Related rates (HL)

As mentioned in Implicit differentiation, the equation should involve variables that change with respect to other variables.

Contents

Hints

Change rule states that

 dy dx⋅ dx dt= dy dt \frac{\d y}{\d x} \cdot \frac{\d x}{\d t} = \frac{\d y}{\d t}

And from implicit differentiation

 dx dy=( dy dx)−1 \frac{\d x}{\d y} = \left(\frac{\d y}{\d x}\right)^{-1}

Steps

  1. Draw a diagram. Label known variables and quantities.
  2. Identify the variable with respect to which to differentiate. This is usually time (tt). If you are given (or asked for) a rate that is not with respect to this variable, you probably need to use chain rule sometime in the problem.
  3. Write down relevant equations involving variables in the problem. This may involve geometry formulas, similar triangles, trigonometry, and/or other equations. Combine your equations into a single equation and simplify (so that it is easy to differentiate).
  4. Implicitly differentiate the equation in 3) with respect to the variable identified in 2).
  5. Substitute in known values and solve. You may need to use the chain rule. You may need to solve a differential equation.

Example

Example: A semispherical bowl of radius 1 dm1 \text{ dm} is filled with water at 0.02 dm3 s−10.02 \text{ dm}^3 \text{ s}^{-1}. Find the radius of the water level when the height is rising at 0.01 dm s−10.01 \text{ dm s}^{-1}.

Let f(x)=−1−x2+1, 0≤x≤1f(x) = -\sqrt{1 - x^2} + 1, \, 0 \leq x \leq 1. Rotating ff a full revolution about yy-axis yields a semispherical bowl of radius 11. We want −- square root so the bowl opens upwards; +1+1 so it is above xx-axis. We need to integrate with respect to the axis of rotation, using

f−1(x)=1−(x−1)2, 0≤x≤1f^{-1}(x) = \sqrt{1 - (x - 1)^2}, \, 0 \leq x \leq 1

The volume of water at depth hh is

V=π∫0h1−(y−1)2 dy\begin{align*} V &= \pi \int_0^h 1 - (y - 1)^2 \d y \\ \end{align*}

Confused by notation? See using definite integral to define a function.

Now implicitly differentiate with respect to time tt,

 dV dt= d dt(π∫0h1−(x−1)2 dy)= d dh(π∫0h1−(x−1)2 dy)⋅ dh dt=π(1−(h−1)2)⋅ dh dt\begin{align*} \frac{\d V}{\d t} &= \frac{\d}{\d t}\left(\pi \int_0^h 1 - (x - 1)^2 \d y\right) \\ &= \frac{\d}{\d h}\left(\pi \int_0^h 1 - (x - 1)^2 \d y\right) \cdot \frac{\d h}{\d t} \\ &= \pi\left(1 - (h - 1)^2\right) \cdot \frac{\d h}{\d t} \end{align*}

by using chain rule and the fundamental theorem of calculus. (Or you can just integrate then differentiate to check.)

We also have radius

r=f−1(h)=1−(h−1)2>0r = f^{-1}(h) = \sqrt{1 - (h - 1)^2} > 0

Then

 dV dt=πr2 dh dtr=1π dV dt÷ dh dt=1π(0.02)÷(0.01)=2π≈0.798■\begin{align*} \frac{\d V}{\d t} &= \pi r^2 \frac{\d h}{\d t} \\ r &= \sqrt{\frac1\pi \frac{\d V}{\d t} \div \frac{\d h}{\d t}} \\ &= \sqrt{\frac1\pi (0.02) \div (0.01)} \\ &= \sqrt{\frac2\pi} \approx 0.798 \qed \end{align*}

The desired radius is 0.798 dm0.798 \text{ dm}.

It can be generalized that for all volumes of revolution (with circular cross sections),

 dV dt=πr2 dh dt\frac{\d V}{\d t} = \pi r^2 \frac{\d h}{\d t}