Related Rates

Two quantities that change together

Blow air into a balloon and two things change at once: its radius r and its volume V. Both change with time t, and they are tied together by a formula, $V=\frac{4\pi r^3}{3}$V = 4πr³/3. That formula is the link between them.

Each quantity has its own rate of change. $\frac{dr}{dt}$dr/dt is how fast the radius grows, in centimeters per second, and $\frac{dV}{dt}$dV/dt is how fast the volume grows, in cubic centimeters per second. Because V is fixed by r, the two rates are tied together too. A related-rates problem gives one of them and asks for the other.

The chain rule links the rates

The volume depends on the radius, and the radius depends on time. That is a chain, so the chain rule gives the rate of the volume with respect to time: $\frac{dV}{dt}=\frac{dV}{dr}\times \frac{dr}{dt}$dV/dt = dV/dr × dr/dt.

Read it as two rates multiplied. $\frac{dV}{dr}$dV/dr says how many cubic centimeters of volume each centimeter of radius carries, and $\frac{dr}{dt}$dr/dt says how many centimeters of radius arrive each second. Their product is cubic centimeters per second. The units are a good check: cm³ per cm, times cm per second, leaves cm³ per second.

A growing cube

A cube has side r centimeters, and r is growing at 2 cm per second. How fast is the volume growing at the moment r = 4?

The link is $V=r^3$V = r³, so $\frac{dV}{dr}=3r^2$dV/dr = 3r². At r = 4 that is $3\times 4^2=48$3 × 4² = 48 cm³ for each centimeter of side. Multiply by the rate the side grows: $\frac{dV}{dt}=48\times 2=96$dV/dt = 48 × 2 = 96 cm³ per second.

Check it against the cube itself. A thousandth of a second later the side is 4 + 2 × 0.001 = 4.002, and the volume has grown from 64 to $4.002^3=64.096024$4.002³ = 64.096024. That is an average rate of $\left(4.002^3-64\right)/0.001=96.048$(4.002³ − 64) / 0.001 = 96.048 cm³ per second over that thousandth, very close to 96.

The volume of the cube against time, when the side is 4 + 2t and t = 0 is the moment the side is 4. The gold tangent at that moment has gradient 96: the volume is growing at 96 cm³ per second there.

A cube of side 2 grows by $\Delta x$Δx. The new volume is the old one plus three gold slabs, three rods and a small corner. As $\Delta x$Δx shrinks, the slabs are nearly all of the growth, so each unit of side adds $3x^2=12$3x² = 12 units of volume. If this side grew at 2 per second, the volume would grow at 12 × 2 = 24 per second.

The balloon, both ways

For the balloon, $V=\frac{4\pi r^3}{3}$V = 4πr³/3, so $\frac{dV}{dr}=4\pi r^2$dV/dr = 4πr². If the radius is 10 cm and growing at 0.5 cm per second, then $\frac{dV}{dt}=4\pi \times 10^2\times 0.5=200\pi$dV/dt = 4π × 10² × 0.5 = 200π, about 628.3 cm³ per second.

The question can run the other way. Air goes in at 100 cm³ per second; how fast is the radius growing when it is 5 cm? Now $\frac{dV}{dt}$dV/dt is known and $\frac{dr}{dt}$dr/dt is wanted. At r = 5, $\frac{dV}{dr}=4\pi \times 25=100\pi$dV/dr = 4π × 25 = 100π, and $100=100\pi \times \frac{dr}{dt}$100 = 100π × dr/dt, so $\frac{dr}{dt}=\frac{100}{100\pi }=\frac{1}{\pi }$dr/dt = 100/(100π) = 1/π, about 0.3183 cm per second.

A ripple spreads on a pond as a circle of area $A=\pi r^2$A = πr², whose derivative with respect to r is $2\pi r$2πr. When the radius is 5 cm and growing at 3 cm per second, the area grows at $2\pi \times 5\times 3=30\pi$2π × 5 × 3 = 30π, about 94.25 cm² per second.

Differentiate first, then substitute

Every problem of this kind has the same shape. Name the quantities and the units. Write the equation that links them. Differentiate it with respect to time. Only then put in the values at the moment the question asks about.

The order matters. Put r = 4 into $V=r^3$V = r³ first and the link becomes V = 64, a constant, whose derivative is 0. The radius is still changing; substituting early has frozen it, and its rate has vanished from the working.

A sliding ladder

Sometimes both quantities in the link are changing. A ladder 5 m long leans against a wall. Its foot is x meters from the wall and its top is y meters up, so $x^2+y^2=25$x² + y² = 25. The foot slides away at 0.5 m per second. How fast is the top moving when the foot is 3 m out?

Differentiate every term with respect to time, using the chain rule on each: $2x\left(\frac{dx}{dt}\right)+2y\left(\frac{dy}{dt}\right)=0$2x(dx/dt) + 2y(dy/dt) = 0. Now substitute. When x = 3, $y=\sqrt{25-9}=4$y = √(25 − 9) = 4, and $\frac{dx}{dt}=0.5$dx/dt = 0.5, so $2\times 3\times 0.5+2\times 4\times \frac{dy}{dt}=0$2 × 3 × 0.5 + 2 × 4 × dy/dt = 0. That gives $3+8\left(\frac{dy}{dt}\right)=0$3 + 8(dy/dt) = 0, so $\frac{dy}{dt}=-\frac{3}{8}=-0.375$dy/dt = −3/8 = −0.375, in meters per second.

The minus sign says y is getting smaller: the top slides down the wall at 0.375 m per second. Check: a thousandth of a second later the foot is at 3.0005 m, the top is at a height of $\sqrt{25-3.0005^2}=3.999625$√(25 − 3.0005²) = 3.999625 meters, and it has dropped at 0.375 m per second over that thousandth, to three decimal places.

The usual mistakes

Adding the rates. With $\frac{dV}{dr}=6$dV/dr = 6 and $\frac{dr}{dt}=3$dr/dt = 3, the answer is 6 × 3 = 18, not 6 + 3 = 9. The two rates are in different units, cm³ per cm and cm per second, and they combine by multiplying.

Dividing the rates. 6 ÷ 3 = 2 runs the chain backward. Divide only when the volume rate is known and the radius rate is wanted, as in the balloon question.

Forgetting to work out $\frac{dV}{dr}$dV/dr at the moment asked about. For $A=r^2$A = r² at r = 5, the derivative 2r is 2 × 5 = 10, so with r growing 3 per second, A grows at 10 × 3 = 30 per second.