Differentiate with respect to time
Two cars leaving a corner
Two cars leave a corner at the same time. Car A drives east at 60 kilometers per hour, and car B drives north at 80 kilometers per hour.
After \(t\) hours, the cars are \(x\) and \(y\) kilometers from the corner and \(z\) kilometers from each other. Move the slider, and all three distances change.
At every time, \(z^2 = x^2 + y^2\), and both sides change with time because \(z\), \(x\) and \(y\) all change.
To find how fast each side changes, you differentiate each term with respect to t, so each term brings the rate of its own quantity.
A rate for each quantity
A quantity \(F\) depends on \(q\), and \(q\) depends on time \(t\). The left graph shows \(F\) against \(q\), the right graph shows \(q\) against \(t\), and the orange lines are their tangents.
A small green step \(dt\) in \(t\) changes \(q\) by about \(\frac{dq}{dt}\,dt\). This change \(dq\) then changes \(F\) by about \(\frac{dF}{dq}\,dq\).
If you make the step smaller, the green steps follow the orange tangents closely, so the estimates become exact.
The two slopes multiply, so at every \(t\) the rate of \(F\) is \(\frac{dF}{dt} = \frac{dF}{dq}\cdot\frac{dq}{dt}\). This is the chain rule.
Notation
Start with the relationship. Every changing quantity is a function of \(t\), so each one has a rate.
\[z^2 = x^2 + y^2\]Differentiate both sides with respect to \(t\). The power rule gives \(2z\), and because \(z\) depends on \(t\), the chain rule adds the factor \(\cy{\frac{dz}{dt}}\).
\[\frac{d}{dt}\left(z^2\right) = 2z\,\cy{\frac{dz}{dt}}\]Each changing quantity brings its own rate, so the right side has one term for \(x\) and one for \(y\).
\[2z\,\cy{\frac{dz}{dt}} = 2x\,\cy{\frac{dx}{dt}} + 2y\,\cy{\frac{dy}{dt}}\]A fixed number has the derivative 0, so it leaves no term.
\[\frac{d}{dt}(25) = 0\]If two changing quantities are multiplied, use the product rule.
\[\frac{d}{dt}(xy) = x\,\frac{dy}{dt} + y\,\frac{dx}{dt}\]
When it fails
If you leave out \(\frac{dx}{dt}\), you treat \(x\) as a fixed number, so the equation becomes \(2z\,\frac{dz}{dt} = 2y\,\frac{dy}{dt}\).
At \(t = 1\), \(x = 60\), \(y = 80\) and \(z = 100\), so this equation gives \(2(100)\frac{dz}{dt} = 2(80)(80)\) and \(\frac{dz}{dt} = 64\) kilometers per hour. But \(z = 100t\) grows by 100 kilometers each hour, as the orange bar shows.
Example 1: A spreading ripple
A circular ripple has area \(A = \pi r^2\), and its radius \(r\) changes with time. Write the rate of change of \(A\) in terms of the rate of change of \(r\).
Both \(A\) and \(r\) change with time. Differentiate both sides with respect to \(t\).
\[\frac{dA}{dt} = \frac{d}{dt}\left(\pi r^2\right)\]The constant \(\pi\) stays, and the chain rule turns \(r^2\) into \(2r\) times the rate of \(r\).
\[\frac{dA}{dt} = 2\pi r\,\cy{\frac{dr}{dt}}\]The factor \(2\pi r\) is the circumference. When the radius grows by \(dr\), the disc gains a thin green ring of about this length and width \(dr\), so \(dA \approx 2\pi r\,dr\). Dividing by \(dt\) gives the rate above.
Example 2: A growing rectangle
A rectangle has length \(\ell\) and width \(w\). Both grow with time, and the area is \(A = \ell w\). Write the rate of change of \(A\) in terms of the rates of \(\ell\) and \(w\).
The area is a product of two changing quantities, so use the product rule.
A tempting wrong move: Differentiate only the width.
\[\cr{\frac{dA}{dt} = \ell\,\frac{dw}{dt}}\]This treats the length as a fixed number. The green strip on top comes from the growth of \(w\), but the length grows too and adds a strip on the right. This equation leaves out that strip, so it is red.
Each changing quantity brings its own term, one for each green strip.
\[\frac{dA}{dt} = \ell\,\frac{dw}{dt} + w\,\frac{d\ell}{dt}\]
Practice
A gas in a cylinder has pressure \(P\) and volume \(V\) that change with time. They satisfy \(PV = 600\). Differentiate with respect to \(t\), and solve for the rate of \(P\).
The product \(PV\) has two changing quantities, and the number 600 is fixed.
Solution
Use the product rule on the left side. The right side is a fixed number, so its derivative is 0.
\[P\,\frac{dV}{dt} + V\,\frac{dP}{dt} = 0\]Solve for the rate of \(P\).
\[\frac{dP}{dt} = -\frac{P}{V}\,\frac{dV}{dt}\]If the volume decreases, \(\frac{dV}{dt}\) is negative. The minus sign then makes \(\frac{dP}{dt}\) positive, so the pressure rises.
Questions
A lighthouse is 3 kilometers from a straight shore. Its beam turns, and the spot of light on the shore is \(x\) kilometers from the point nearest to the lighthouse. The beam makes the angle \(\theta\) with the line to that point, so \(x = 3\tan\theta\). The figure shows the same small turn of the beam at \(\theta = 0\) and at \(\theta = \pi/3\). Each green piece is the stretch of shore that the spot crosses.
The beam turns at the same speed. Where on the shore does the spot move fastest?
- Far from the nearest point, where the angle is large
- Near the nearest point, where the angle is 0
- At the same speed everywhere
- The picture cannot show this
Answer
Far from the nearest point, where the angle is large
The beam turns at the same rate at both angles in the picture. How many times faster does the spot move at \(\theta = \pi/3\) than at \(\theta = 0\)?
- 4 times
- 2 times
- \(\sqrt{3}\) times
- The same speed
Answer
4 times
A square plate has side \(s\) and area \(A = s^2\). The side is 5 centimeters and grows at 0.2 centimeter per minute. How fast does the area grow?
- 2 square centimeters per minute
- 10 square centimeters per minute
- 0.2 square centimeter per minute
- 1 square centimeter per minute
Answer
2 square centimeters per minute
Summary
You differentiate both sides of the relationship with respect to t, and each changing quantity brings a factor of its own rate.
This works because every quantity depends on t, so the chain rule applies to each one, while a fixed number has the derivative 0.
When two changing quantities are multiplied, you use the product rule, and you keep the rates as symbols until a later step.
This is one of four skills in topic 4.4 of Which & Why: AP Calculus.