Corrected exercises: related rates (MATH 140)

a) The length $6.5$ m is fixed, and so is the rate $\frac{dx}{dt} = 0.3$ m/s; the distance $x$, the height $y$ and the angle $\theta$ all change. The wall is vertical and the ground horizontal, so the ladder is the hypotenuse of a right triangle, and by Pythagoras $x^2 + y^2 = 6.5^2 = 42.25$ at every instant. Differentiate both sides with respect to $t$. By the chain rule, with inner function $x(t)$, $\frac{d}{dt}\left(x^2\right) = 2x\frac{dx}{dt}$; likewise $\frac{d}{dt}\left(y^2\right) = 2y\frac{dy}{dt}$, and the constant $42.25$ has derivative $0$. So $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$, that is $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$. No number of the instant has been used yet, and that is the point: this equation is true at every moment of the slide.

b) Now, and only now, freeze the instant. When $x = 2.5$, the relation gives $y^2 = 42.25 - 6.25 = 36$, so $y = 6$ (a height is positive). Then $\frac{dy}{dt} = -\frac{2.5}{6}(0.3) = -\frac{0.75}{6} = -\frac{1}{8}$ m/s. The minus sign is part of the answer: $y$ decreases, the top slides DOWN at $\frac{1}{8}$ m/s, that is $12.5$ cm/s. A student who writes $\frac{dy}{dt} = \frac{1}{8}$ has described a ladder that climbs the wall, and loses the sign mark. The other classic error is to substitute $x = 2.5$ into $x^2 + y^2 = 42.25$ BEFORE differentiating: the equation becomes $6.25 + y^2 = 42.25$, whose derivative is $2y\frac{dy}{dt} = 0$, and the top would not move at all.

c) When $x = 6$: $y^2 = 42.25 - 36 = 6.25$, so $y = 2.5$, and $\frac{dy}{dt} = -\frac{6}{2.5}(0.3) = -0.72 = -\frac{18}{25}$ m/s. The foot still moves at $0.3$ m/s, but the top now falls at $0.72$ m/s, $5.76$ times faster than in b). The factor $\frac{x}{y}$ explains it: when the ladder is almost flat, $y$ is small and a small move of the foot costs the top a lot of height. The rate of $y$ depends on the INSTANT, which is exactly why the relation must be differentiated before any value is substituted. (As $y \to 0$ the formula would give a speed without bound; in reality the top leaves the wall long before, and the model stops applying.)

d) The angle at the foot satisfies $\cos\theta = \frac{x}{6.5}$ at every instant. Differentiate with respect to $t$, chain rule with inner function $\theta(t)$: $-\sin\theta\,\frac{d\theta}{dt} = \frac{1}{6.5}\frac{dx}{dt}$. At the instant, $\sin\theta = \frac{y}{6.5} = \frac{6}{6.5}$, so $-\frac{6}{6.5}\frac{d\theta}{dt} = \frac{0.3}{6.5}$, and $\frac{d\theta}{dt} = -\frac{0.3}{6} = -\frac{1}{20}$ rad/s. The unit is the RADIAN per second, because the derivative of $\cos$ is $-\sin$ only when the angle is in radians. The sign agrees with the picture: the ladder is tipping over, the angle with the ground shrinks.

e) $A = \frac{1}{2}xy$, a product of two functions of $t$. By the product rule, $\frac{dA}{dt} = \frac{1}{2}\left(\frac{dx}{dt}\,y + x\frac{dy}{dt}\right)$. At $x = 2.5$: $\frac{1}{2}\left(0.3 \times 6 + 2.5 \times \left(-\frac{1}{8}\right)\right) = \frac{1}{2}(1.8 - 0.3125) = \frac{1.4875}{2} = \frac{119}{160}$ m²/s, positive: the area grows. At $x = 6$: $\frac{1}{2}\left(0.3 \times 2.5 + 6 \times (-0.72)\right) = \frac{1}{2}(0.75 - 4.32) = -\frac{357}{200}$ m²/s, negative: the area shrinks. The two factors move in opposite directions and the product rule decides which wins; writing $\frac{dA}{dt} = \frac{1}{2}\frac{dx}{dt}\frac{dy}{dt}$ would give $-\frac{3}{160}$, the wrong sign in the first case, because the derivative of a product is not the product of the derivatives.

a) The relation $V = \frac{4}{3}\pi r^3$ holds at every instant. Differentiate with respect to $t$, chain rule with inner function $r(t)$: $\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}$. The data are $\frac{dV}{dt} = 72\pi$ and, at the instant, $r = 6$: $72\pi = 4\pi(36)\frac{dr}{dt} = 144\pi\frac{dr}{dt}$, so $\frac{dr}{dt} = \frac{1}{2}$ cm/s. The $\pi$ cancels, which is why the rate was given as $72\pi$: no calculator is needed, and none is allowed.

b) Same differentiated relation, new instant: $72\pi = 4\pi(144)\frac{dr}{dt} = 576\pi\frac{dr}{dt}$, so $\frac{dr}{dt} = \frac{72}{576} = \frac{1}{8}$ cm/s. The radius grows four times more slowly although the air arrives at the same rate. The factor $4\pi r^2$ in the derivative is the area of the skin: the $72\pi$ cm³ that arrive in one second form a thin shell over the whole surface, and a surface four times larger (radius doubled, area multiplied by $2^2$) gets a shell four times thinner. That is what the two circles of the figure show. It also says why nothing could have been substituted before differentiating: the answer depends on $r$.

c) $S = 4\pi r^2$ gives, chain rule again, $\frac{dS}{dt} = 8\pi r\frac{dr}{dt}$. At $r = 6$ with $\frac{dr}{dt} = \frac{1}{2}$ from a): $\frac{dS}{dt} = 8\pi(6)\left(\frac{1}{2}\right) = 24\pi$ cm²/s. For the check, eliminate $\frac{dr}{dt}$ between the two differentiated relations: $\frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}$, so $\frac{dS}{dt} = 8\pi r \cdot \frac{1}{4\pi r^2}\frac{dV}{dt} = \frac{2}{r}\frac{dV}{dt}$. At $r = 6$: $\frac{2}{6}(72\pi) = 24\pi$. Two routes, one answer. At $r = 12$ the same formula gives $12\pi$ cm²/s: the skin also grows more slowly.

d) The diameter is $D = 2r$ at every instant, so $\frac{dD}{dt} = 2\frac{dr}{dt}$. The instant is given by the DIAMETER: $D = 24$ means $r = 12$, and by b) $\frac{dr}{dt} = \frac{1}{8}$, so $\frac{dD}{dt} = \frac{1}{4}$ cm/s. The trap is to put $r = 24$ into the formula of a), which gives $\frac{72\pi}{4\pi(576)} = \frac{1}{32}$ cm/s, eight times too small. Before substituting, translate the instant into the variable of your relation.

e) For the tank, $V = \pi(2)^2 h = 4\pi h$ at every instant, where $h$ is the depth in metres. Differentiating: $\frac{dV}{dt} = 4\pi\frac{dh}{dt}$, so $\frac{dh}{dt} = \frac{3}{4\pi}$ m/min, about $0.24$ m/min with $\pi \approx 3.14$, the exact answer being $\frac{3}{4\pi}$. The depth never appears: the answer is the same at $h = 1$ and at $h = 4$. The reason is geometric: the free surface of the water has the same area $4\pi$ m² at every depth, so each cubic metre raises the level by the same amount. In the balloon, the surface over which the new air spreads grows with $r$. In general, the level rises at the rate of the inflow divided by the area of the free surface, and it is constant exactly when that area is.

a) Cut the cone by a vertical plane through its axis. The water forms a small triangle with height $h$ and half-width $r$, the whole tank a large one with height $6$ and half-width $2$; they share the angle at the vertex and both have a right angle on the axis, so they are similar: $\frac{r}{h} = \frac{2}{6}$, that is $r = \frac{h}{3}$ at every instant. Substitute into the volume: $V = \frac{1}{3}\pi\left(\frac{h}{3}\right)^2 h = \frac{\pi h^3}{27}$. This substitution is allowed BEFORE differentiating, and it is the only one: $r = \frac{h}{3}$ is a relation valid at every instant, not the value of one instant.

b) Differentiate $V = \frac{\pi h^3}{27}$ with respect to $t$, chain rule with inner function $h(t)$: $\frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2\frac{dh}{dt} = \frac{\pi h^2}{9}\frac{dh}{dt}$. Now freeze the instant, $h = 3$ and $\frac{dV}{dt} = \frac{\pi}{2}$: $\frac{\pi}{2} = \frac{9\pi}{9}\frac{dh}{dt} = \pi\frac{dh}{dt}$, so $\frac{dh}{dt} = \frac{1}{2}$ m/min, positive since the tank fills. At $h = 6$, the top, the same computation gives $\frac{\pi}{2} = 4\pi\frac{dh}{dt}$, so $\frac{1}{8}$ m/min: the level slows down as the cone widens, although the pump never changes.

c) Differentiate $r = \frac{h}{3}$: $\frac{dr}{dt} = \frac{1}{3}\frac{dh}{dt} = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$ m/min. A linear relation between two quantities gives the same linear relation between their rates, with the same constant factor, at every instant.

d) $A = \pi r^2$, so by the chain rule $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$. At the instant, $r = \frac{3}{3} = 1$, so $\frac{dA}{dt} = 2\pi(1)\left(\frac{1}{6}\right) = \frac{\pi}{3}$ m²/min. For the check: $A = \pi(1)^2 = \pi$ at the instant, and $A\frac{dh}{dt} = \pi \cdot \frac{1}{2} = \frac{\pi}{2} = \frac{dV}{dt}$. This is no coincidence: with $r^2 = \frac{h^2}{9}$, the derivative of b) reads $\frac{dV}{dt} = \pi r^2\frac{dh}{dt} = A\frac{dh}{dt}$. A thin new layer of water has the area of the free surface and the thickness $dh$, exactly as for the cylinder of Exercise 2.

e) With $r$ frozen at $1$, the student's volume is $V = \frac{\pi}{3}h$, whose derivative is $\frac{dV}{dt} = \frac{\pi}{3}\frac{dh}{dt}$; with $\frac{dV}{dt} = \frac{\pi}{2}$ he finds $\frac{dh}{dt} = \frac{3}{2}$. But $r$ is NOT constant: it grows with $h$, and freezing it removes its contribution to the derivative. The correct derivative of d) is $\pi r^2\frac{dh}{dt}$, while the student's is $\frac{1}{3}\pi r^2\frac{dh}{dt}$: the factor $\frac{1}{3}$ of the cone formula survives in his computation, whereas in the true one the two thirds contributed by the growing radius restore it to $1$. Hence his answer is exactly $3$ times too large. The rule: eliminate the extra variable by a relation that holds at every instant (here the similar triangles), never by its value at the instant, and only then differentiate. This error costs the whole question on an exam, since every later line inherits it.

a) The large triangle has the post as vertical side ($4.5$) and the ground from the post to the tip as horizontal side ($x + s$). The small one has the person as vertical side ($1.8$) and the shadow as horizontal side ($s$). Both are right-angled on the ground and share the angle $\theta$ at the tip, so they are similar and their sides are proportional: $\frac{s}{1.8} = \frac{x + s}{4.5}$. Cross-multiplying, $4.5s = 1.8x + 1.8s$, so $2.7s = 1.8x$ and $s = \frac{1.8}{2.7}x = \frac{2}{3}x$. The most common error is to write $\frac{s}{1.8} = \frac{x}{4.5}$, matching the shadow with the distance to the post instead of with the WHOLE base $x + s$ of the large triangle; the figure prevents it only if both triangles are traced from the same tip.

b) Differentiate $s = \frac{2}{3}x$ with respect to $t$: $\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt} = \frac{2}{3}(1.2) = 0.8$ m/s. The relation is linear, so the rate of the shadow is a fixed multiple of the rate of the walker: $0.8$ m/s at $1$ m from the post as at $10$ m. Here no instant is needed at all, and a question that gives one (the person is $5$ m from the post) is testing whether you notice that the datum is useless.

c) The tip is at the distance $x + s$ from the post. Its rate is $\frac{dx}{dt} + \frac{ds}{dt} = 1.2 + 0.8 = 2$ m/s, or directly $\frac{d}{dt}\left(\frac{5}{3}x\right) = \frac{5}{3}(1.2) = 2$ m/s. The tip runs ahead of the walker. Answering $0.8$ m/s gives the speed at which the shadow LENGTHENS, not the speed at which its tip MOVES; the two questions differ by the speed of the person, and an exam asks both precisely to see who separates them.

d) Walking toward the post, $x$ decreases: $\frac{dx}{dt} = -1.2$ m/s. The same relations give $\frac{ds}{dt} = \frac{2}{3}(-1.2) = -0.8$ m/s (the shadow shortens) and the tip moves at $-2$ m/s (toward the post). The signs are not decoration: a rate is the derivative of a named quantity, and when the quantity decreases the derivative is negative. Answering $1.2$, $0.8$ and $2$ here, the same numbers as before, describes a person walking away.

e) In the large triangle, $\tan\theta = \frac{4.5}{x + s} = \frac{4.5}{\frac{5}{3}x} = \frac{2.7}{x}$ at every instant. Differentiate, chain rule on the left with inner function $\theta(t)$, power rule on the right: $\sec^2\theta\,\frac{d\theta}{dt} = -\frac{2.7}{x^2}\frac{dx}{dt}$. At $x = 2.7$: $\tan\theta = 1$, so $\theta = \frac{\pi}{4}$ and $\sec^2\theta = 1 + \tan^2\theta = 2$. Then $2\frac{d\theta}{dt} = -\frac{2.7}{7.29}(1.2) = -\frac{3.24}{7.29} = -\frac{4}{9}$, and $\frac{d\theta}{dt} = -\frac{2}{9}$ rad/s. The angle decreases, as it should: the farther the tip, the lower the lamp appears. Note that $\sec^2\theta$ was obtained from $\tan\theta$ by the identity, without ever computing $\theta$ in degrees.

a) The camera, the pad and the rocket form a right triangle with the right angle at the pad; the side adjacent to $\theta$ is the fixed $4$ km and the opposite side is $h$. So $\tan\theta = \frac{h}{4}$ at every instant. Differentiate with respect to $t$: on the left, chain rule with inner function $\theta(t)$, $\frac{d}{dt}\tan\theta = \sec^2\theta\,\frac{d\theta}{dt}$; on the right, $\frac{1}{4}\frac{dh}{dt}$. Hence $\sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{4}\frac{dh}{dt}$. The fixed distance is the only number in the relation: it is the one quantity that does not change.

b) At $h = 4$: $\tan\theta = 1$, so $\theta = \frac{\pi}{4}$ and $\sec^2\theta = 1 + \tan^2\theta = 2$. Then $2\frac{d\theta}{dt} = \frac{0.8}{4} = 0.2$, and $\frac{d\theta}{dt} = \frac{1}{10}$ rad/s. The formula $\frac{d}{d\theta}\tan\theta = \sec^2\theta$ holds only for $\theta$ in radians, so the derivative comes out in rad/s. To convert, multiply by $\frac{180}{\pi}$: $\frac{1}{10} \cdot \frac{180}{\pi} = \frac{18}{\pi}$ degrees per second, about $5.7$ with $\pi \approx 3.14$. A student who writes $\theta = 45$ and differentiates in degrees mixes two units in the same line and gets a number off by the factor $\frac{180}{\pi}$.

c) Pythagoras: $z^2 = 16 + h^2$ at every instant, so $2z\frac{dz}{dt} = 2h\frac{dh}{dt}$ (chain rule on both sides), that is $\frac{dz}{dt} = \frac{h}{z}\frac{dh}{dt}$. At $h = 4$: $z = \sqrt{32} = 4\sqrt 2$, and $\frac{dz}{dt} = \frac{4}{4\sqrt 2}(0.8) = \frac{0.8}{\sqrt 2} = \frac{2\sqrt 2}{5}$ km/s, about $0.57$ km/s. The distance grows more slowly than the altitude, because part of the rocket's motion is across the line of sight: $\frac{h}{z} = \sin\theta < 1$.

d) At $h = 4\sqrt 3$: $\tan\theta = \sqrt 3$, so $\theta = \frac{\pi}{3}$ and $\sec^2\theta = 1 + 3 = 4$. Then $4\frac{d\theta}{dt} = \frac{1.2}{4} = 0.3$, and $\frac{d\theta}{dt} = \frac{3}{40}$ rad/s. The rocket climbs $1.5$ times faster than in b), yet the camera turns more slowly, $0.075$ rad/s against $0.1$. The factor $\sec^2\theta$ explains it: high in the sky, a kilometre of altitude corresponds to a smaller angle. The rate of an angle depends on the instant through $\theta$ itself, which is one more reason never to substitute before differentiating.

e) $\theta = \arctan\frac{h}{4}$ for $h \ge 0$, and the derivative of $\arctan u$ is $\frac{1}{1 + u^2}$. Chain rule with inner function $u = \frac{h}{4}$: $\frac{d\theta}{dt} = \frac{1}{1 + \frac{h^2}{16}} \cdot \frac{1}{4}\frac{dh}{dt} = \frac{4}{16 + h^2}\frac{dh}{dt}$. At $h = 4$, $\frac{dh}{dt} = 0.8$: $\frac{4(0.8)}{32} = \frac{1}{10}$. At $h = 4\sqrt 3$, $\frac{dh}{dt} = 1.2$: $\frac{4(1.2)}{16 + 48} = \frac{4.8}{64} = \frac{3}{40}$. Same values, as they must be: $\frac{4}{16 + h^2} = \frac{\cos^2\theta}{4}$, since $\cos\theta = \frac{4}{z}$. Two methods that agree are the best check available without a calculator.

a) A drives toward $O$, so its distance to $O$ decreases: $\frac{dy}{dt} = -60$ km/h. B drives away, so $\frac{dx}{dt} = 45$ km/h. The speed is a positive number; the rate of a DISTANCE carries the sign of its variation, and this is where half of the marks of the problem are won or lost. The roads are perpendicular, so by Pythagoras $x^2 + y^2 = z^2$ at every instant. Differentiating with respect to $t$, chain rule on each square: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$, that is $\frac{dz}{dt} = \frac{1}{z}\left(x\frac{dx}{dt} + y\frac{dy}{dt}\right)$.

b) At the instant: $x = 8$, $y = 6$, so $z = \sqrt{64 + 36} = 10$. Then $\frac{dz}{dt} = \frac{1}{10}\left(8 \times 45 + 6 \times (-60)\right) = \frac{360 - 360}{10} = 0$. Both cars move at highway speed, yet the distance between them is not changing at that instant: B's move away lengthens the segment exactly as much as A's approach shortens it. A zero rate means stationary AT THAT INSTANT, not constant: just before and just after, the distance does change, as c) shows. Adding or subtracting the two speeds ($105$ or $15$ km/h) makes no sense here, since the cars do not move along the segment AB.

c) Six minutes is $\frac{1}{10}$ h. Going back in time, A was $6$ km farther from $O$ and B was $4.5$ km closer: $y = 6 + 60 \times \frac{1}{10} = 12$ and $x = 8 - 45 \times \frac{1}{10} = 3.5$. Then $z = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5$, and $\frac{dz}{dt} = \frac{1}{12.5}\left(3.5 \times 45 + 12 \times (-60)\right) = \frac{157.5 - 720}{12.5} = -\frac{562.5}{12.5} = -45$ km/h. The cars were getting closer at $45$ km/h. Between c) and b) the rate went from $-45$ to $0$: the relation is the same, the instant changes everything, which is why the numbers of an instant are substituted only after the differentiation.

d) With $\frac{dy}{dt} = +60$ the student computes $\frac{1}{10}(360 + 360) = 72$ km/h. His equation is right and his arithmetic too; his car A is wrong. A positive $\frac{dy}{dt}$ means that $y$ increases, a car driving north, AWAY from $O$: in that situation both cars would indeed separate at $72$ km/h. The sign of each rate is read from the figure and the direction of motion before any number is written, and a marker who sees $72$ gives nothing for a problem whose answer is $0$.

e) In the right triangle $OBA$, the angle at B satisfies $\tan\theta = \frac{y}{x}$ at every instant. Differentiate: on the left, chain rule, $\sec^2\theta\,\frac{d\theta}{dt}$; on the right, quotient rule, $\frac{x\frac{dy}{dt} - y\frac{dx}{dt}}{x^2}$. At the instant, $\sec^2\theta = 1 + \frac{36}{64} = \frac{100}{64}$, and the right side is $\frac{8(-60) - 6(45)}{64} = \frac{-750}{64}$. So $\frac{100}{64}\frac{d\theta}{dt} = -\frac{750}{64}$ and $\frac{d\theta}{dt} = -\frac{15}{2}$ rad/h. Divide by $60$ for minutes: $-\frac{1}{8}$ rad/min. Although the distance AB is stationary, the segment AB is turning: the line of sight from B sinks toward the road, which is exactly what a stationary length with moving ends must do.