Corrected exercises: related rates (MATH 203)

a) The top slides down, so $y$ decreases and $\frac{dy}{dt} = -0.25$ m/s: the minus sign belongs to the datum, before any computation. The ladder is the hypotenuse of a right triangle, so by Pythagoras $x^2 + y^2 = 2.6^2 = 6.76$ at every instant. Differentiate with respect to $t$, chain rule on each square with inner functions $x(t)$ and $y(t)$: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$. The unknown rate $\frac{dx}{dt}$ appears to the first power: isolate it exactly as the $x$ of a linear equation, $2x\frac{dx}{dt} = -2y\frac{dy}{dt}$, then divide by $2x$: $\frac{dx}{dt} = -\frac{y}{x}\frac{dy}{dt}$.

b) Now freeze the instant. The height $y = 2.4$ is given; the COMPANION value $x$ comes from the relation: $x^2 = 6.76 - 5.76 = 1$, so $x = 1$ m. This is the algebra of the instant, and it is where MATH 203 marks go: $\sqrt{6.76 - 5.76}$ is $\sqrt 1 = 1$, NOT $2.6 - 2.4 = 0.2$, because the square root of a difference is not the difference of the square roots. With $x = 0.2$ the student finds $3$ m/s, five times too fast. Correctly: $\frac{dx}{dt} = -\frac{2.4}{1}(-0.25) = 0.6$ m/s. Two minus signs make a plus: the foot moves AWAY from the wall, as the figure says it must.

c) The number $2.4$ is the same, but it now names $x$, not $y$. Translate the instant into the variables of the relation: $x = 2.4$ gives $y^2 = 6.76 - 5.76 = 1$, so $y = 1$. Then $\frac{dx}{dt} = -\frac{1}{2.4}(-0.25) = \frac{0.25}{2.4} = \frac{5}{48} \approx 0.1042$ m/s. The two configurations are mirror images, top high and foot close in b), top low and foot far in c), and the factor $\frac{y}{x}$ is $2.4$ in one and $\frac{1}{2.4}$ in the other: the answers differ by a factor $2.4^2 = 5.76$. Reading which variable the instant names is part of the question.

d) The speeds are $\left|\frac{dx}{dt}\right| = \frac{y}{x}\left|\frac{dy}{dt}\right|$ and $\left|\frac{dy}{dt}\right|$, equal exactly when $\frac{y}{x} = 1$, that is $y = x$. Substitute into the relation, which holds at that instant too: $y^2 + y^2 = 6.76$, so $2y^2 = 6.76$, $y^2 = 3.38$ and $y = \sqrt{3.38} = \frac{2.6}{\sqrt 2} = 1.3\sqrt 2 \approx 1.84$ m (a height is positive). The algebra step to watch is $2y^2 = 6.76$: divide by $2$ BEFORE taking the root; $\sqrt{6.76} = 2.6$ followed by a division by $2$ would give $1.3$, a different number.

e) Choose the ratio that contains the fixed side and the given rate: $\sin\theta = \frac{y}{2.6}$ at every instant. Differentiate, chain rule with inner function $\theta(t)$: $\cos\theta\,\frac{d\theta}{dt} = \frac{1}{2.6}\frac{dy}{dt}$. At the instant, $\cos\theta = \frac{x}{2.6} = \frac{1}{2.6}$, so $\frac{1}{2.6}\frac{d\theta}{dt} = \frac{-0.25}{2.6}$ and $\frac{d\theta}{dt} = -0.25$ rad/s. The derivative of $\sin$ is $\cos$ only in radians, so the answer is in rad/s; to convert, multiply by $\frac{180}{\pi}$: $-0.25 \times \frac{180}{\pi} \approx -14.3$ degrees per second. The angle shrinks, as it must for a ladder that slides toward the ground.

a) The area of the slick is $A = \pi r^2$ at every instant. Differentiate with respect to $t$, chain rule with inner function $r(t)$: $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$. At $r = 15$ with $\frac{dr}{dt} = 0.4$: $\frac{dA}{dt} = 2\pi(15)(0.4) = 12\pi \approx 37.70$ m²/s. The rate grows with $r$: the same $0.4$ m of radius per second adds a ring along a longer and longer edge.

b) The instant is now given by the AREA, and the relation needs the radius. Translate: $\pi r^2 = 900$, so $r^2 = \frac{900}{\pi}$ and $r = \frac{30}{\sqrt\pi} \approx 16.93$ m. Then $\frac{dA}{dt} = 2\pi\left(\frac{30}{\sqrt\pi}\right)(0.4) = 24\sqrt\pi \approx 42.54$ m²/s, since $\frac{\pi}{\sqrt\pi} = \sqrt\pi$. Two algebra slips cost this part: putting $r = 900$, the area in the place of the radius, and solving $\pi r^2 = 900$ as $r = \sqrt{900} = 30$, forgetting to divide by $\pi$ first, which gives $24\pi \approx 75.40$.

c) The question speaks of the diameter, so write the formula in $D$ BEFORE differentiating: $r = \frac{D}{2}$ gives $S = 4\pi\left(\frac{D}{2}\right)^2 = 4\pi\frac{D^2}{4} = \pi D^2$. The square applies to the whole fraction: $\left(\frac{D}{2}\right)^2 = \frac{D^2}{4}$, not $\frac{D^2}{2}$. Differentiate: $\frac{dS}{dt} = 2\pi D\frac{dD}{dt}$. The surface decreases, so $\frac{dS}{dt} = -3$; at $D = 12$: $-3 = 24\pi\frac{dD}{dt}$ and $\frac{dD}{dt} = -\frac{3}{24\pi} = -\frac{1}{8\pi} \approx -0.0398$ cm/min. The diameter shrinks by about $0.4$ mm per minute.

d) In the diameter again: $V = \frac{4}{3}\pi\left(\frac{D}{2}\right)^3 = \frac{4}{3}\pi\frac{D^3}{8} = \frac{\pi D^3}{6}$. Differentiate: $\frac{dV}{dt} = \frac{\pi D^2}{2}\frac{dD}{dt}$. At $D = 12$: $\frac{dV}{dt} = \frac{144\pi}{2}\left(-\frac{1}{8\pi}\right) = -\frac{72}{8} = -9$ cm³/min. Keeping the exact $-\frac{1}{8\pi}$ lets the $\pi$ cancel; with the rounded $-0.0398$ you would find $-9.00$ only by luck of rounding.

e) From c) and d), $\frac{dD}{dt} = \frac{1}{2\pi D}\frac{dS}{dt}$, so $\frac{dV}{dt} = \frac{\pi D^2}{2} \cdot \frac{1}{2\pi D}\frac{dS}{dt} = \frac{D}{4}\frac{dS}{dt}$, after simplifying $\frac{\pi D^2}{2\pi D} = \frac{D}{2}$ and dividing by the remaining $2$. Check: $\frac{12}{4}(-3) = -9$ ✓. With $\frac{dS}{dt} = -3$ fixed, the volume is lost at $\frac{3D}{4}$ cm³/min, a quantity that decreases with $D$: the snowball loses volume more and more SLOWLY, although its surface keeps shrinking at the same rate.

a) Cut the cone by a vertical plane through its axis. The water forms a triangle of height $h$ and half-width $r$; the whole tank one of height $4$ and half-width $3$, the RADIUS, half of the $6$ m diameter. The two share the angle at the vertex and have a right angle on the axis, so they are similar: $\frac{r}{h} = \frac{3}{4}$ and $r = \frac{3h}{4}$ at every instant. Substitute into the volume, squaring the WHOLE fraction: $r^2 = \left(\frac{3h}{4}\right)^2 = \frac{9h^2}{16}$, so $V = \frac{1}{3}\pi\cdot\frac{9h^2}{16}\cdot h = \frac{3\pi h^3}{16}$. Check at the brim: $\frac{3\pi(64)}{16} = 12\pi$, and $\frac{1}{3}\pi(3^2)(4) = 12\pi$ ✓.

b) Differentiate $V = \frac{3\pi h^3}{16}$, chain rule with inner function $h(t)$: $\frac{dV}{dt} = \frac{9\pi h^2}{16}\frac{dh}{dt}$. The tank loses water, so $\frac{dV}{dt} = -0.3$. At $h = 2$: $-0.3 = \frac{9\pi(4)}{16}\frac{dh}{dt} = \frac{9\pi}{4}\frac{dh}{dt}$, so $\frac{dh}{dt} = -\frac{1.2}{9\pi} = -\frac{2}{15\pi} \approx -0.0424$ m/min. Multiply by $100$ for centimetres: the level falls at about $4.24$ cm/min. Isolate the rate first, round last: $\frac{dh}{dt} = \frac{16}{9\pi h^2}\frac{dV}{dt}$ is one exact expression, and the calculator comes in only at the end.

c) Differentiate $r = \frac{3h}{4}$: $\frac{dr}{dt} = \frac{3}{4}\frac{dh}{dt} = \frac{3}{4}\left(-\frac{2}{15\pi}\right) = -\frac{1}{10\pi} \approx -0.0318$ m/min. A linear relation between two quantities carries over to their rates with the same factor.

d) Student A squared only the numerator: $V = \frac{1}{3}\pi\cdot\frac{3h^2}{16}\cdot h = \frac{\pi h^3}{16}$, three times too small, so $\frac{dV}{dt} = \frac{3\pi h^2}{16}\frac{dh}{dt}$ and at $h = 2$, $\frac{dh}{dt} = -\frac{0.3 \cdot 16}{12\pi} = -\frac{0.4}{\pi} \approx -0.1273$ m/min: THREE times too fast. Student B used the diameter: $V = \frac{1}{3}\pi(1.5h)^2h = 0.75\pi h^3$, four times too large, so $\frac{dh}{dt} = -\frac{0.3}{2.25\pi(4)} = -\frac{0.3}{9\pi} \approx -0.0106$ m/min: FOUR times too slow. Both errors are in the algebra of a), not in the calculus, and both cost the whole question, since every later line inherits the wrong volume.

e) Now $\frac{dV}{dt}$ is the NET rate, pump minus valve: $\frac{dV}{dt} = p - 0.3$, where $p$ is the pump rate. The level rate must be in metres per minute, like the lengths of the formula: $2$ cm/min $= 0.02$ m/min. At $h = 3$: $\frac{dV}{dt} = \frac{9\pi(9)}{16}(0.02) = \frac{81\pi}{16}(0.02) = 0.10125\pi \approx 0.318$ m³/min. So $p - 0.3 = 0.318$ and $p \approx 0.618$ m³/min. Keeping $2$ in place of $0.02$ gives a pump of $32.1$ m³/min, a hundred times too strong: the units of the formula decide the units of the rates.

a) The ray from the spotlight grazes the top of the actor's head and hits the wall at the top of the shadow. The large right triangle has legs $u$ (along the floor to the wall) and $H$ (up the wall); the small one has legs $u - 2$ (along the floor to the actor) and $1.8$. They share the angle at the spotlight, so $\frac{H}{u} = \frac{1.8}{u - 2}$ and $H = \frac{1.8u}{u - 2}$ at every instant. Now the algebra that makes the derivative easy: write $u = (u - 2) + 2$ in the numerator, $H = \frac{1.8(u - 2) + 3.6}{u - 2} = 1.8 + \frac{3.6}{u - 2}$. The variable now appears only once, in a denominator. At $u = 8$: $H = 1.8 + \frac{3.6}{6} = 2.4$ m.

b) The spotlight moves toward the wall, so $u$ decreases: $\frac{du}{dt} = -0.5$ m/s. Rewrite $\frac{3.6}{u - 2} = 3.6(u - 2)^{-1}$ and differentiate, chain rule with inner function $u(t)$: $\frac{dH}{dt} = -3.6(u - 2)^{-2}\frac{du}{dt} = -\frac{3.6}{(u - 2)^2}\frac{du}{dt}$. Translate the instant: $6$ m from the ACTOR means $u - 2 = 6$, so $u = 8$. Then $\frac{dH}{dt} = -\frac{3.6}{36}(-0.5) = 0.05$ m/s, positive: the shadow GROWS as the light comes closer to the actor. Taking $u = 6$ instead, the distance to the actor put in the place of the distance to the wall, gives $0.1125$ m/s.

c) One metre from the actor, $u - 2 = 1$: $H = 1.8 + \frac{3.6}{1} = 5.4$ m and $\frac{dH}{dt} = -\frac{3.6}{1}(-0.5) = 1.8$ m/s, thirty-six times faster than in b) although the spotlight rolls at the same speed. The relation is not linear, so the rate depends on the instant through $(u - 2)^2$.

d) Set the rate equal to $0.2$: $\frac{1.8}{(u - 2)^2} = 0.2$, so $(u - 2)^2 = \frac{1.8}{0.2} = 9$. Taking the square root of both sides gives $u - 2 = 3$ or $u - 2 = -3$; the second gives $u = -1$, which is not a distance, and is rejected. The spotlight is $3$ m from the actor, $5$ m from the wall.

e) Quotient rule on $H = \frac{1.8u}{u - 2}$, both $u$'s being functions of $t$: $\frac{dH}{dt} = \frac{1.8\frac{du}{dt}(u - 2) - 1.8u\frac{du}{dt}}{(u - 2)^2} = \frac{1.8(u - 2) - 1.8u}{(u - 2)^2}\frac{du}{dt} = \frac{-3.6}{(u - 2)^2}\frac{du}{dt}$. Simplify the numerator BEFORE substituting: $1.8u - 3.6 - 1.8u = -3.6$. At $u = 8$ it gives $0.05$ m/s, as in b). The order of the quotient rule is (derivative of the top) times (bottom) MINUS (top) times (derivative of the bottom); the reversed numerator $1.8u - 1.8(u - 2) = +3.6$ flips the sign, gives $-0.05$ m/s, and the student concludes that the shadow shrinks, which the figure contradicts.

a) The station, the point below the plane and the plane form a right triangle with the fixed vertical side $3$ km. By Pythagoras, $x^2 + 3^2 = D^2$ at every instant. Differentiate with respect to $t$, chain rule on both squares, the constant $9$ giving $0$: $2x\frac{dx}{dt} = 2D\frac{dD}{dt}$. The unknown is now $\frac{dx}{dt}$: divide by $2x$, $\frac{dx}{dt} = \frac{D}{x}\frac{dD}{dt}$. Same relation as for a forward problem, isolated for the other rate. Read the other way, $\frac{dD}{dt} = \frac{x}{D}\frac{dx}{dt}$, and $x < D$ because the hypotenuse is the longest side: the radar reading is always SMALLER than the speed of the plane, and equal to $0$ overhead.

b) Companion value from the relation: $x^2 = 25 - 9 = 16$, so $x = 4$ km (the plane is on the approach side; distances are positive). The radar distance decreases, so $\frac{dD}{dt} = -480$. Then $\frac{dx}{dt} = \frac{5}{4}(-480) = -600$ km/h. The sign says $x$ decreases, the plane is coming toward the station, and its speed, a positive number, is $600$ km/h. The radar reads LESS than the true speed because only part of the motion is along the line of sight: $\frac{dD}{dt} = \frac{x}{D}\frac{dx}{dt} = \frac{4}{5}\frac{dx}{dt}$.

c) Flying away, $\frac{dx}{dt} = +600$ km/h. At $D = 6$: $x^2 = 36 - 9 = 27$, so $x = \sqrt{27} = 3\sqrt 3 \approx 5.196$ km. Then $\frac{dD}{dt} = \frac{x}{D}\frac{dx}{dt} = \frac{3\sqrt 3}{6}(600) = 300\sqrt 3 \approx 519.6$ km/h. Keep $\sqrt{27}$ exact until the last line: $\frac{3\sqrt 3}{6} = \frac{\sqrt 3}{2}$ simplifies by hand, and the calculator is needed only once.

d) Directly overhead, $x = 0$ and $D = 3$, so $\frac{dD}{dt} = \frac{x}{D}\frac{dx}{dt} = 0$. No contradiction: at that instant the plane moves perpendicular to the line of sight, and the distance, which was decreasing and is about to increase, is at its minimum. A rate of zero means the distance is stationary at that instant, not that the plane has stopped.

e) $\sin\theta = \frac{3}{D} = 3D^{-1}$ at every instant. Differentiate, chain rule on both sides: $\cos\theta\,\frac{d\theta}{dt} = -\frac{3}{D^2}\frac{dD}{dt}$. At the instant of b), $\cos\theta = \frac{x}{D} = \frac{4}{5}$ and $\frac{dD}{dt} = -480$: $\frac{4}{5}\frac{d\theta}{dt} = -\frac{3}{25}(-480) = 57.6$, so $\frac{d\theta}{dt} = 72$ rad/h. The rate is in radians PER HOUR, because $t$ is in hours: divide by $60$ for $1.2$ rad/min, by $60$ again for $0.02$ rad/s, then multiply by $\frac{180}{\pi}$: about $1.15$ degrees per second. The antenna tilts upward, $\theta$ increases, as the plane approaches.

a) Divide by $3.6$ to go from km/h to m/s: $72$ km/h $= 20$ m/s and $108$ km/h $= 30$ m/s. Mixing km/h with lengths in metres is the first error the problem invites. For the relation, drop the car vertically onto the ground: its foot is $x$ from $C$ along the direction of the road, the train is $y$ from $C$ along the track, and these two directions are perpendicular, so the horizontal distance between the foot and the train is $\sqrt{x^2 + y^2}$. The car is $12$ m above its foot, which gives a second right triangle, vertical this time: $D^2 = \left(\sqrt{x^2 + y^2}\right)^2 + 12^2 = x^2 + y^2 + 144$. Differentiate, chain rule on each square, the constant giving $0$: $2D\frac{dD}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$, that is $D\frac{dD}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}$.

b) Companion value first: $D = \sqrt{81 + 64 + 144} = \sqrt{289} = 17$ m. Both move away, so $\frac{dx}{dt} = 20$ and $\frac{dy}{dt} = 30$: $17\frac{dD}{dt} = 9(20) + 8(30) = 420$, and $\frac{dD}{dt} = \frac{420}{17} \approx 24.71$ m/s. This is far less than $20 + 30 = 50$ m/s: speeds add only along the segment that joins the two vehicles, and neither moves along it.