Related rates (HL)

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

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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. Identify the variable(s) with respect to which to differentiate. This is often time.
  2. Write an equation involving variables that depend on the variable from previous step. That variable does not need to appear in the equation. You may need to combine multiple formulas, use similar triangles, and use trigonometry.
  3. Implicitly differentiate the equation in 2) with respect to the variable identified in 1).
  4. Substitute in known quantities and rates. You may need to use the chain rule. And solve for the missing value (or rate).

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. (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}
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