Use a composite derivative in context
A ripple spreads
A ripple spreads on a pond as a teal circle with an orange radius. After \(t\) seconds, the radius is \(r = 3t\) centimeters and the area is \(A = \pi r^2\) square centimeters.
At \(t = 2\), the radius is 6 centimeters. The orange tangent on the left graph has slope 3, so the radius grows 3 centimeters each second.
At \(r = 6\), the tangent on the middle graph has slope \(2\pi(6) = 12\pi\), so the area grows \(12\pi\) square centimeters for each centimeter of radius.
Multiply the two rates. The centimeters cancel, so the area grows \(36\pi\) square centimeters each second, about 113.
Drag the time to the right. The slope of the middle graph, \(2\pi r\), grows with the radius, so the area grows faster each second.
Any linked quantities
A quantity \(Q\) depends on \(x\), and \(x\) depends on the time \(t\).
The slope of the left graph is the rate of \(x\) per unit of time. The slope of the middle graph is the rate of \(Q\) per unit of \(x\).
Drag the time. At every time, the product of these two slopes equals the slope of the right graph, the rate of \(Q\) per unit of time.
The unit of \(x\) is in one numerator and one denominator, so it cancels.
Notation
The quantity \(Q\) depends on \(x\), and \(x\) depends on \(t\).
\[Q = f(x), \qquad x = g(t)\]The rate of \(Q\) with respect to time is the product of two rates.
\[\frac{dQ}{dt} = \cy{\frac{dQ}{dx}} \cdot \cy{\frac{dx}{dt}}\]The units multiply in the same way.
\[\begin{aligned} &\frac{\text{units of } Q}{\text{units of } x} \cdot \frac{\text{units of } x}{\text{units of } t} \\ &= \frac{\text{units of } Q}{\text{units of } t} \end{aligned}\]To answer at a given time, first find \(x\) at that time. Then read \(\frac{dQ}{dx}\) at that value of \(x\), not at \(t\).
State the answer in a sentence, with its unit. A positive rate means the quantity grows, and a negative rate means it shrinks.
When it fails
The rule \(\frac{d}{dx}\sin x = \cos x\) needs \(x\) in radians. If \(t\) counts degrees instead, the angle reaches only about 3 degrees, so the graph of \(Q\) against \(t\) barely rises.
An angle of \(t\) degrees is \(x = \frac{\pi}{180}t\) radians, so the left graph has slope \(\frac{\pi}{180}\). The chain rule then gives a rate of about \(\frac{\pi}{180} \approx 0.0175\) per degree, not about 1.
Example 1: time inside a sine
The height of the tide at a harbor is \(h(t) = 2 + 1.5\sin\!\left(\frac{\pi t}{6}\right)\) meters, where \(t\) is the number of hours after midnight. Find how fast the tide rises at \(t = 2\).
The inner function is \(u = \frac{\pi t}{6}\), in radians. Its rate is \(\frac{\pi}{6}\) radians per hour.
\[\frac{du}{dt} = \cy{\frac{\pi}{6}}\]The outer function is \(h = 2 + 1.5\sin u\). Its rate is \(1.5\cos u\) meters per radian.
\[\frac{dh}{du} = \cy{1.5\cos u}\]Multiply the rates. The radians cancel, so the unit is meters per hour.
\[\begin{aligned} \frac{dh}{dt} &= 1.5\cos\!\left(\frac{\pi t}{6}\right) \cdot \frac{\pi}{6} \\ &= \frac{\pi}{4}\cos\!\left(\frac{\pi t}{6}\right) \end{aligned}\]At \(t = 2\), the inner function gives \(u = \frac{\pi}{3}\), and \(\cos\frac{\pi}{3} = \frac12\). The rate is positive, so the tide rises \(\frac{\pi}{8}\) meters each hour, about 0.39.
\[\frac{dh}{dt}\bigg|_{t=2} = \cy{\frac{\pi}{8}} \approx 0.39 \text{ m/h}\]
Example 2: linked sizes
An ice cube melts. Its edge is \(x(t) = 10 - 0.5t\) centimeters, where \(t\) is in minutes. Its volume is \(V = x^3\). Find \(\frac{dV}{dt}\) at \(t = 4\).
The edge shrinks at a steady rate. At \(t = 4\), the edge is \(10 - 0.5(4) = 8\) centimeters.
\[\begin{aligned} \frac{dx}{dt} &= \cy{-0.5} \text{ cm/min} \\ x(4) &= \cy{8} \text{ cm} \end{aligned}\]A tempting wrong move: The volume changes by \(3x^2\) cubic centimeters for each centimeter of edge. Put in \(t = 4\).
\[\begin{aligned} \frac{dV}{dx} &= 3(4)^2 = 48 \\ \frac{dV}{dt} &= 48 \cdot (-0.5) = -24 \end{aligned}\]The number 4 is a time in minutes, but \(3x^2\) needs an edge in centimeters. The red line on the middle graph is the wrong tangent, at an edge of 4. The edge at \(t = 4\) is 8 centimeters, so the orange tangent has slope 192, not 48.
Put in the edge, 8 centimeters.
\[\frac{dV}{dx} = 3(8)^2 = \cy{192} \text{ cm}^3\text{ per cm}\]Multiply the rates. The centimeters cancel.
\[\begin{aligned} \frac{dV}{dt} &= 192 \cdot (-0.5) \\ &= \cy{-96} \text{ cm}^3\text{/min} \end{aligned}\]The rate is negative, so the volume shrinks by 96 cubic centimeters each minute.
Practice
A square metal sheet has side \(s = 20 + 0.05T\) centimeters at a temperature of \(T\) degrees Celsius. Its area is \(A = s^2\). Find \(\frac{dA}{dT}\) at \(T = 100\), with its unit.
Find the side at T = 100 first. Read the area rate dA/ds at that side, not at 100.
Solution
At \(T = 100\), the side is \(20 + 0.05(100) = 25\) centimeters. The side grows at \(\frac{ds}{dT} = 0.05\) centimeters per degree.
\[s = 25, \qquad \frac{ds}{dT} = 0.05\]At \(s = 25\), the area rate is \(2s = 50\) square centimeters per centimeter.
\[\frac{dA}{ds} = 2s = 50\]Multiply. The centimeters cancel, so the area grows 2.5 square centimeters for each degree.
\[\frac{dA}{dT} = 50 \cdot 0.05 = 2.5 \text{ cm}^2\text{ per }^\circ\text{C}\]
Questions
A snowball melts. Its radius is \(r(t) = 6 - 0.5t\) centimeters, where \(t\) is in minutes. Its volume is \(V = \frac43\pi r^3\). The figure shows \(t = 2\).
The orange tangent on the radius graph has slope \(-0.5\). What does this number mean?
- The radius shrinks 0.5 centimeters each minute.
- The radius is 0.5 centimeters.
- The volume shrinks 0.5 cubic centimeters each minute.
Answer
The radius shrinks 0.5 centimeters each minute.
The middle graph, volume against radius, is read at the radius of the snowball at \(t = 2\). Which radius is that?
- 5 centimeters
- 2 centimeters
- 6 centimeters
Answer
5 centimeters
Find \(\frac{dV}{dt}\) at \(t = 2\), in cubic centimeters per minute.
- \(-50\pi\)
- \(100\pi\)
- \(-8\pi\)
- \(50\pi\)
Answer
\(-50\pi\)
Summary
The ripple's area grows at \(\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}\), where \(\frac{dA}{dr}\) is read at the radius of that moment.
This works because the centimeters of radius cancel, which leaves square centimeters per second.
You use this product when one quantity depends on another that changes in time, and you give the answer with its unit.
This is one of four skills in topic 3.1 of Which & Why: AP Calculus.