Calculus 1

Related rates

A sphere of ice melts, losing volume at a known rate. How fast is its radius shrinking, when nobody has a formula for the radius as a function of time?

You need

Implicit differentiationCalculus 1 Volume of prisms and cylindersMath 7

Where you are 11 in Applications of derivatives

Calculus8

Applications of derivatives11

▴ 5 above Position, velocity and acceleration The mean value theorem

When you reach for it

Some quantities change together because an equation ties them at every instant: the radius and the area of a spreading circle, the depth and the volume of liquid in a glass, the two legs and the hypotenuse of a right triangle, an angle and the height it looks up to. One rate of change is known or easy to measure, and another is wanted. Usually neither quantity has a formula in terms of time, and none is needed.

The move is to differentiate the equation with respect to time t. Every changing quantity is treated as an unnamed function of t, so each term picks up a factor such as \dfrac{dr}{dt} by the chain rule, exactly as y^2 gave 2y\,\dfrac{dy}{dx} in implicit differentiation. Related rates is implicit differentiation with t in the place of x. In the notation of The chain rule, the area of a circle changes at

\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt}

so a rate in centimeters per second becomes a rate in square centimeters per second.

The lesson runs over three days. Day 1 sets up the idea with circles, spheres, boxes and the Pythagorean theorem. Day 2 uses similar triangles to remove a variable whose rate is unknown. Day 3 brings in angles, two moving objects, and ways to check an answer.

The procedure

Day 1: an equation that holds at every instant

Start by naming every quantity that changes as a function of time, and keep every quantity that stays fixed as a number. Then find an equation that ties the changing quantities together at every instant. For geometry this is usually a formula already known: A = \pi r^2 and C = 2\pi r for a circle, \displaystyle V = \frac{4}{3}\pi r^3 for a sphere (Volume of a sphere) and S = 4\pi r^2 for its surface (Surface area of a sphere), V = Bh for a prism or cylinder with base area B (Volume of prisms and cylinders), and x^2 + y^2 = z^2 for a right triangle (The Pythagorean theorem).

Differentiate both sides with respect to t. Two functions that are equal at every instant of an interval have equal derivatives there, which is what makes the step allowed. For the sphere,

\frac{d}{dt}\Big[\tfrac{4}{3}\pi r^3\Big] = 4\pi r^2\,\frac{dr}{dt}

and the factor 4\pi r^2 is the sphere’s surface area: a thin new layer of thickness dr adds about 4\pi r^2\,dr of volume.

Only after differentiating are the values at the instant substituted. A value such as r = 6 holds at one moment, not over an interval, so it is not an equation between functions of t and cannot be differentiated. Substituting it first turns a changing quantity into a constant, whose derivative is 0. A notation such as \dfrac{dr}{dt}\Big|_{r=6} marks the rate at the moment when r = 6.

Every rate carries units, the units of the quantity divided by the units of time: an area changing in square centimeters per second, an angle in radians per second. Checking that each term of the differentiated equation has the same units catches many slips. Every rate also carries a sign. A quantity that grows has a positive derivative, and one that shrinks has a negative derivative, so “the volume decreases at 2 cubic centimeters per minute” means \dfrac{dV}{dt} = -2. The sign of the answer then says whether the wanted quantity grows or shrinks.

Day 2: similar triangles remove a variable

The volume of a cone, \displaystyle V = \frac{1}{3}\pi r^2 h (Volume of pyramids and cones), has three changing quantities. If only \dfrac{dV}{dt} is known, differentiating leaves two unknown rates, \dfrac{dr}{dt} and \dfrac{dh}{dt}, and one equation cannot give both. The shape of the container supplies a second equation. Liquid in a cone, point down, always forms a smaller cone of the same shape, so the triangles in its cross-section are similar (Proportions in similar triangles) and \dfrac{r}{h} equals the ratio for the full container at every instant.

An equation that holds at every instant may be substituted at any time, and doing it before differentiating is shorter: V becomes a function of h alone, and one use of the chain rule finishes the work. The two kinds of fact behave differently, and this is the point to hold on to: a relation true for all t may be used before differentiating, a value true at one instant only after. Similar triangles also give the length of a shadow, since the light ray, the object and the ground make one triangle and the light, its post or the screen make a similar one.

Day 3: angles, two moving objects, and checking the answer

When an angle changes, a trigonometric ratio ties it to the lengths: an angle of elevation \theta seen from a fixed distance a satisfies \tan\theta = \dfrac{h}{a}. Differentiating uses \dfrac{d}{d\theta}\tan\theta = \sec^2\theta from Derivatives of the trigonometric functions, which holds only when \theta is measured in radians, so a rate of turning comes out in radians per unit of time. Differentiating the tangent relation avoids needing the derivative of \arctan; once Derivatives of inverse trigonometric functions is available, writing \theta = \arctan\dfrac{h}{a} and differentiating gives the same result.

When two objects move, measure each position from a fixed point, such as the crossing of two roads. Each distance from that point gets its own rate, positive if that distance grows and negative if it shrinks, and the Pythagorean theorem ties the two distances to the distance between the objects.

An answer can be checked three ways without working it again. The sign should match the picture: a shadow that must shrink should have a negative rate. The size should fit the speeds given: two objects moving at 5 and 4 meters per second cannot change their distance faster than 9 meters per second. And a short step forward in time gives a numerical check: move each given quantity forward by its rate for one short time step, recompute the wanted quantity directly, and compare its change per unit of time with the answer.

Worked examples

Example 1

Day 1: batter spreading on a griddle

Meruyert pours batter onto a griddle, and it spreads in a circle. At the moment the radius is 6 cm, it is growing at 0.5 cm per second. How fast is the area growing at that moment? How fast would it grow at the moment the radius is 12 cm, still growing at 0.5 cm per second?

Solution

The radius r and the area A are both functions of time t in seconds, tied by A = \pi r^2 at every instant. Differentiate both sides with respect to t. The right side is \pi times the square of a function of t, so the chain rule gives

\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}

Only now substitute r = 6 and \dfrac{dr}{dt} = 0.5:

\frac{dA}{dt}\Big|_{r=6} = 2\pi(6)(0.5) = 6\pi \approx 18.85

The area grows at 6\pi \approx 18.85 square centimeters per second. The units agree: centimeters times centimeters per second is square centimeters per second.

At r = 12 the same radius rate gives 2\pi(12)(0.5) = 12\pi \approx 37.70 square centimeters per second, twice as fast. The factor 2\pi r is the circumference: the area added each moment is a thin ring around the edge, and a longer edge adds more area for the same outward spread.

Substituting r = 6 first would give A = 36\pi, a constant with derivative 0, as if the area were not changing at all. The value r = 6 holds at one instant; the equation A = \pi r^2 holds at every instant, and only that kind of equation may be differentiated.

Example 2

Day 1: a melting ice sphere, and a negative rate

Daniyar drops an ice sphere into a drink. As it melts it stays a sphere, and its volume decreases at 2 cubic centimeters per minute. How fast is its radius changing when the radius is 2.5 cm? How fast is its surface area changing then?

Solution

A decreasing volume has a negative rate: \dfrac{dV}{dt} = -2. The volume and the radius are tied by \displaystyle V = \frac{4}{3}\pi r^3 at every instant, and differentiating with respect to t, with the chain rule on r^3, gives

\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}

Substituting r = 2.5 and \dfrac{dV}{dt} = -2 gives -2 = 4\pi(6.25)\,\dfrac{dr}{dt} = 25\pi\,\dfrac{dr}{dt}, so

\frac{dr}{dt} = -\frac{2}{25\pi} \approx -0.0255

centimeters per minute. The sign is negative, so the radius shrinks, as it must when the volume shrinks. The units check: cubic centimeters per minute divided by square centimeters is centimeters per minute.

The surface area is S = 4\pi r^2, so \dfrac{dS}{dt} = 8\pi r\,\dfrac{dr}{dt} = 8\pi(2.5)\left(-\dfrac{2}{25\pi}\right) = -1.6 square centimeters per minute. Dividing the two rate equations gives \dfrac{dS}{dt} = \dfrac{2}{r}\cdot\dfrac{dV}{dt} for any sphere, and indeed \dfrac{2}{2.5}(-2) = -1.6.

Example 3

Day 1: a resizing rectangle, its diagonal and its area

In a drawing program, Daniyar drags a corner of a rectangular selection box. The width x grows at 3 cm per second while the height y shrinks at 2 cm per second. At the moment x = 24 cm and y = 7 cm, how fast are the diagonal and the area changing?

Solution

The known rates are \dfrac{dx}{dt} = 3 and \dfrac{dy}{dt} = -2, negative because the height shrinks. The diagonal z is the hypotenuse of a right triangle with legs x and y, so x^2 + y^2 = z^2 at every instant. Differentiating each term by the chain rule,

2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 2z\,\frac{dz}{dt}

At this moment \displaystyle z = \sqrt{576 + 49} = \sqrt{625} = 25. Dividing by 2 and substituting, 24(3) + 7(-2) = 25\,\dfrac{dz}{dt}, so 58 = 25\,\dfrac{dz}{dt} and \dfrac{dz}{dt} = \dfrac{58}{25} = 2.32 cm per second.

The area is A = xy, a product of two functions of t, so the product rule gives

\frac{dA}{dt} = \frac{dx}{dt}\,y + x\,\frac{dy}{dt} = 3(7) + 24(-2) = -27

square centimeters per second. The diagonal grows while the area shrinks, and nothing contradicts: in the diagonal’s equation each rate is weighted by its own side, so the long growing width wins, while in the area’s equation each rate is weighted by the other side, so the shrinking height is multiplied by the long width. Each rate carries its own sign.

Example 4

Day 2: filling a parfait glass

Lyazzat pours juice into a parfait glass shaped like a cone, point down, 12 cm deep with a rim of radius 4 cm, at a steady 8 cubic centimeters per second. How fast is the level rising when the juice is 6 cm deep? How fast is the radius of the surface growing then?

Figure 1The glass of Example 4. The juice and the whole glass make similar triangles, so \dfrac{r}{h} = \dfrac{4}{12} at every instant.

Solution

Let h be the depth, r the radius of the surface and V the volume of juice at time t. The cone formula \displaystyle V = \frac{1}{3}\pi r^2 h has three changing quantities, and only \dfrac{dV}{dt} = 8 is known. The glass supplies the missing link. In the cross-section, the triangle made by the axis, the juice’s surface radius and the side of the glass is similar to the triangle made by the axis, the rim radius and the side: they share the angle at the point, and both have a right angle at the axis. So \dfrac{r}{h} = \dfrac{4}{12} at every instant, that is, r = \dfrac{h}{3}. Because this holds at every instant, substitute it before differentiating:

V = \frac{1}{3}\pi\left(\frac{h}{3}\right)^2 h = \frac{\pi h^3}{27}

Differentiating with respect to t gives \dfrac{dV}{dt} = \dfrac{\pi h^2}{9}\,\dfrac{dh}{dt}. At h = 6, 8 = \dfrac{36\pi}{9}\,\dfrac{dh}{dt} = 4\pi\,\dfrac{dh}{dt}, so the level rises at \dfrac{dh}{dt} = \dfrac{2}{\pi} \approx 0.637 cm per second. Differentiating r = \dfrac{h}{3} gives \dfrac{dr}{dt} = \dfrac{1}{3}\cdot\dfrac{dh}{dt} = \dfrac{2}{3\pi} \approx 0.212 cm per second.

Solving the rate equation for \dfrac{dh}{dt} before putting in any depth gives \dfrac{dh}{dt} = \dfrac{72}{\pi h^2}. With the pour steady, the level rises more slowly as it climbs, because each centimeter of height near the wide rim holds more juice: at h = 9 it rises at \dfrac{8}{9\pi} \approx 0.283 cm per second. Keeping r instead would have meant the product rule and an unknown \dfrac{dr}{dt}, which only r = \dfrac{h}{3} could supply anyway.

Example 5

Day 2: a puppet in a shadow theater

In a shadow theater, a small lamp sits on the table and the screen stands 90 cm from it. Meruyert moves an upright puppet 15 cm tall from the lamp toward the screen at 6 cm per second. How fast is the height of the shadow on the screen changing when the puppet is 30 cm from the lamp?

Figure 2The shadow theater of Example 5. The triangle with legs x and 15 is similar to the triangle with legs 90 and S.

Solution

Let x be the distance from the lamp to the puppet and S the height of the shadow; the 90 and the 15 stay fixed. The ray from the lamp over the puppet’s top ends at the top of the shadow. The right triangle with legs x and 15 and the right triangle with legs 90 and S share the angle at the lamp, so they are similar and \dfrac{S}{90} = \dfrac{15}{x}, that is,

S = \frac{1350}{x}

Here the similar triangles do not remove an extra variable; they are the whole relation. Differentiating with respect to t by the chain rule, \dfrac{dS}{dt} = -\dfrac{1350}{x^2}\cdot\dfrac{dx}{dt}. At x = 30 with \dfrac{dx}{dt} = 6,

\frac{dS}{dt} = -\frac{1350}{900}\cdot 6 = -9

At that moment the shadow is \dfrac{1350}{30} = 45 cm tall and shrinking at 9 cm per second. The same result comes implicitly from xS = 1350: the product rule gives \dfrac{dx}{dt}\,S + x\,\dfrac{dS}{dt} = 0, so \dfrac{dS}{dt} = -\dfrac{S}{x}\cdot\dfrac{dx}{dt} = -\dfrac{45}{30}\cdot 6 = -9. The sign fits the picture: as the puppet nears the screen its shadow shrinks toward the puppet’s own height, reaching 15 cm at x = 90.

Example 6

Day 3: watching a window-washing platform

A window-washing platform descends the face of a glass tower at 0.4 meters per second. Lyazzat stands 24 m from the base of the tower and watches it; take her eye to be at ground level. How fast is the angle of elevation of her line of sight changing when the platform is 32 m up?

Figure 3Example 6. As the platform moves, h and \theta change together, tied by \tan\theta = \dfrac{h}{24}.

Solution

Let \theta be the angle of elevation in radians and h the platform’s height; the 24 stays fixed, and \dfrac{dh}{dt} = -0.4 because the platform descends. The right triangle gives \tan\theta = \dfrac{h}{24} at every instant. Differentiating with respect to t, with the chain rule and \dfrac{d}{d\theta}\tan\theta = \sec^2\theta,

\sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{24}\cdot\frac{dh}{dt}

At h = 32 the line of sight is \displaystyle \sqrt{24^2 + 32^2} = 40 m long, so \sec\theta = \dfrac{40}{24} = \dfrac{5}{3} and \sec^2\theta = \dfrac{25}{9}. The angle itself is never needed, only its secant, read from the triangle. Then \dfrac{25}{9}\cdot\dfrac{d\theta}{dt} = -\dfrac{0.4}{24}, so

\frac{d\theta}{dt} = -\frac{0.4}{24}\cdot\frac{9}{25} = -0.006

radians per second, about 0.344 degrees per second, or about 20.6 degrees per minute.

Checking: the sign is negative, and the angle should fall as the platform comes down. For a numerical check, one second later the platform is at 31.6 m. A calculator gives \theta = \arctan\dfrac{32}{24} \approx 0.92730 and then \arctan\dfrac{31.6}{24} \approx 0.92125, a change of about -0.00605 radians in that second, which agrees with -0.006.

Example 7

Day 3: two skaters, and a check on the answer

Two straight paths cross at right angles. Meruyert, on rollerblades, is 600 m north of the crossing, skating south toward it at 5 meters per second. At the same moment Daniyar is 800 m east of the crossing, skating east away from it at 4 meters per second. How fast is the distance between them changing? Is the answer reasonable?

Figure 4Example 7. Both legs change at once, and the Pythagorean theorem ties them to the distance z.

Solution

Measure each position from the crossing, a fixed point. Let y be Meruyert’s distance north of it, x Daniyar’s distance east, and z the distance between them. Meruyert’s distance shrinks, so \dfrac{dy}{dt} = -5; Daniyar’s grows, so \dfrac{dx}{dt} = 4. The paths meet at a right angle, so x^2 + y^2 = z^2 at every instant, and differentiating and dividing by 2 gives

x\,\frac{dx}{dt} + y\,\frac{dy}{dt} = z\,\frac{dz}{dt}

At this moment \displaystyle z = \sqrt{800^2 + 600^2} = 1000. Substituting, 800(4) + 600(-5) = 1000\,\dfrac{dz}{dt}, so 3200 - 3000 = 1000\,\dfrac{dz}{dt} and \dfrac{dz}{dt} = 0.2 meters per second. The distance is growing, slowly.

Is that reasonable? Daniyar’s motion pulls them apart and Meruyert’s brings them together, and dividing the equation by z shows each speed weighted by its own leg over z: \dfrac{800}{1000}\cdot 4 = 3.2 apart against \dfrac{600}{1000}\cdot 5 = 3 together, which nearly cancel. The size fits: in one second each moves at most at its own speed, so by the Triangle inequality the distance between them changes by at most 5 + 4 = 9 meters in a second, and 0.2 is well inside that. Finally a step forward: one second later Daniyar is 804 m east and Meruyert 595 m north, \displaystyle \sqrt{804^2 + 595^2} = \sqrt{1{,}000{,}441} \approx 1000.22 m apart. The distance grew by about 0.22 m in that second, close to 0.2; the small excess is because the rate itself is growing during the second.

Boundaries

The quantities must change smoothly. The method differentiates functions of t, so each quantity must be differentiable at the instant in question. A container with a corner in its side, such as a glass whose shape changes partway up, needs a separate equation on each side of the corner.
The equation must hold over an interval of time. A relation true only at one instant, such as r = 6 or h = 32, cannot be differentiated. Similar triangles in a container hold only while the liquid stays inside the shape that gives them; once the glass overflows, r = \dfrac{h}{3} stops describing anything.
Angles in radians. The derivatives \dfrac{d}{d\theta}\tan\theta = \sec^2\theta and the like assume radians. A rate given in degrees per second is converted first, by the factor \dfrac{\pi}{180}.
The coefficient of the wanted rate must not be zero. Solving for \dfrac{dr}{dt} in \dfrac{dV}{dt} = 4\pi r^2\,\dfrac{dr}{dt} divides by 4\pi r^2, and solving for \dfrac{dz}{dt} divides by z. At an instant where that coefficient is 0 (a sphere of radius 0, two objects at the same spot) the equation says nothing about the wanted rate, and the model behind it may fail nearby (exercise 23).
The answer is a rate at one instant. It usually changes from moment to moment, as the parfait glass fills more slowly near its rim. Multiplying it by an elapsed time gives only an estimate, good over a short time.

Where this goes next

The same chain-rule link between changes, read for small changes instead of rates, gives linear approximation and differentials, where a small error in a measured radius becomes an estimated error in a volume. The equations of Day 1 and Day 2, built from a picture and a geometry formula, return in optimization, where the aim is the largest or smallest value instead of a rate. Fixed sightings from two places, as in Angles of elevation and depression, are the still pictures that Day 3 sets in motion. Later, integration runs the idea backward: from a rate known at every instant, it recovers the total change.

Exercises

Day 1: circles, spheres, boxes and right triangles

The radius of a circle grows at 3 cm per second. How fast are its area and its circumference growing when the radius is 10 cm?
The area of a circle decreases at 5 square centimeters per minute. How fast is the radius changing when the radius is 4 cm?
The radius of a sphere grows at 0.2 cm per second. Find the rates of change of its volume and of its surface area when the radius is 3 cm.
The volume of a sphere grows at 50 cubic centimeters per second. How fast is the radius growing when the diameter is 10 cm?
A cube of gelatin dissolves so that each edge shrinks at 0.1 cm per minute. Find the rates of change of its volume and of its surface area when the edge is 3 cm.
A cylindrical tank of radius 25 cm drains at 2500 cubic centimeters per minute. How fast does the water level fall? Explain why the answer does not depend on the depth.
A model airplane flies level at an altitude of 30 m, directly away from Lyazzat, at 8 meters per second. Taking her to be at ground level, how fast is its distance from her growing when its horizontal distance from her is 40 m?
Functions x and y of t satisfy x^2 + y^2 = 169 for all t. At an instant when x = 12, y = 5 and \dfrac{dx}{dt} = 5, find \dfrac{dy}{dt}.

Day 2: similar triangles

A paper cone cup, point down, is 10 cm deep with a rim of radius 5 cm. Water leaks from it at 2 cubic centimeters per second. How fast is the level falling when the water is 4 cm deep?
Road salt poured from a chute forms a cone whose base radius is always twice its height. Salt arrives at 3 cubic meters per minute. How fast are the height and the radius of the heap growing when the heap is 1.5 m tall?
Daniyar, 1.8 m tall, walks away from a light on top of a 6 m pole at 1.4 meters per second. How fast does his shadow lengthen, and how fast does the tip of his shadow move?
A closed V-shaped rain gutter is 3 m long, and its ends are triangles 0.6 m across the top and 0.4 m deep, point down. Water flows in at 0.06 cubic meters per minute. How fast does the level rise when the water is 0.2 m deep?
In the parfait glass of Example 4, with the same steady pour, at what depth is the level rising at 1 cm per second? How fast is it rising at the instant the juice reaches the rim?
An inverted conical cistern 6 m deep with a top radius of 3 m leaks at 0.5 cubic meters per minute while being filled at a constant rate. When the water is 2 m deep the level is rising at 0.2 meters per minute. Find the rate at which it is being filled.

Day 3: angles and two moving objects

As the sun sinks, its angle of elevation \theta decreases at 0.2 radian per hour. How fast does the shadow of a 24 m minaret on level ground lengthen when \theta = \dfrac{\pi}{6}?
A coastal radar dish stands 9 km from a straight shipping lane. A ferry travels along the lane at 1.5 km per minute, and the dish turns to track it. Let \theta be the angle between the dish’s line to the nearest point of the lane and its line to the ferry. How fast is \theta changing when the ferry is 12 km past that nearest point?
A parking lot gate arm 4 m long swings up about its hinge at 0.5 radian per second. When the arm makes an angle of \dfrac{\pi}{3} with the horizontal, how fast is its tip rising, and how fast is the tip’s horizontal distance from the hinge changing?
Daniyar and Lyazzat leave the same corner of a rectangular field at the same moment, jogging along the two sides that meet there at 3 and 4 meters per second. How fast is the distance between them growing after 10 seconds? Explain why the answer is the same at every time.
A car is 300 m east of an intersection, driving east at 10 meters per second. A bus is 400 m north of the intersection, driving south toward it at 15 meters per second. Is the distance between them growing or shrinking, and how fast?

Reasoning

Someone works exercise 1 by putting r = 10 into A = \pi r^2 first and then differentiating. What do they get, and exactly which step is not allowed?
A spherical raindrop evaporates so that its volume decreases at a rate proportional to its surface area: \dfrac{dV}{dt} = -kS for a constant k > 0. Prove that its radius shrinks at the constant rate k.
For Example 6, show that \dfrac{d\theta}{dt} = \dfrac{24}{576 + h^2}\cdot\dfrac{dh}{dt} at every height. With the platform’s speed fixed, at what height is the angle changing fastest, and why does it change slowly when the platform is high up?
A rod of length L slides with its ends on the positive x axis and the positive y axis, the bottom end moving away from the origin at a constant speed v. Show that \dfrac{dy}{dt} = -\dfrac{xv}{y}, and describe what happens to the speed of the top end as y \to 0^+. What does this say about whether the bottom end can keep a constant speed all the way to the floor?
In Example 7, someone reports that \dfrac{dz}{dt} = 12 meters per second. Without redoing the calculation, explain why this must be wrong.

Answers

\dfrac{dA}{dt} = 60\pi \approx 188.5 square centimeters per second and \dfrac{dC}{dt} = 6\pi \approx 18.85 cm per second. From \dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt} = 2\pi(10)(3); and C = 2\pi r gives \dfrac{dC}{dt} = 2\pi\,\dfrac{dr}{dt} at every radius.

\dfrac{dr}{dt} = -\dfrac{5}{8\pi} \approx -0.199 cm per minute. From -5 = 2\pi(4)\,\dfrac{dr}{dt}; negative because the circle shrinks.

\dfrac{dV}{dt} = 7.2\pi \approx 22.6 cubic centimeters per second and \dfrac{dS}{dt} = 4.8\pi \approx 15.1 square centimeters per second. From \dfrac{dV}{dt} = 4\pi(3)^2(0.2) and \dfrac{dS}{dt} = 8\pi(3)(0.2).

\dfrac{dr}{dt} = \dfrac{1}{2\pi} \approx 0.159 cm per second. The radius is 5, so 50 = 4\pi(25)\,\dfrac{dr}{dt}.

\dfrac{dV}{dt} = -2.7 cubic centimeters per minute and \dfrac{dS}{dt} = -3.6 square centimeters per minute. With edge s, V = s^3 gives 3s^2\,\dfrac{ds}{dt} = 3(9)(-0.1), and S = 6s^2 gives 12s\,\dfrac{ds}{dt} = 12(3)(-0.1).

It falls at \dfrac{4}{\pi} \approx 1.27 cm per minute, that is, \dfrac{dh}{dt} = -\dfrac{4}{\pi}. The volume is V = Bh with the fixed base area B = 625\pi, so \dfrac{dV}{dt} = 625\pi\,\dfrac{dh}{dt} and -2500 = 625\pi\,\dfrac{dh}{dt}. No h is left in the rate equation because every centimeter of depth holds the same volume.

6.4 meters per second. With horizontal distance x and distance z, x^2 + 30^2 = z^2 and z = 50; then 2x\,\dfrac{dx}{dt} = 2z\,\dfrac{dz}{dt} gives \dfrac{dz}{dt} = \dfrac{40 \cdot 8}{50}.

\dfrac{dy}{dt} = -12. From 2x\,\dfrac{dx}{dt} + 2y\,\dfrac{dy}{dt} = 0: 12(5) + 5\,\dfrac{dy}{dt} = 0.

It falls at \dfrac{1}{2\pi} \approx 0.159 cm per second, that is, \dfrac{dh}{dt} = -\dfrac{1}{2\pi}. Similar triangles give r = \dfrac{h}{2}, so V = \dfrac{\pi h^3}{12} and \dfrac{dV}{dt} = \dfrac{\pi h^2}{4}\cdot\dfrac{dh}{dt}; at h = 4, -2 = 4\pi\,\dfrac{dh}{dt}.

\dfrac{dh}{dt} = \dfrac{1}{3\pi} \approx 0.106 meters per minute and \dfrac{dr}{dt} = \dfrac{2}{3\pi} \approx 0.212 meters per minute. With r = 2h, V = \dfrac{1}{3}\pi(2h)^2 h = \dfrac{4\pi h^3}{3} and \dfrac{dV}{dt} = 4\pi h^2\,\dfrac{dh}{dt}, so 3 = 4\pi(2.25)\,\dfrac{dh}{dt} = 9\pi\,\dfrac{dh}{dt}.

The shadow lengthens at 0.6 meters per second and its tip moves at 2 meters per second. With x his distance from the pole and s the shadow’s length, similar triangles give \dfrac{s}{1.8} = \dfrac{x + s}{6}, so 4.2s = 1.8x and s = \dfrac{3x}{7}. Then \dfrac{ds}{dt} = \dfrac{3}{7}(1.4) = 0.6, and the tip, at distance x + s, moves at 1.4 + 0.6. Neither rate depends on x.

\dfrac{1}{15} \approx 0.0667 meters per minute. By similar triangles the water’s width at depth h is \dfrac{0.6}{0.4}h = 1.5h. The gutter is a prism, so V = 3 \cdot \dfrac{1}{2}(1.5h)h = 2.25h^2, and \dfrac{dV}{dt} = 4.5h\,\dfrac{dh}{dt}; at h = 0.2, 0.06 = 0.9\,\dfrac{dh}{dt}.

At \displaystyle h = \sqrt{\dfrac{72}{\pi}} \approx 4.79 cm; at the rim it rises at \dfrac{1}{2\pi} \approx 0.159 cm per second. Both come from \dfrac{dh}{dt} = \dfrac{72}{\pi h^2} in Example 4: setting it equal to 1 gives h^2 = \dfrac{72}{\pi}, and h = 12 gives \dfrac{72}{144\pi}.

0.5 + 0.2\pi \approx 1.13 cubic meters per minute. Similar triangles give r = \dfrac{h}{2}, so V = \dfrac{\pi h^3}{12} and \dfrac{dV}{dt} = \dfrac{\pi h^2}{4}\cdot\dfrac{dh}{dt} = \dfrac{4\pi}{4}(0.2) = 0.2\pi at h = 2. The net rate is the fill rate minus 0.5, so the fill rate is 0.2\pi + 0.5.

19.2 meters per hour. The shadow is s = 24\cot\theta, so \dfrac{ds}{dt} = -24\csc^2\theta\,\dfrac{d\theta}{dt} = -24(4)(-0.2), since \csc\dfrac{\pi}{6} = 2. Positive: the shadow lengthens as the sun sinks.

0.06 radian per minute. With x the ferry’s distance past the nearest point, \tan\theta = \dfrac{x}{9} and \sec^2\theta\,\dfrac{d\theta}{dt} = \dfrac{1}{9}\cdot\dfrac{dx}{dt}. The line to the ferry is \displaystyle \sqrt{81 + 144} = 15 km, so \sec^2\theta = \dfrac{225}{81} = \dfrac{25}{9} and \dfrac{d\theta}{dt} = \dfrac{1.5}{9}\cdot\dfrac{9}{25}.

The tip rises at 1 meter per second, and its horizontal distance changes at \displaystyle -\sqrt{3} \approx -1.73 meters per second. The tip’s height is 4\sin\theta, with rate 4\cos\theta\,\dfrac{d\theta}{dt} = 4\left(\dfrac{1}{2}\right)(0.5); its horizontal distance is 4\cos\theta, with rate \displaystyle -4\sin\theta\,\dfrac{d\theta}{dt} = -4\left(\dfrac{\sqrt{3}}{2}\right)(0.5).

5 meters per second. At t = 10 the legs are 30 and 40 and the distance is 50, so 50\,\dfrac{dz}{dt} = 30(3) + 40(4) = 250. In general the legs are 3t and 4t, so z = 5t and \dfrac{dz}{dt} = 5 at every time.

Shrinking, at 6 meters per second: \dfrac{dz}{dt} = -6. The distance is \displaystyle \sqrt{300^2 + 400^2} = 500, and 500\,\dfrac{dz}{dt} = 300(10) + 400(-15) = -3000.

They get A = 100\pi and then \dfrac{dA}{dt} = 0, which says a growing circle’s area is not changing. The equation A = \pi r^2 holds at every instant, so both sides are equal functions of t and have equal derivatives. The value r = 10 holds at one instant only, so after substituting it the equation no longer holds over an interval of time and may not be differentiated. Values go in after differentiating.

Differentiating V = \dfrac{4}{3}\pi r^3 gives \dfrac{dV}{dt} = 4\pi r^2\,\dfrac{dr}{dt}, and -kS = -4\pi k r^2. Setting these equal and dividing by 4\pi r^2, which is not 0 while the drop exists, gives \dfrac{dr}{dt} = -k.

From \sec^2\theta\,\dfrac{d\theta}{dt} = \dfrac{1}{24}\cdot\dfrac{dh}{dt} and \displaystyle \cos\theta = \dfrac{24}{\sqrt{576 + h^2}}, \dfrac{d\theta}{dt} = \dfrac{\cos^2\theta}{24}\cdot\dfrac{dh}{dt} = \dfrac{24}{576 + h^2}\cdot\dfrac{dh}{dt}. With \dfrac{dh}{dt} fixed, the factor \dfrac{24}{576 + h^2} is largest at h = 0, so the angle turns fastest when the platform is level with her eye, and the factor goes to 0 as h grows. High up, the line of sight is nearly vertical, and moving the platform a meter barely turns it. At h = 32 the formula gives \dfrac{24(-0.4)}{1600} = -0.006, as in Example 6.

The rod gives x^2 + y^2 = L^2, so 2x\,\dfrac{dx}{dt} + 2y\,\dfrac{dy}{dt} = 0, and with \dfrac{dx}{dt} = v, \dfrac{dy}{dt} = -\dfrac{xv}{y}. As y \to 0^+, x \to L, so \left|\dfrac{dy}{dt}\right| grows without bound. No real top end moves infinitely fast, so the top must leave the wall, or the bottom must slow, before the rod lies flat: the model with a constant v and the top on the wall breaks down near the floor. At y = 0 itself the equation cannot be solved for \dfrac{dy}{dt} at all.

In any one second Meruyert moves at most 5 m and Daniyar at most 4 m, so by the triangle inequality the distance between them changes by at most 9 m. Every average rate of change of z is then at most 9 in size, and the instantaneous rate, a limit of those averages, satisfies \left|\dfrac{dz}{dt}\right| \le 9. A rate of 12 is impossible.