7.8 Connected rates of change and inverse functions — Learn

Prerequisites: the chain rule from Section 7.7; differentiating $x^n$ and $e^{kx}$; rearranging a formula.

By the end of this section you can: connect two rates of change with the chain rule; find one rate from another by differentiating the linking formula; differentiate an inverse function using $\dfrac{dy}{dx} = \dfrac{1}{dx/dy}$; and explain how a rate can change as the quantity grows.

1. Connecting two rates

When two quantities are linked, the chain rule connects their rates. If $V$ depends on $r$, and $r$ depends on $t$, then

To use it: write the formula that links the two quantities, differentiate it to find the middle rate, then multiply by the known rate.

Example 1. A spherical balloon is inflated so that its volume increases at $100$ cm$^3$ per second. Find the rate at which the radius grows when the radius is $5$ cm.

The volume of a sphere is $V = \dfrac{4}{3}\pi r^3$, so

The chain rule gives $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$. Substitute $\dfrac{dV}{dt} = 100$ and $r = 5$:

So

The radius grows at about $0.318$ cm/s when the radius of the balloon is $5$ cm.

Example 2. A cube is heated and each edge grows at $0.5$ cm/s. Find the rate at which the volume grows when the edge is $4$ cm.

The volume is $V = x^3$, so $\dfrac{dV}{dx} = 3x^2$. Then

At $x = 4$:

Example 3. The area of a circle grows at $2$ cm$^2$ per second. Find the rate at which the radius grows when $r = 3$ cm.

The area is $A = \pi r^2$, so $\dfrac{dA}{dr} = 2\pi r$. The chain rule gives

Substitute $\dfrac{dA}{dt} = 2$ and $r = 3$:

2. The derivative of an inverse function

If $y = f(x)$ and the inverse function exists, then near a point

The gradient of the inverse function is the reciprocal of the gradient of the original function, measured at the matching point.

Example 4. The function $f(x) = x^3$ has inverse $f^{-1}(y) = y^{1/3}$. Find the gradient of the inverse at $y = 8$.

The original function has $f(2) = 8$, so the matching point is $x = 2$.

Therefore

Example 5. The function $f(x) = x^3 + 2x$ is increasing, so it has an inverse. Find $\left(f^{-1}\right)'(0)$.

First find the matching point. $f(0) = 0$, so the point $y = 0$ on the inverse matches $x = 0$ on the original function.

Therefore

Watch out. Three mistakes are common with connected rates.

Substituting the value before differentiating. In Example 1, put the expression $4\pi r^2$ into the chain rule first, then substitute $r = 5$. If you substitute too early you differentiate a number and get zero.

Inverting the chain rule the wrong way. The product is $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$, so the unknown rate is found by dividing, not multiplying: $\dfrac{dr}{dt} = \dfrac{dV/dt}{dV/dr}$.

Expecting the rate to stay constant. The same volume added to a larger balloon spreads over a larger surface, so the radius grows more slowly. In Example 1 the rate falls as $r$ grows. Re-check: after finding a rate, decide whether the answer should grow or shrink as the quantity grows, and compare with your expression.

Progress check

Return to the balloon in the chapter problem. Its volume grows at $\dfrac{dV}{dt} = 100$ cm$^3$/s.

1. Write down $\dfrac{dV}{dr}$ for $V = \dfrac{4}{3}\pi r^3$.
2. Find $\dfrac{dr}{dt}$ when $r = 5$ cm.
3. Find $\dfrac{dr}{dt}$ when $r = 10$ cm.
4. Explain in one sentence why the radius grows more slowly as the balloon gets bigger.

You should now be able to

• write the chain rule that connects two rates;
• differentiate the formula that links two quantities;
• find an unknown rate by substituting the known rate and the given value;
• use $\dfrac{dy}{dx} = \dfrac{1}{dx/dy}$ to differentiate an inverse function;
• explain how a rate can change as the quantity grows.