Exercises — Related rates — setting up and solving

Draw: a sphere with radius arrow $r$ pointing outward, and a small "air-in" arrow labelling $\frac{dV}{dt}$. Link: $V=\frac{4}{3}\pi r^3$. Differentiate in $t$ (chain rule on $r^3$, since $r=r(t)$): $\frac{dV}{dt}=\frac{4}{3}\pi \cdot 3r^2\cdot \frac{dr}{dt}=4\pi r^2\frac{dr}{dt}.$ That is the relation. Nothing has been substituted — exactly right for a Level‑1 "set up" task.

Draw: a right triangle with legs labelled $x$, $y$ (letters, both changing) and hypotenuse the fixed $13$. The right side is a constant, so its rate is $0$: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.$ Divide by $2$: $x\frac{dx}{dt}+y\frac{dy}{dt}=0.$ This is Implicit differentiation applied with $t$ as the hidden variable.

Draw: a circle with an inward radius arrow (shrinking) labelled $\frac{dr}{dt}<0$. $r>0$ and $2\pi >0$, so $2\pi r>0$. Multiplying a positive number by a negative $\frac{dr}{dt}$ gives $\frac{dA}{dt}<0$: area shrinks. Sign in, sign out — the rate equation carries physical direction.

Draw: a circle with an outward radius arrow labelled $\frac{dr}{dt}=0.5$; the area $A$ is the shaded interior. Link: $A=\pi r^2$. Differentiate: $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$. Substitute (now safe): $r=4,\frac{dr}{dt}=0.5$: $\frac{dA}{dt}=2\pi (4)(0.5)=4\pi \approx 12.57{\text{m}}^2/\text{s}.$

Draw: a cube with one edge labelled $s$ (a letter) and a tiny outward arrow for $\frac{ds}{dt}$. Link: $V=s^3$. Differentiate: $\frac{dV}{dt}=3s^2\frac{ds}{dt}$. Substitute: $3(5{)}^2(0.02)=3(25)(0.02)=1.5{\text{cm}}^3/\text{s}.$

Draw: the right triangle above — base $x$, height $y$ (both letters), hypotenuse the fixed $10$, base-arrow $\frac{dx}{dt}=2$ pointing away from the wall. Link (Pythagorean theorem): $x^2+y^2=100$. Find $y$ now: $y=\sqrt{100-36}=8$ m. Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$. Substitute: $\frac{dy}{dt}=-\frac{6}{8}(2)=-1.5\text{m/s}.$ Negative $\Rightarrow$ top moves down, as expected.

Draw: a shrinking sphere with an inward radius arrow labelled $\frac{dr}{dt}$ and an "air/volume out" note for $\frac{dV}{dt}<0$. Link: $V=\frac{4}{3}\pi r^3$. Differentiate: $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. Solve for $\frac{dr}{dt}$: $\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}$. Substitute: $\frac{dr}{dt}=\frac{-8}{4\pi (9)}=\frac{-8}{36\pi }=-\frac{2}{9\pi }\approx -0.0707\text{cm/min}.$

Draw: the cone cross-section above — fixed $R=3$, $H=6$ at the top; the water triangle with depth $h$ and surface radius $r$ (both letters). Link: $V=\frac{1}{3}\pi r^{2}h$, but $r$ also changes. Use Similar triangles: $\frac{r}{h}=\frac{R}{H}=\frac{3}{6}=\frac{1}{2}\Rightarrow r=\frac{h}{2}$. Substitute $r$ FIRST: $V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h=\frac{\pi }{12}{h}^3.$ Differentiate: $\frac{dV}{dt}=\frac{\pi }{12}\cdot 3h^2\frac{dh}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}.$ Substitute $h=2$: $4=\frac{\pi (4)}{4}\frac{dh}{dt}=\pi \frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{4}{\pi }\approx 1.273\text{m/min}.$

Draw: the intersection at the origin, A's path north (distance $a$), B's path east (distance $b$), and the separation $D$ as the slanted hypotenuse joining the two cars. Let $a$ = A's distance north, $b$ = B's distance east, $D$ = separation. Link: $D^2=a^2+b^2$. Differentiate: $2D\frac{dD}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}\Rightarrow D\frac{dD}{dt}=a\frac{da}{dt}+b\frac{db}{dt}.$ At $t=2$ h: $a=120,b=160,D=\sqrt{120^2+160^2}=\sqrt{14400+25600}=\sqrt{40000}=200$ km. Substitute: $200\frac{dD}{dt}=120(60)+160(80)=7200+12800=20000\Rightarrow \frac{dD}{dt}=\frac{20000}{200}=100\text{km/h}.$

Draw: a $6$ m lamppost, the $2$ m person at distance $x$ from it, and the shadow of length $s$ stretching past the person to a tip; two nested right triangles share that tip. Similar triangles: big triangle (lamp, ground, tip) has height $6$ and base $x+s$; small triangle (head, ground, tip) has height $2$ and base $s$: $\frac{6}{x+s}=\frac{2}{s}\Rightarrow 6s=2(x+s)\Rightarrow 6s=2x+2s\Rightarrow 4s=2x\Rightarrow s=\frac{x}{2}.$ Tip position $T=x+s=x+\frac{x}{2}=\frac{3x}{2}$. Differentiate: $\frac{dT}{dt}=\frac{3}{2}\frac{dx}{dt}=\frac{3}{2}(1.5)=2.25\text{m/s}.$ Note: the tip's speed does not depend on $x$ — it moves at a constant $2.25$ m/s.

Draw: the same ladder triangle as Ex 2.3, now with its interior shaded — that shaded region is the area $A$ we track. We need both $\frac{dx}{dt}$ and $\frac{dy}{dt}$. From Ex 2.3: at $x=6$, $y=8$, $\frac{dx}{dt}=2$, $\frac{dy}{dt}=-1.5$. Link: $A=\frac{1}{2}xy$. Differentiate (product rule): $\frac{dA}{dt}=\frac{1}{2}\left(\frac{dx}{dt}y+x\frac{dy}{dt}\right).$ Substitute: $\frac{1}{2}((2)(8)+(6)(-1.5))=\frac{1}{2}(16-9)=\frac{1}{2}(7)=3.5{\text{m}}^2/\text{s}.$ Positive: the triangle is still growing at that instant even though the top is dropping.

Draw: the observer at ground level $500$ m from the pad (a fixed horizontal leg), the rocket at height $y$ (vertical leg, a letter), and the line of sight making angle $\theta$ at the observer. Link: $\tan \theta =\frac{y}{500}$ (opposite over adjacent of the observer's right triangle). Differentiate: ${\sec }^2\theta \frac{d\theta }{dt}=\frac{1}{500}\frac{dy}{dt}.$ (Left side uses $\frac{d}{d\theta }\tan \theta ={\sec }^2\theta$ then the chain rule $\times \frac{d\theta }{dt}$.) At $y=500$: $\tan \theta =\frac{500}{500}=1\Rightarrow \theta =45^{\circ }$, and by the identity ${\sec }^2\theta =1+{\tan }^2\theta =1+1=2$. Solve: $\frac{d\theta }{dt}=\frac{1}{500{\sec }^2\theta }\frac{dy}{dt}=\frac{1}{500(2)}(100)=\frac{100}{1000}=0.1\text{rad/s}.$

Draw: the cone cross-section with fixed $R=2$, $H=4$; water depth $h$ falling (downward arrow on the level line), surface radius $r$ shrinking with it. Similar triangles: $r=\frac{R}{H}h=\frac{2}{4}h=\frac{h}{2}$, so $V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h=\frac{\pi }{12}{h}^3.$ Differentiate: $\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}$. Substitute $h=3$: $\frac{dV}{dt}=\frac{\pi (9)}{4}(-0.1)=-\frac{0.9\pi }{4}=-0.225\pi \approx -0.7069{\text{m}}^3/\text{min}.$

Draw: the pulley $4$ m up (fixed vertical leg), the boat at horizontal distance $x$ (a letter), and the rope $L$ as the slanted hypotenuse from boat to pulley, with a reel-in arrow shrinking $L$. Link (Pythagorean theorem): $L^2=x^2+4^2=x^2+16.$ Reeling in means $\frac{dL}{dt}=-2$. Differentiate: $2L\frac{dL}{dt}=2x\frac{dx}{dt}\Rightarrow \frac{dx}{dt}=\frac{L}{x}\frac{dL}{dt}.$ At $x=3$: $L=\sqrt{9+16}=5$. So $\frac{dx}{dt}=\frac{5}{3}(-2)=-\frac{10}{3}\approx -3.33\text{m/s}.$ Limit $x\to 0$: $L\to 4$, and $\frac{dx}{dt}=\frac{L}{x}\frac{dL}{dt}\to \frac{4}{x}(-2)\to -\infty$. The boat's horizontal speed blows up as it reaches the dock — even though rope comes in at a steady $2$ m/s. Physically, near the dock almost all the rope-shortening converts into horizontal motion.

Draw: a sphere with a steady outward radius arrow $\frac{dr}{dt}=1$; imagine equal-time snapshots — the shells added get thicker-looking because area $\propto r^2$ grows. Link: $V=\frac{4}{3}\pi r^3\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi r^2$ (since $\frac{dr}{dt}=1$). So $\frac{dV}{dt}$ grows like $r^2$ — it increases. To find its rate, differentiate again in $t$ (this yields the second derivative $\frac{d^{2}V}{d{t}^2}$): $\frac{d^{2}V}{d{t}^2}=\frac{d}{dt}\left(\frac{dV}{dt}\right)=4\pi \cdot 2r\frac{dr}{dt}=8\pi r\frac{dr}{dt}.$ At $r=5$: $8\pi (5)(1)=40\pi \approx 125.66{\text{cm}}^3/{\text{s}}^2.$ (Units: cm³ per second, per second.)

Draw: a cone pile with height $h$ equal to base radius $r$ (mark $h=r$), and a "sand in" arrow labelling $\frac{dV}{dt}=6$. Link: with $r=h$, $V=\frac{1}{3}\pi h^{2}h=\frac{\pi }{3}{h}^3.$ Differentiate: $\frac{dV}{dt}=\pi h^2\frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{1}{\pi h^2}\frac{dV}{dt}=\frac{6}{\pi h^2}.$ (a) $h=2$: $\frac{dh}{dt}=\frac{6}{\pi (4)}=\frac{6}{4\pi }=\frac{3}{2\pi }\approx 0.477\text{m/min}.$ (b) Set $\frac{6}{\pi h^2}=0.5$: $h^2=\frac{6}{0.5\pi }=\frac{12}{\pi }\Rightarrow h=\sqrt{\frac{12}{\pi }}\approx 1.954\text{m}.$

Every problem above followed the same skeleton — can you recite it? The six steps in order ::: Draw, Link, Differentiate, Substitute, Solve, and State units. What "Draw" must contain ::: a labelled picture using variables (letters) for moving quantities and fixed numbers only for constants, plus rate-arrows. Where a product like $A=\frac{1}{2}xy$ needs care ::: use the product rule when two time-functions multiply. How a limiting case ($x\to 0$) reveals a blow-up ::: a variable in a denominator drives the rate to $\pm \infty$. When to eliminate a variable ::: before differentiating, using fixed geometry (similar triangles / Pythagoras).