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Related rates: differentiate the relation, then substitute

Ten related-rates problems solved by differentiating an equation in time: a growing circle and sphere, a sliding ladder, a filling cone and trough, two cars, a shadow, a rising balloon's angle, the length of a moving vector, and the distance between clock hands, with worked solutions and the mistakes of substituting too early and dropping the chain rule.

Before you start

A related-rates problem is the chain rule with time as the input. Several quantities change together, an equation ties them, and differentiating that equation with respect to tt ties their rates. The same move gives the rate a loss changes along a training trajectory, ddtL(w(t))=∇L⋅dwdt\tfrac{d}{dt}L(w(t)) = \nabla L\cdot\tfrac{dw}{dt}. The work goes wrong in a few places: a value substituted before differentiating, which freezes a quantity that is changing, a missing dxdt\tfrac{dx}{dt} from the chain rule, a rate given the wrong sign, and an answer to a different question from the one asked. These ten problems cover each of them.

  • Every changing quantity is a function of time tt. Write x′x' for dxdt\tfrac{dx}{dt}. A rate is positive when the quantity is increasing and negative when it is decreasing.
  • The method: (1) name the quantities and write an equation that relates them and holds at every time; (2) differentiate both sides with respect to tt, using the chain rule on each changing quantity; (3) only then substitute the values at the instant asked about, and solve for the unknown rate.
  • Chain rule in time: ddt x2=2x x′\tfrac{d}{dt}\,x^2 = 2x\,x', ddt tan⁡θ=sec⁡2θ θ′\tfrac{d}{dt}\,\tan\theta = \sec^2\theta\,\theta', and for a product ddt(xy)=x′y+xy′\tfrac{d}{dt}(xy) = x'y + xy'. A constant differentiates to 0.
  • Formulas used: circle area A=πr2A = \pi r^2; sphere volume V=43πr3V = \tfrac43\pi r^3; cone volume V=13πr2hV = \tfrac13\pi r^2h; Pythagoras; and the law of cosines c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C.
  • Angles are in radians, so an angle's rate is in radians per unit time.

Builds on: Differentiation rules: chain, product, quotient

Problems

  1. ·

    The radius of a circle grows at 22 cm/s. How fast is its area growing when the radius is 55 cm?

  2. ·

    A 55 m ladder leans against a wall. Its foot slides away from the wall at 0.30.3 m/s. How fast is the top sliding down the wall when the foot is 33 m from the wall?

  3. ··

    Air is pumped into a spherical balloon at 100100 cm³/s. How fast is the radius increasing when the radius is 55 cm?

  4. ··

    Water flows at 22 m³/min into a tank shaped like an inverted cone, 66 m tall with a top radius of 33 m. How fast is the water level rising when the water is 44 m deep?

  5. ··

    Car A is 33 km north of an intersection, driving south toward it at 6060 km/h. Car B is 44 km east of the intersection, driving east at 8080 km/h. Is the distance between them increasing or decreasing, and how fast?

  6. ··

    A person 1.81.8 m tall walks away from a 66 m lamppost at 1.41.4 m/s. How fast is their shadow lengthening, and how fast is the tip of the shadow moving?

  7. ···

    A balloon is released 4040 m from an observer and rises straight up at 33 m/s. How fast is the angle of elevation from the observer to the balloon increasing when the balloon is 3030 m up?

  8. ··

    A parameter vector moves with w(t)∈R2w(t) \in \mathbb R^2. At one instant w=(3,4)w = (3, 4) and w′=(1,−2)w' = (1, -2). How fast is its length ∥w∥\lVert w\rVert changing?

  9. ··

    A trough is 44 m long, and its cross-section is an inverted isosceles triangle 11 m across the top and 0.50.5 m deep. Water flows in at 0.10.1 m³/min. How fast is the level rising when the water is 0.250.25 m deep?

  10. ···

    A clock's hour hand is 44 cm long and its minute hand 66 cm. How fast is the distance between their tips changing at 3:00?

Answers

  1. 20π≈62.8320\pi \approx 62.83 cm²/s
  2. The top slides down at 0.2250.225 m/s
  3. r′=1π≈0.318r' = \dfrac1\pi \approx 0.318 cm/s
  4. h′=12π≈0.159h' = \dfrac1{2\pi} \approx 0.159 m/min
  5. Increasing, at 2828 km/h
  6. The shadow lengthens at 0.60.6 m/s; the tip moves at 2.02.0 m/s
  7. 0.0480.048 rad/s
  8. ∥w∥′=−1\lVert w\rVert' = -1: the length is shrinking at 1 unit per unit time
  9. h′=0.05h' = 0.05 m/min
  10. Decreasing, at 11π3013≈0.319\dfrac{11\pi}{30\sqrt{13}} \approx 0.319 cm/min

Worked solutions

Problem 1

The radius of a circle grows at 22 cm/s. How fast is its area growing when the radius is 55 cm?

  1. A=πr2A = \pi r^2Holds at every time.
  2. A′=2πr r′A' = 2\pi r\,r'Differentiate with respect to tt; the chain rule gives the factor r′r'.
  3. A′=2π⋅5⋅2A' = 2\pi\cdot5\cdot2Now substitute r=5r = 5 and r′=2r' = 2.
  4. 20π≈62.8320\pi \approx 62.83 cm²/sThe area grows at the circumference times the speed of the edge.

Problem 2

A 55 m ladder leans against a wall. Its foot slides away from the wall at 0.30.3 m/s. How fast is the top sliding down the wall when the foot is 33 m from the wall?

  1. x2+y2=25x^2 + y^2 = 25, with xx the distance of the foot from the wall and yy the height of the top.Pythagoras; the ladder's length is constant.
  2. 2x x′+2y y′=02x\,x' + 2y\,y' = 0Differentiate in tt; the constant 25 gives 0.
  3. At the instant: x=3x = 3, y=25−9=4y = \sqrt{25 - 9} = 4, x′=0.3x' = 0.3.x′x' is positive because xx is increasing.
  4. y′=−x x′y=−3⋅0.34=−0.225y' = -\dfrac{x\,x'}{y} = -\dfrac{3\cdot0.3}{4} = -0.225Solve step 2 for y′y'.
  5. The top slides down at 0.2250.225 m/sy′<0y' < 0: the height is decreasing.

Problem 3

Air is pumped into a spherical balloon at 100100 cm³/s. How fast is the radius increasing when the radius is 55 cm?

  1. V=43πr3V = \tfrac43\pi r^3Sphere volume, at every time.
  2. V′=4πr2 r′V' = 4\pi r^2\,r'Chain rule. 4πr24\pi r^2 is the surface area: the volume grows at the surface area times the speed of the surface.
  3. 100=4π⋅25⋅r′100 = 4\pi\cdot25\cdot r'Substitute V′=100V' = 100 and r=5r = 5.
  4. r′=1π≈0.318r' = \dfrac1\pi \approx 0.318 cm/s100100π\tfrac{100}{100\pi}. The same inflow makes the radius grow more slowly as the balloon gets bigger.

Problem 4

Water flows at 22 m³/min into a tank shaped like an inverted cone, 66 m tall with a top radius of 33 m. How fast is the water level rising when the water is 44 m deep?

  1. r=h2r = \tfrac h2 at every depth hhSimilar triangles: the water forms a smaller cone of the same shape, with rh=36\tfrac rh = \tfrac36.
  2. V=13πr2h=13π(h2)2h=πh312V = \tfrac13\pi r^2h = \tfrac13\pi\big(\tfrac h2\big)^2h = \dfrac{\pi h^3}{12}Use step 1 to remove rr before differentiating, so that only one changing quantity is left.
  3. V′=πh24 h′V' = \dfrac{\pi h^2}{4}\,h'Chain rule.
  4. 2=π⋅164 h′2 = \dfrac{\pi\cdot16}{4}\,h'Substitute V′=2V' = 2 and h=4h = 4, now and not before.
  5. h′=12π≈0.159h' = \dfrac1{2\pi} \approx 0.159 m/min24π\tfrac{2}{4\pi}. Substituting r=2r = 2 before differentiating gives three times this (mistake 1).

Problem 5

Car A is 33 km north of an intersection, driving south toward it at 6060 km/h. Car B is 44 km east of the intersection, driving east at 8080 km/h. Is the distance between them increasing or decreasing, and how fast?

  1. s2=x2+y2s^2 = x^2 + y^2, with xx B's distance east and yy A's distance north of the intersectionThe roads are perpendicular.
  2. 2s s′=2x x′+2y y′2s\,s' = 2x\,x' + 2y\,y', so s′=x x′+y y′ss' = \dfrac{x\,x' + y\,y'}{s}Differentiate in tt.
  3. At the instant: x=4x = 4, x′=80x' = 80, y=3y = 3, y′=−60y' = -60, s=5s = 5.A is driving toward the intersection, so yy is decreasing and y′y' is negative.
  4. s′=4⋅80+3⋅(−60)5=320−1805s' = \dfrac{4\cdot80 + 3\cdot(-60)}{5} = \dfrac{320 - 180}{5}Substitute.
  5. Increasing, at 2828 km/h1405\tfrac{140}5. B moving away outweighs A closing in.

Problem 6

A person 1.81.8 m tall walks away from a 66 m lamppost at 1.41.4 m/s. How fast is their shadow lengthening, and how fast is the tip of the shadow moving?

  1. 6x+s=1.8s\dfrac{6}{x + s} = \dfrac{1.8}{s}, with xx the person's distance from the post and ss the shadow's lengthSimilar triangles: the lamp and the person cast the same ray to the tip, which is x+sx + s from the post.
  2. 6s=1.8x+1.8s6s = 1.8x + 1.8s, so 4.2s=1.8x4.2s = 1.8x and s=37xs = \tfrac37xCross-multiply.
  3. s′=37x′=37⋅1.4=0.6s' = \tfrac37x' = \tfrac37\cdot1.4 = 0.6Differentiate; here the rate does not depend on where the person is.
  4. Tip: ddt(x+s)=1.4+0.6=2.0\tfrac{d}{dt}(x + s) = 1.4 + 0.6 = 2.0The tip is at distance x+sx + s from the post.
  5. The shadow lengthens at 0.60.6 m/s; the tip moves at 2.02.0 m/sTwo different questions, two different rates (mistake 4).

Problem 7

A balloon is released 4040 m from an observer and rises straight up at 33 m/s. How fast is the angle of elevation from the observer to the balloon increasing when the balloon is 3030 m up?

  1. tan⁡θ=h40\tan\theta = \dfrac h{40}Opposite over adjacent; the 40 m is constant.
  2. sec⁡2θ θ′=h′40\sec^2\theta\,\theta' = \dfrac{h'}{40}Chain rule on the left: ddttan⁡θ=sec⁡2θ θ′\tfrac{d}{dt}\tan\theta = \sec^2\theta\,\theta'.
  3. At h=30h = 30: the line of sight is 302+402=50\sqrt{30^2 + 40^2} = 50 m, so cos⁡θ=4050=0.8\cos\theta = \tfrac{40}{50} = 0.8 and sec⁡2θ=10.64=1.5625\sec^2\theta = \tfrac{1}{0.64} = 1.5625.Read the cosine from the triangle instead of finding θ\theta.
  4. θ′=3/401.5625=0.0751.5625\theta' = \dfrac{3/40}{1.5625} = \dfrac{0.075}{1.5625}Substitute h′=3h' = 3 and solve.
  5. 0.0480.048 rad/sAbout 2.75∘2.75^\circ per second.

Problem 8

A parameter vector moves with w(t)∈R2w(t) \in \mathbb R^2. At one instant w=(3,4)w = (3, 4) and w′=(1,−2)w' = (1, -2). How fast is its length ∥w∥\lVert w\rVert changing?

  1. ∥w∥2=w⋅w\lVert w\rVert^2 = w\cdot wAt every time. Work with the square to avoid differentiating a square root.
  2. 2∥w∥ ∥w∥′=2 w⋅w′2\lVert w\rVert\,\lVert w\rVert' = 2\,w\cdot w', so ∥w∥′=w⋅w′∥w∥\lVert w\rVert' = \dfrac{w\cdot w'}{\lVert w\rVert}The product rule for a dot product: ddt(w⋅w)=w′⋅w+w⋅w′\tfrac{d}{dt}(w\cdot w) = w'\cdot w + w\cdot w'.
  3. w⋅w′=3⋅1+4⋅(−2)=−5w\cdot w' = 3\cdot1 + 4\cdot(-2) = -5 and ∥w∥=5\lVert w\rVert = 5Substitute.
  4. ∥w∥′=−1\lVert w\rVert' = -1: the length is shrinking at 1 unit per unit timeOnly the component of w′w' along ww changes the length. Problem 5 is the same formula with ww the cars' position difference.

Problem 9

A trough is 44 m long, and its cross-section is an inverted isosceles triangle 11 m across the top and 0.50.5 m deep. Water flows in at 0.10.1 m³/min. How fast is the level rising when the water is 0.250.25 m deep?

  1. At depth hh the water's surface is 2h2h wide.Similar triangles: the cross-section is twice as wide as it is deep, 10.5=2\tfrac{1}{0.5} = 2.
  2. V=4⋅12⋅2h⋅h=4h2V = 4\cdot\tfrac12\cdot2h\cdot h = 4h^2Length times the area of the triangle of water.
  3. V′=8h h′V' = 8h\,h'Differentiate in tt.
  4. 0.1=8⋅0.25⋅h′0.1 = 8\cdot0.25\cdot h'Substitute V′=0.1V' = 0.1 and h=0.25h = 0.25.
  5. h′=0.05h' = 0.05 m/min0.12\tfrac{0.1}{2}.

Problem 10

A clock's hour hand is 44 cm long and its minute hand 66 cm. How fast is the distance between their tips changing at 3:00?

  1. d2=42+62−2⋅4⋅6cos⁡θ=52−48cos⁡θd^2 = 4^2 + 6^2 - 2\cdot4\cdot6\cos\theta = 52 - 48\cos\theta, with θ\theta the angle between the handsLaw of cosines in the triangle formed by the hands and the segment between their tips.
  2. 2d d′=48sin⁡θ θ′2d\,d' = 48\sin\theta\,\theta'Differentiate in tt; ddtcos⁡θ=−sin⁡θ θ′\tfrac{d}{dt}\cos\theta = -\sin\theta\,\theta'.
  3. θ′=2π720−2π60=−11π360\theta' = \dfrac{2\pi}{720} - \dfrac{2\pi}{60} = -\dfrac{11\pi}{360} rad/minJust after 3:00 the hour hand is θ\theta ahead of the minute hand. The minute hand turns 2π2\pi in 60 min and the hour hand in 720 min, so the gap closes at the difference of their rates.
  4. At 3:00: θ=π2\theta = \tfrac\pi2, sin⁡θ=1\sin\theta = 1, d=52=213d = \sqrt{52} = 2\sqrt{13}.cos⁡π2=0\cos\tfrac\pi2 = 0.
  5. d′=48⋅1⋅(−11π360)2⋅213=−11π3013d' = \dfrac{48\cdot1\cdot\big(-\tfrac{11\pi}{360}\big)}{2\cdot2\sqrt{13}} = -\dfrac{11\pi}{30\sqrt{13}}Solve step 2 for d′d'.
  6. Decreasing, at 11π3013≈0.319\dfrac{11\pi}{30\sqrt{13}} \approx 0.319 cm/minThe minute hand is catching up with the hour hand, so the tips are getting closer.

Where this goes wrong

1. Substituting before differentiating

The numbers are given, and putting them in early makes the equation simpler.

  1. Cone tank, h=4h = 4, so the water's surface radius is r=2r = 2Right so far: Problem 4, step 1 at the instant.
  2. “V=13πr2h=13π⋅4⋅hV = \tfrac13\pi r^2h = \tfrac13\pi\cdot4\cdot h, then differentiate.”The habit that causes the mistake: substituting values first, as in a problem with no rates.
  3. V′=43π h′V' = \tfrac43\pi\,h', so h′=2⋅34π=32π≈0.477h' = \dfrac{2\cdot3}{4\pi} = \dfrac{3}{2\pi} \approx 0.477 m/minr=2r = 2 only at this instant; rr is changing with hh, and fixing it freezes the cone's widening. Differentiate the relation that holds at every time, V=πh312V = \tfrac{\pi h^3}{12}, first: h′=12π≈0.159h' = \tfrac{1}{2\pi} \approx 0.159 m/min, a third of this.

2. Dropping the chain rule's factor

Differentiating πr2\pi r^2 gives 2πr2\pi r in every exercise where the variable is rr.

  1. A=πr2A = \pi r^2, r=5r = 5, r′=2r' = 2Right so far: Problem 1.
  2. “The derivative of πr2\pi r^2 is 2πr2\pi r.”The habit that causes the mistake: differentiating with respect to rr when the question is about time.
  3. A′=2πr=10π≈31.4A' = 2\pi r = 10\pi \approx 31.4 cm²/sThat is dAdr\tfrac{dA}{dr}, the rate per centimeter of radius, not per second. The chain rule multiplies it by drdt=2\tfrac{dr}{dt} = 2: A′=20πA' = 20\pi cm²/s. A rate that does not depend on how fast the radius grows cannot be right.

3. The wrong sign on an approaching rate

Speeds are given as positive numbers, so every rate gets entered as positive.

  1. s′=x x′+y y′ss' = \dfrac{x\,x' + y\,y'}{s}, with x=4x = 4, y=3y = 3, s=5s = 5Right so far: Problem 5, step 2.
  2. “Car A drives at 60 km/h, so y′=60y' = 60.”The habit that causes the mistake: treating a speed as the rate of the variable.
  3. s′=4⋅80+3⋅605=100s' = \dfrac{4\cdot80 + 3\cdot60}{5} = 100 km/hA drives toward the intersection, so its distance yy is decreasing and y′=−60y' = -60. The correct rate is 320−1805=28\tfrac{320 - 180}{5} = 28 km/h. Before substituting, ask of each variable whether it is growing or shrinking.

4. Answering a different question

Problems with two moving things often ask about the one that is not the obvious unknown.

  1. s=37xs = \tfrac37x and s′=0.6s' = 0.6 m/sRight so far: Problem 6, step 3.
  2. “The shadow's rate is found, so that is how fast the tip moves.”The habit that causes the mistake: stopping at the first rate computed.
  3. The tip of the shadow moves at 0.60.6 m/s0.60.6 m/s is how fast the shadow gets longer, measured from the person's feet, which are themselves moving at 1.41.4 m/s. The tip is x+sx + s from the post, so it moves at 1.4+0.6=2.01.4 + 0.6 = 2.0 m/s. Write down which distance the question is about before differentiating.

5. Differentiating with respect to x instead of t

Implicit differentiation is usually practiced with dydx\tfrac{dy}{dx}, and the habit carries over.

  1. x2+y2=25x^2 + y^2 = 25, x=3x = 3, y=4y = 4Right so far: Problem 2, step 1.
  2. “Differentiate: 2x+2y dydx=02x + 2y\,\tfrac{dy}{dx} = 0.”The habit that causes the mistake: xx treated as the independent variable when both xx and yy depend on time.
  3. dydx=−34\dfrac{dy}{dx} = -\dfrac34, so the top slides down at 0.750.75 m/s−34-\tfrac34 is the change in height per meter the foot moves, not per second, and it never used the foot's speed. The chain rule in tt gives y′=dydx x′=−34⋅0.3=−0.225y' = \tfrac{dy}{dx}\,x' = -\tfrac34\cdot0.3 = -0.225 m/s.

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Read the guide: Related rates: write the equation, differentiate in time, then plug in

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