Substitute information at the correct stage
Pumping up a balloon
Air is pumped into a spherical balloon. At the moment of the question the radius is 10 centimeters, and it grows at 0.3 centimeter per second. How fast does the volume grow at that moment?
The teal disc is the balloon with the orange radius \(r\), and its volume is \(V = \frac{4}{3}\pi r^3\). Move the time slider. The graph below shows the volume against time, and the dashed line marks the moment of the question.
Press the first button, which puts \(r = 10\) into \(V\) first. The volume becomes a fixed number, so its graph is the flat red line with the rate 0, which is wrong because the balloon grows.
Press the second button, which differentiates first. The rate of \(V\) is \(4\pi r^2\) times the rate of \(r\) at every time, so the values go in only at the end.
A value holds at one moment
A quantity q changes with time, and the curve shows q against t. The orange point marks the moment where the question gives the value of q.
Press the first button. If you put this value in first, q becomes a constant, so its graph is the flat red line. This line has the slope 0, so the rate of q is lost.
Press the second button. If you differentiate first, q is still a function of t, so the orange tangent keeps the slope of q. The value holds only at that moment, so it goes in at the end.
Notation
Write the relationship in symbols. It holds at every time.
\[V = \frac{4}{3}\pi r^3\]Differentiate both sides with respect to t. Keep every changing quantity as a symbol.
\[\cy{\frac{dV}{dt}} = 4\pi r^2\,\cy{\frac{dr}{dt}}\]Put the values of the moment into the result. All the values must belong to the same moment.
\[4\pi(10)^2(0.3) = \cy{120\pi}\]If the wanted rate is not alone, solve for it. Then give its units, here cubic centimeters per second.
When it fails
If \(r = 10\) goes in first, \(V\) becomes the fixed number \(\frac{4000\pi}{3}\), whose derivative is 0.
The balloon is growing, so this rate is wrong. Only a length that is fixed in the whole problem can go in at the start.
Example 1: A melting ice cube
An ice cube has the edge \(s\) centimeters. The edge shrinks at 0.2 centimeter per minute. Find the rate of change of the volume when \(s = 6\).
Write the relationship. It holds at every time.
\[V = s^3\]Differentiate with respect to \(t\). Keep \(s\) as a symbol, because it changes.
\[\frac{dV}{dt} = 3s^2\,\frac{ds}{dt}\]Put in the values of the moment. The edge shrinks, so its rate is negative.
\[3(6)^2(-0.2) = -21.6\]The volume decreases at 21.6 cubic centimeters per minute.
Example 2: A plane over a radar
A plane flies level 3 kilometers above a radar station. Its horizontal distance \(x\) from the point above the station decreases at 300 kilometers per hour. Find the rate of its distance \(s\) from the station when \(x = 4\).
Write the relationship. The height 3 is fixed, so it can go in now.
\[s^2 = x^2 + 3^2\]A tempting wrong move: At this moment \(x = 4\) and \(s = 5\). Put these values in first.
\[\cr{25 = 16 + 9}\]The red equation has only fixed numbers, so differentiating it gives \(0 = 0\) and no rates. But the orange arrow shows that \(x\) and \(s\) change.
Differentiate the relationship with respect to \(t\). Keep \(x\) and \(s\) as symbols.
\[2s\,\frac{ds}{dt} = 2x\,\frac{dx}{dt}\]Substitute \(x = 4\), \(s = 5\) and \(\frac{dx}{dt} = -300\), which is negative because \(x\) decreases. Then solve for the rate of \(s\).
\[2(5)\,\frac{ds}{dt} = 2(4)(-300)\]The distance to the station decreases at 240 kilometers per hour.
\[\frac{ds}{dt} = -240\]
Practice
Sand falls into a pile that has the shape of a cone. The volume and the height of the pile satisfy \(V = \frac{\pi}{12}h^3\). The sand falls at 8 cubic meters per minute. How fast does the height grow when \(h = 4\) meters?
The height in the picture grows, so differentiate \(V\) with respect to \(t\) before you use \(h = 4\).
Solution
Differentiate with respect to \(t\). Keep \(h\) as a symbol.
\[\frac{dV}{dt} = \frac{\pi}{4}h^2\,\frac{dh}{dt}\]Substitute \(h = 4\) and \(\frac{dV}{dt} = 8\).
\[8 = \frac{\pi}{4}(4)^2\,\frac{dh}{dt} = 4\pi\,\frac{dh}{dt}\]Solve for the rate of \(h\). The height grows at \(\frac{2}{\pi}\) meters per minute.
\[\frac{dh}{dt} = \frac{2}{\pi}\]
Questions
A hailstone is a sphere of radius \(r\) centimeters with the volume \(V = \frac{4}{3}\pi r^3\). The graph shows \(V\) against \(r\), and the slope of the orange tangent is the rate of \(V\) with respect to \(r\).
At \(r = 2\), the radius grows at 0.1 centimeter per minute. Which value is the rate of \(V\) with respect to time?
- \(1.6\pi\) cubic centimeters per minute
- \(\frac{32\pi}{30}\) cubic centimeters per minute
- 0 cubic centimeters per minute
- \(16\pi\) cubic centimeters per minute
Answer
\(1.6\pi\) cubic centimeters per minute
For the same rate of \(r\), how does the rate of \(V\) at \(r = 3\) compare with the rate at \(r = 2\)?
- \(\frac{9}{4}\) times as large
- \(\frac{3}{2}\) times as large
- \(\frac{27}{8}\) times as large
- The same
Answer
\(\frac{9}{4}\) times as large
Two quantities satisfy \(x^2 + y^2 = 169\). At one moment \(x = 5\), \(y = 12\) and \(\frac{dx}{dt} = 3\). Find \(\frac{dy}{dt}\) at that moment.
- \(-\frac{5}{4}\)
- \(\frac{5}{4}\)
- \(-\frac{36}{5}\)
- 0
Answer
\(-\frac{5}{4}\)
Summary
You differentiate the relationship while its changing quantities are still symbols, and then you put in the values of the moment.
A value put in first turns a changing quantity into a fixed number with the rate 0, so the rate you need is lost.
In every related rates problem, only a quantity that is fixed throughout, such as the height of the plane, can go in before you differentiate.
This is one of four skills in topic 4.4 of Which & Why: AP Calculus.