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Applications of the Derivative · Topic 13 of 23

Related Rates

In a related-rates problem every quantity is a function of time: write an equation linking them that holds on an interval of times, differentiate it with respect to tt (chain rule on every letter), then substitute the instant's values. Substituting first turns a moving quantity into a constant and destroys the rate you were asked for.

5 min readCore — on every final8 formulas6 quick checks
01

Key ideas

5 things to remember
  1. 1

    Every letter is a function of time

    Write x=x(t)x=x(t), V=V(t)V=V(t), θ=θ(t)\theta=\theta(t) and never suppress the tt. Differentiating then attaches a rate factor to each letter: ddtr3=3r2drdt\frac{d}{dt}r^3=3r^2\frac{dr}{dt}, never 3r23r^2.

  2. 2

    Differentiate first, substitute last

    The constraint must hold on an interval of times before it can be differentiated. Putting the instant's numbers in first freezes a moving quantity into a constant, whose derivative is 00.

  3. 3

    Only genuine constants may go in early

    A ladder's length, a cone's shape ratio r/hr/h, a fixed angle at a junction: these never change, so use them straight away. Anything that moves stays a letter until after the ddt\frac{d}{dt}.

  4. 4

    Reduce before you differentiate

    If the relation carries a variable whose rate is neither given nor wanted — the rr in V=13πr2hV=\frac13\pi r^2h — eliminate it with similar triangles first. One equation cannot pin down two unknown rates.

  5. 5

    Report the sign and the units

    Draining, falling, shrinking and approaching all mean a negative rate. Report dydt=−0.75\frac{dy}{dt}=-0.75 m/s, or "down at 0.750.75 m/s" — not both. Angles must be in radians.

L = 10 m (fixed)x(t)y(t)dx/dt = +1 m/s →↓ dy/dt = −0.75 m/sx² + y² = 100 for all tx dx/dt + y dy/dt = 0instant: x = 6, y = 8
The ladder's length is the only number you may use before differentiating. From x2+y2=100x^2+y^2=100, true at every instant, comes xdxdt+ydydt=0x\frac{dx}{dt}+y\frac{dy}{dt}=0; only then put in x=6x=6, y=8y=8 to get dydt=−68(1)=−0.75\frac{dy}{dt}=-\frac68(1)=-0.75 m/s.
tank: 6 m deep, top radius 2 mhrr = h/3 (constant ratio)V = πh³/27dV/dt = −2 m³/mindh/dt = −2/π m/min at h = 3
The water cone stays similar to the tank, so r=h3r=\frac h3 is a genuine constant ratio and may be substituted first: V=13πr2h=πh327V=\frac13\pi r^2h=\frac{\pi h^3}{27}, hence dVdt=πh29dhdt\frac{dV}{dt}=\frac{\pi h^2}{9}\frac{dh}{dt} with no unknown drdt\frac{dr}{dt} left in it.
02

Formulas

What to have memorised

  • Chain rule in the time variable

    ddtf(x(t))=f′(x(t))⋅dxdt\frac{d}{dt}f\big(x(t)\big)=f'\big(x(t)\big)\cdot\frac{dx}{dt}

    The whole topic is this one rule applied to a constraint.

  • Ladder of fixed length

    x2+y2=L2⟹xdxdt+ydydt=0x^2+y^2=L^2\quad\Longrightarrow\quad x\frac{dx}{dt}+y\frac{dy}{dt}=0

    So dydt=−xydxdt\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}, valid while y≠0y\neq0.

  • Separation at right angles

    z2=x2+y2⟹zdzdt=xdxdt+ydydtz^2=x^2+y^2\quad\Longrightarrow\quad z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}

    dzdt<0\frac{dz}{dt}<0 means closing; distance is not the sum of the two rates.

  • Circle and sphere

    dAdt=2πrdrdt,dVdt=4πr2drdt,dSdt=8πrdrdt\frac{dA}{dt}=2\pi r\frac{dr}{dt},\qquad \frac{dV}{dt}=4\pi r^2\frac{dr}{dt},\qquad \frac{dS}{dt}=8\pi r\frac{dr}{dt}

    Circle area, sphere volume, sphere surface — each is one chain rule.

  • Cone with a fixed shape ratio

    r=kh ⇒ V=13πk2h3 ⇒ dVdt=πk2h2dhdtr=kh\ \Rightarrow\ V=\tfrac13\pi k^2h^3\ \Rightarrow\ \frac{dV}{dt}=\pi k^2h^2\frac{dh}{dt}

    kk is the tank's radius over its depth. Eliminate rr before differentiating.

  • Shadow from a lamp

    dsdt=pH−pdxdt,d(x+s)dt=HH−pdxdt\frac{ds}{dt}=\frac{p}{H-p}\frac{dx}{dt},\qquad \frac{d(x+s)}{dt}=\frac{H}{H-p}\frac{dx}{dt}

    Lamp height HH, walker height pp: the shadow's length, then its tip.

  • Angle of elevation

    tan⁡θ=hx⟹sec⁡2θ dθdt=ddt ⁣(hx)\tan\theta=\frac{h}{x}\quad\Longrightarrow\quad \sec^2\theta\,\frac{d\theta}{dt}=\frac{d}{dt}\!\left(\frac{h}{x}\right)

    Radians only, and sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta saves finding θ\theta.

  • A non-geometric constraint

    PV=C⟹VdPdt+PdVdt=0PV=C\quad\Longrightarrow\quad V\frac{dP}{dt}+P\frac{dV}{dt}=0

    Boyle's law: any equation between moving quantities works the same way.

03

Solve a related-rates problem

The steps, in order
  1. 1

    Draw a general instant, not the special one. Number every length that never changes; put a letter on every length that moves.

  2. 2

    Name the moving quantities as functions of tt with units, then record what is given, what is wanted, and at which instant.

  3. 3

    Relate them by one equation valid for all nearby tt: Pythagoras, similar triangles, a volume formula, trigonometry, or a physical law.

  4. 4

    Eliminate any variable whose rate is neither given nor wanted, usually through a fixed ratio such as r=khr=kh.

  5. 5

    Differentiate both sides with respect to tt, attaching the chain-rule factor d(⋅)dt\frac{d(\cdot)}{dt} to every letter.

  6. 6

    Only now substitute the instant's values and solve for the rate you want.

  7. 7

    State it with units and a direction, then check the sign against the picture.

04

Watch out

The mistakes that cost marks

  • ✗ Wrong

    At x=5x=5: 25+y2=16925+y^2=169 so y=12y=12; then ddt(12)=0\frac{d}{dt}(12)=0, so dydt=0\frac{dy}{dt}=0.

    ✓ Right

    Differentiate x2+y2=169x^2+y^2=169 first, then substitute x=5x=5, y=12y=12.

    Why: y=12y=12 holds at one instant, not on an interval, so it cannot be differentiated.

  • ✗ Wrong

    ddt(r3)=3r2\frac{d}{dt}\big(r^3\big)=3r^2 and ddt(xy)=dxdt⋅dydt\frac{d}{dt}(xy)=\frac{dx}{dt}\cdot\frac{dy}{dt}

    ✓ Right

    3r2drdt3r^2\frac{dr}{dt} and dxdt y+x dydt\frac{dx}{dt}\,y+x\,\frac{dy}{dt}

    Why: Every letter moves with tt, so chain rule and product rule both apply.

  • ✗ Wrong

    "Water drains at 22" recorded as dVdt=+2\frac{dV}{dt}=+2

    ✓ Right

    dVdt=−2\frac{dV}{dt}=-2. Draining, falling, shrinking and approaching are all negative.

    Why: One sign error flips the entire answer, and markers look for it.

  • ✗ Wrong

    Differentiating V=13πr2hV=\frac13\pi r^2h with both rr and hh still varying

    ✓ Right

    Substitute the shape ratio r=khr=kh first, then differentiate V=13πk2h3V=\frac13\pi k^2h^3.

    Why: One equation cannot determine two unknown rates.

  • ✗ Wrong

    Putting dθdt=4\frac{d\theta}{dt}=4 rev/min straight into sec⁡2θ dθdt\sec^2\theta\,\frac{d\theta}{dt}

    ✓ Right

    Convert first: dθdt=4(2π)=8π\frac{d\theta}{dt}=4(2\pi)=8\pi rad/min.

    Why: ddθtan⁡θ=sec⁡2θ\frac{d}{d\theta}\tan\theta=\sec^2\theta is false unless θ\theta is in radians.

  • ✗ Wrong

    dzdt=dxdt+dydt\frac{dz}{dt}=\frac{dx}{dt}+\frac{dy}{dt} for two vehicles separating

    ✓ Right

    From z2=x2+y2z^2=x^2+y^2: zdzdt=xdxdt+ydydtz\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}.

    Why: A distance is a hypotenuse, not the sum of the legs.

05

Quick check

Commit to an answer before you reveal one
  1. Q1easy

    Air is pumped into a spherical balloon at 100100 cm3^3/s. (a) How fast is the radius increasing at the moment when the diameter is 5050 cm? (b) How fast is the surface area increasing then? Use V=43πr3V=\frac43\pi r^3 and S=4πr2S=4\pi r^2.

    Hint

    Halve the diameter first, and differentiate VV with respect to tt before any number goes in.

    Show answer

    Answer

    (a) drdt=125π≈0.0127\frac{dr}{dt}=\frac{1}{25\pi}\approx0.0127 cm/s. (b) dSdt=8\frac{dS}{dt}=8 cm2^2/s.

    Steps

    Diameter 5050 cm means r=25r=25 cm at that instant.

    dVdt=4πr2drdt ⟹ 100=2500πdrdt ⟹ drdt=125π\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\ \Longrightarrow\ 100=2500\pi\frac{dr}{dt}\ \Longrightarrow\ \frac{dr}{dt}=\frac{1}{25\pi}

    Then dSdt=8πrdrdt=8π(25)⋅125π=8\frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi(25)\cdot\frac{1}{25\pi}=8 cm2^2/s.

    Do not put r=25r=25 into VV first: that makes VV the constant 62500π3\frac{62500\pi}{3}, forcing dVdt=0\frac{dV}{dt}=0 and contradicting the data.

  2. Q2easy

    A 1010 m ladder leans against a vertical wall and its foot is pulled away at 11 m/s. (a) How fast is the top sliding down when the foot is 66 m from the wall? (b) A student sets x=6x=6, gets y=8y=8, differentiates y=8y=8 and concludes the top is not moving. Which step is invalid?

    Show answer

    Answer

    (a) dydt=−0.75\frac{dy}{dt}=-0.75 m/s: the top slides down at 0.750.75 m/s. (b) Differentiating y=8y=8 — that equation is true at one instant, not on an interval of times.

    Steps

    x2+y2=100x^2+y^2=100 holds for every tt while the ladder touches both wall and floor, so

    2xdxdt+2ydydt=0 ⟹ dydt=−xy⋅dxdt2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\ \Longrightarrow\ \frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}

    At the instant, y=100−36=8y=\sqrt{100-36}=8, so dydt=−68(1)=−0.75\frac{dy}{dt}=-\frac68(1)=-0.75 m/s.

    (b) Substituting x=6x=6 first replaces a moving yy by a constant, and a derivative depends on nearby times. Only the fixed 100100 may go in before ddt\frac{d}{dt}.

  3. Q3medium

    Sand falls at 1212 ft3^3/min into a conical pile whose height always equals the diameter of its base. How fast is the height increasing when the pile is 66 ft high? Use V=13πr2hV=\frac13\pi r^2h.

    Hint

    "Height equals base diameter" gives h=2rh=2r; use it to write VV in terms of hh alone.

    Show answer

    Answer

    dhdt=43π≈0.42\frac{dh}{dt}=\frac{4}{3\pi}\approx0.42 ft/min.

    Steps

    The shape condition h=2rh=2r, i.e. r=h2r=\frac h2, is a constant ratio, so substitute it now:

    V=13π(h2)2h=πh312 ⟹ dVdt=πh24⋅dhdtV=\frac13\pi\left(\frac h2\right)^2h=\frac{\pi h^3}{12}\ \Longrightarrow\ \frac{dV}{dt}=\frac{\pi h^2}{4}\cdot\frac{dh}{dt}

    At h=6h=6: 12=π(36)4dhdt=9πdhdt12=\frac{\pi(36)}{4}\frac{dh}{dt}=9\pi\frac{dh}{dt}, so dhdt=129π=43π≈0.42\frac{dh}{dt}=\frac{12}{9\pi}=\frac{4}{3\pi}\approx0.42 ft/min.

    Sensible: the rate goes like 1/h21/h^2, so a taller pile rises more slowly.

  4. Q4medium

    A lamp sits at the top of a 66 m pole. A person 22 m tall walks away from the pole on level ground at 1.51.5 m/s. (a) How fast is the shadow lengthening? (b) How fast is the tip of the shadow moving away from the pole? (c) Why is the person's distance from the pole never needed?

    Hint

    Cross-multiply the similar-triangle proportion before differentiating, and keep the shadow's length and its tip as two different quantities.

    Show answer

    Answer

    (a) 0.750.75 m/s. (b) 2.252.25 m/s. (c) Because s=x2s=\frac x2 is linear, so xx vanishes on differentiating.

    Steps

    Let xx be the distance walked and ss the shadow's length. Similar triangles give

    6x+s=2s ⟹ 6s=2x+2s ⟹ s=x2\frac{6}{x+s}=\frac{2}{s}\ \Longrightarrow\ 6s=2x+2s\ \Longrightarrow\ s=\frac{x}{2}

    (a) dsdt=12(1.5)=0.75\frac{ds}{dt}=\frac12(1.5)=0.75 m/s.

    (b) The tip is at x+s=3x2x+s=\frac{3x}{2}, so it moves at 32(1.5)=2.25\frac32(1.5)=2.25 m/s — and 1.5+0.75=2.251.5+0.75=2.25 checks it.

    (c) Both answers are constant multiples of dxdt\frac{dx}{dt}, so no position survives the differentiation.

  5. Q5medium

    Two roads meet at right angles. Car AA drives west toward the crossing at 5050 km/h; car BB drives north toward it at 6060 km/h. At one moment AA is 0.30.3 km from the crossing and BB is 0.40.4 km from it. (a) At what rate is the distance between them changing? (b) Redo it if AA has already passed 0.30.3 km beyond the crossing, still heading west.

    Hint

    Let xx and yy be distances from the crossing, not positions, and fix each sign before substituting.

    Show answer

    Answer

    (a) dzdt=−78\frac{dz}{dt}=-78 km/h — closing at 7878 km/h. (b) dzdt=−18\frac{dz}{dt}=-18 km/h — still closing, at 1818 km/h.

    Steps

    With zz the gap, z2=x2+y2z^2=x^2+y^2 for all nearby tt, so zdzdt=xdxdt+ydydtz\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}, and z=0.09+0.16=0.5z=\sqrt{0.09+0.16}=0.5.

    (a) Both distances shrink: dxdt=−50\frac{dx}{dt}=-50, dydt=−60\frac{dy}{dt}=-60.

    0.5dzdt=0.3(−50)+0.4(−60)=−39 ⟹ dzdt=−780.5\frac{dz}{dt}=0.3(-50)+0.4(-60)=-39\ \Longrightarrow\ \frac{dz}{dt}=-78

    (b) Now AA's distance grows: dxdt=+50\frac{dx}{dt}=+50, giving 0.5dzdt=15−24=−90.5\frac{dz}{dt}=15-24=-9, so dzdt=−18\frac{dz}{dt}=-18 km/h. One sign changed the answer fourfold.

  6. Q6hard

    A lighthouse stands 33 km from a long straight shore and its lamp turns at 44 revolutions per minute. Let xx be the distance along the shore from the nearest point PP to the lit spot. (a) How fast is the spot moving when x=1x=1 km? (b) How fast at PP itself? (c) Is (b) the slowest the spot ever moves?

    Hint

    Revolutions per minute is not radians per minute — convert before using sec⁡2θ\sec^2\theta.

    Show answer

    Answer

    (a) 80π3≈83.8\frac{80\pi}{3}\approx83.8 km/min. (b) 24π≈75.424\pi\approx75.4 km/min. (c) Yes.

    Steps

    Convert: dθdt=4(2π)=8π\frac{d\theta}{dt}=4(2\pi)=8\pi rad/min. With θ\theta measured from the perpendicular, x=3tan⁡θx=3\tan\theta for all tt, so

    dxdt=3sec⁡2θ dθdt=24πsec⁡2θ=24π(1+x29)\frac{dx}{dt}=3\sec^2\theta\,\frac{d\theta}{dt}=24\pi\sec^2\theta=24\pi\left(1+\frac{x^2}{9}\right)

    (a) At x=1x=1: 24π⋅109=80π3≈83.824\pi\cdot\frac{10}{9}=\frac{80\pi}{3}\approx83.8 km/min.

    (b) At x=0x=0: 24π≈75.424\pi\approx75.4 km/min.

    (c) 1+x291+\frac{x^2}{9} is least at x=0x=0, so yes — and the speed grows without bound as the beam turns toward the shore.

06

On the exam

How this topic is marked

  • Marks sit in the setup: a labelled picture, the constraint written for a general tt, the differentiated equation, then the substitution. A bare number with no differentiated constraint rarely earns full credit.

  • Finish with units and a direction — "the level falls at 2π\frac{2}{\pi} m/min" — and never put a minus sign and the word "down" in the same phrase.

  • If two letters vary and you have only one equation, you have skipped a reduction: hunt for similar triangles or a fixed ratio before differentiating.

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