} \sec^2\theta\,\frac{d\theta}{dt} &= \frac{1}{ {60} }\,\frac{dy}{dt} \end{aligned}

At the instant $y = 80$ , $\tan\theta = \frac{ {80} }{ {60} } = {\frac{4}{3}}$ , and $\sec^2\theta = 1 + \tan^2\theta = {\frac{25}{9}}$ . With $\frac{dy}{dt} = {4}$ :

\begin{aligned} {\frac{25}{9}}\,\frac{d\theta}{dt} &= \frac{ {4} }{ {60} } \\ \frac{d\theta}{dt} &= \frac{3}{125} \end{aligned}

The angle of elevation is increasing at $\frac{3}{125}$ radians per second. An answer of $\frac{ {4} }{ {60} } = {\frac{1}{15}}$ forgets the factor $\sec^2\theta$ that the chain rule puts on the angle. (The rate of an angle is in radians per unit of time, because the derivative of $\tan\theta$ is $\sec^2\theta$ only for $\theta$ in radians.)

Example.

5. Two dimensions changing at once. The length of a rectangle is increasing at 2 centimeters per second while its width is decreasing at 3 centimeters per second. When the length is 10 centimeters and the width 6 centimeters, is the area increasing or decreasing, and how fast? How fast is the perimeter changing?

Let $x$ be the length and $y$ the width. The width is decreasing, so $\frac{dy}{dt} = {-3}$ . The area is $A = xy$ , a product of two changing quantities, so the product rule gives

\begin{aligned} \frac{dA}{dt} &= \frac{dx}{dt}\,y + x\,\frac{dy}{dt} \end{aligned}

At the instant, with $x = 10$ , $y = 6$ and $\frac{dx}{dt} = {2}$ :

\begin{aligned} \frac{dA}{dt} &= ({2})({6}) + ({10})({-3}) = {-18} \end{aligned}

The rate is negative, so the area is decreasing, at $18$ square centimeters per second: the shrinking width wins over the growing length at this instant. Two slips to avoid: dropping one of the two terms (that gives $12$ or $-30$ ), and taking $\frac{dy}{dt}$ as $+3$ because the problem says "3" (that gives $42$ ).

The perimeter is $P = 2x + 2y$ , so $\frac{dP}{dt} = 2\,\frac{dx}{dt} + 2\,\frac{dy}{dt} = 2({2}) + 2({-3}) = {-2}$ centimeters per second: the perimeter is decreasing too, and at the same rate at every instant.

Practice

Answer AP-style questions on related rates. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise related rates with new numbers