Unit 4, lesson 39

Related Rates

How an equation linking several changing quantities turns into an equation linking their rates, and how to use it to find one rate from the others.

  • Theory part 77
  • Practice part 78
  • Modelling and Applications

Theory, part 77

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Practice, part 78

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Blow air into a balloon at a steady pace and the radius grows fast at first, then more and more slowly. The air goes in at one rate, the radius changes at another, and the two are tied together by the formula for the volume of a sphere. Related rates problems use such a link to find a rate that is hard to measure from one that is easy to measure.

Key terms

Quantities that change with time

Definition In a related rates problem each varying quantity is a differentiable function of time tt on some interval, such as x=x(t)x = x(t) or V=V(t)V = V(t). Its rate of change at time tt is the derivative

dxdt=x′(t).\frac{dx}{dt} = x'(t).

In plain words Every changing length, area, volume or angle in the problem is secretly a function of time, even when the formula does not show tt.

Example A spherical balloon has radius r(t)r(t) and volume

V(t)=43πr(t)3.V(t) = \frac{4}{3}\pi r(t)^3.

Differentiating with the chain rule gives

dVdt=4πr2drdt.\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.

If air goes in at 100100 cubic centimeters per second, then at the moment r=50r = 50 centimeters

drdt=1004π⋅2500=1100π≈0.0032,\frac{dr}{dt} = \frac{100}{4\pi \cdot 2500} = \frac{1}{100\pi} \approx 0.0032,

so the radius grows by about 0.00320.0032 centimeters per second.

Constraint equations

Definition A constraint is an equation F(x(t),y(t),…)=0F(x(t), y(t), \ldots) = 0 between the quantities that holds for every tt in an interval, usually because of the geometry of the situation (the Pythagorean theorem, similar triangles, a volume formula). Differentiating both sides with respect to tt, using the chain rule, gives an equation between the rates that also holds for every tt in that interval.

In plain words First find the equation that ties the quantities together at every moment. Its derivative ties their rates together.

Example A ladder 1010 meters long leans against a wall. Let xx be the distance from the wall to the foot of the ladder and yy the height of its top. At every moment

x2+y2=100,so2xdxdt+2ydydt=0.x^2 + y^2 = 100, \qquad \text{so} \qquad 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0.

Differentiate first, then substitute

Definition Values that hold only at one instant, such as x=6x = 6, may be put into the equation only after it has been differentiated. Substituting them earlier replaces a function of tt by a constant, and the derivative of a constant is 00, which makes the resulting rate wrong.

In plain words Differentiate the general relationship. Plug in the numbers for the moment you care about only at the very end.

Example In the ladder problem the foot slides away from the wall at 11 meter per second. At the moment x=6x = 6, the constraint gives y=8y = 8. From the differentiated equation,

dydt=−xydxdt=−68⋅1=−0.75,\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{6}{8} \cdot 1 = -0.75,

so the top slides down at 0.750.75 meters per second. Had you substituted x=6x = 6 first, you would get 36+y2=10036 + y^2 = 100, a statement about one instant, and differentiating it would give the false answer dydt=0\frac{dy}{dt} = 0.

Consistent units

Definition The rate dxdt\frac{dx}{dt} is measured in units of xx per unit of tt. Before combining rates in one equation, express all lengths, volumes and times in the same system of units, and convert where needed.

In plain words Liters and cubic meters, or minutes and seconds, cannot share an equation without a conversion.

Example Water flows at 22 liters per minute into a cylindrical tank of radius 0.50.5 meters. Since 11 liter is 0.0010.001 cubic meters, dVdt=0.002\frac{dV}{dt} = 0.002 cubic meters per minute. The volume is V=π(0.5)2h=0.25πhV = \pi (0.5)^2 h = 0.25\pi h, so

dhdt=0.0020.25π≈0.00255 meters per minute,\frac{dh}{dt} = \frac{0.002}{0.25\pi} \approx 0.00255 \text{ meters per minute},

about 0.250.25 centimeters per minute.

Signs of rates

Definition A rate dxdt\frac{dx}{dt} is positive when xx is increasing at that instant and negative when xx is decreasing. A quantity that is shrinking must be given a negative rate in the equations, whatever words the problem uses.

In plain words “Leaking at 30” means the rate is −30-30. Get the sign right going in and the answer’s sign will tell you the direction.

Example Air leaks from a spherical balloon at 3030 cubic centimeters per second, so dVdt=−30\frac{dV}{dt} = -30. When r=10r = 10 centimeters,

drdt=−304π⋅100=−340π≈−0.024,\frac{dr}{dt} = \frac{-30}{4\pi \cdot 100} = -\frac{3}{40\pi} \approx -0.024,

so the radius is shrinking by about 0.0240.024 centimeters per second.

Reading the answer

Definition The answer to a related rates problem is the value of a derivative at one instant. Interpreting it means stating which quantity changes, whether it increases or decreases, how fast, in which units and at which moment.

In plain words End with a sentence a person could act on, not just a number.

Example Car A is 0.30.3 kilometers east of an intersection and drives toward it at 5050 kilometers per hour, so x=0.3x = 0.3 and dxdt=−50\frac{dx}{dt} = -50. Car B is 0.40.4 kilometers north of it and drives toward it at 6060 kilometers per hour, so y=0.4y = 0.4 and dydt=−60\frac{dy}{dt} = -60. The distance between them satisfies z2=x2+y2z^2 = x^2 + y^2, so z=0.5z = 0.5 and

dzdt=xdxdt+ydydtz=0.3(−50)+0.4(−60)0.5=−78.\frac{dz}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{z} = \frac{0.3(-50) + 0.4(-60)}{0.5} = -78.

At that moment the cars are getting closer to each other at 7878 kilometers per hour.

Check yourself

  1. The radius of a circle grows at 22 centimeters per second. How fast is its area growing when the radius is 55 centimeters?
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The area is A=πr2A = \pi r^2, so

dAdt=2πrdrdt=2π⋅5⋅2=20π≈62.8.\frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi \cdot 5 \cdot 2 = 20\pi \approx 62.8.

The area grows at about 62.862.8 square centimeters per second.

  1. A 1313 meter ladder leans on a wall. Its foot is pulled away at 0.50.5 meters per second. How fast does the top move when the foot is 55 meters from the wall?
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When x=5x = 5 the height is y=169−25=12y = \sqrt{169 - 25} = 12. Solving 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 for the rate of the top gives

dydt=−xydxdt=−5⋅0.512=−524≈−0.21.\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{5 \cdot 0.5}{12} = -\frac{5}{24} \approx -0.21.

The top slides down at about 0.210.21 meters per second.

  1. In the leaking balloon example, why is it wrong to write dVdt=30\frac{dV}{dt} = 30?
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The volume is decreasing, so its rate must be negative. With +30+30 the computation would report a radius growing at 0.0240.024 centimeters per second, the opposite of what happens.

How this lesson connects

It also uses geometry and rates.