Solve for the requested rate
A kite carried by the wind
A kite flies 12 meters above the hand of a person. The wind carries it sideways, away from the person, at 3 meters per second. The first green arrow shows this rate, and green marks a positive rate.
The orange segment \(x\) is the horizontal distance from the person to the point below the kite. The string \(s\) is the longest side of a right triangle, so \(s^2 = x^2 + 144\) at every moment.
Differentiate both sides with respect to \(t\) and divide by 2, which gives \(s\,ds/dt = x\,dx/dt\).
The requested rate is \(ds/dt\), how fast the string lengthens, which the second green arrow shows. Divide by \(s\) to isolate it: \(ds/dt = (x/s)\,dx/dt\).
At \(x = 16\), \(s = 20\) because \(256 + 144 = 400\), so \(ds/dt = (16/20)(3) = 2.4\) meters per second.
Drag the kite to other distances. The readout shows that \(ds/dt\) changes, although \(dx/dt\) stays 3.
The given rate times a ratio
The string lengthens at the speed of the kite times the ratio \(x/s\). The dashed line shows that the second arrow is the part of the first arrow that points along the string.
Drag the kite to the right. The angle \(\theta\) shrinks and the second arrow grows, because \(x/s\) is the cosine of \(\theta\).
This cosine is between 0 and 1, so the string lengthens more slowly than the kite moves.
The ratio changes from moment to moment, so you substitute the values for the moment of the question.
Notation
Differentiate \(s^2 = x^2 + 144\) and divide by 2. Each rate is multiplied only by a length, so one division isolates it.
\[\cy{s}\,\frac{ds}{dt} = \cy{x}\,\frac{dx}{dt}\]Isolate the requested rate before you substitute. Divide both sides by \(s\), which is not 0.
\[\frac{ds}{dt} = \frac{x}{s}\,\frac{dx}{dt}\]Substitute the values for one moment, all in meters and seconds.
\[\frac{ds}{dt} = \frac{16}{20}\cdot 3\]Calculate the result. Write the answer with its unit.
\[\frac{ds}{dt} = \cy{2.4}\ \text{m/s}\]Keep a fraction such as \(12/5\) exact until the last step.
When it fails
Suppose the problem gives \(ds/dt\) and asks for \(dx/dt\). To isolate \(dx/dt\), you divide by \(x\), so you need \(x \ne 0\).
Move the kite directly above the person, where \(x = 0\). The string is at its shortest, 12 meters, so \(s\) does not change at that moment. Then \(ds/dt = 0\) for any speed of the kite, so \(ds/dt\) cannot give \(dx/dt\).
Example 1: A heated sheet
A square metal sheet expands when heated. Its side \(s\) and its area \(A\) both grow. At one moment \(s = 8\) cm, and the area grows at 4 cm²/min. How fast does the side grow?
The area of a square is its side squared.
\[\cb{A} = \cy{s}^2\]Differentiate with respect to \(t\). The green strip is the area added as \(s\) grows: it is about \(2s\) long and as wide as the growth of \(s\).
\[\frac{dA}{dt} = 2\cy{s}\,\frac{ds}{dt}\]Isolate the requested rate, \(ds/dt\), by dividing by \(2s\).
\[\frac{ds}{dt} = \frac{1}{2s}\,\frac{dA}{dt}\]Substitute \(s = 8\) and \(dA/dt = 4\).
\[\frac{ds}{dt} = \frac{4}{16} = \cy{\tfrac14}\ \text{cm/min}\]
Example 2: Two cars
Car A drives toward an intersection at 40 mph, on a road from the west. Car B drives away from it at 50 mph, on a road to the north. At one moment A is 0.9 mi and B is 1.2 mi from the intersection. How fast does the distance between the cars change?
Let \(x\) and \(y\) be the distances of the cars from the intersection, and \(z\) the distance between them. Then \(z^2 = x^2 + y^2\). Differentiate and divide by \(2z\).
\[\frac{dz}{dt} = \frac{x\,\frac{dx}{dt} + y\,\frac{dy}{dt}}{z}\]At this moment \(z = \sqrt{0.81 + 1.44} = 1.5\) mi. The arrows show the motion: red where a distance decreases, green where it increases.
A tempting wrong move: Substitute \(dx/dt = 40\) and \(dy/dt = 50\), the speeds of the cars.
\[\frac{dz}{dt} = \frac{0.9(40) + 1.2(50)}{1.5} = 64\]Car A drives toward the intersection, so \(x\) decreases and \(dx/dt\) is negative. The solid red arrow on car A points toward the intersection, and the dashed arrow shows the motion that \(+40\) would describe.
Give each rate its sign. Car A has \(dx/dt = -40\) because \(x\) decreases, and car B has \(dy/dt = +50\) because \(y\) increases.
\[\begin{aligned} \frac{dz}{dt} &= \frac{0.9(-40) + 1.2(50)}{1.5} \\ &= \cy{16}\ \text{mph} \end{aligned}\]
Practice
A camera is 400 m from a launch pad. A rocket rises straight up at 100 m/s. When the rocket is 300 m high, how fast does the angle \(\theta\) of the line of sight above the ground increase?
In the picture, \(h\) is the side opposite \(\theta\) and 400 m is the side next to it, so the tangent of \(\theta\) links them.
Solution
The camera, the pad and the rocket form a right triangle, so the tangent of \(\theta\) is the height over 400.
\[\tan\theta = \frac{h}{400}\]Differentiate with respect to \(t\). The derivative of \(\tan\theta\) is \(\sec^2\theta\,d\theta/dt\), and \(\sec^2\theta = 1/\cos^2\theta\).
\[\frac{1}{\cos^2\theta}\,\frac{d\theta}{dt} = \frac{1}{400}\,\frac{dh}{dt}\]Multiply both sides by \(\cos^2\theta\).
\[\frac{d\theta}{dt} = \frac{\cos^2\theta}{400}\,\frac{dh}{dt}\]At \(h = 300\), the line of sight is 500 m long, so \(\cos\theta = 400/500 = 4/5\). Substitute this and \(dh/dt = 100\).
\[\frac{d\theta}{dt} = \frac{(4/5)^2}{400}\cdot 100 = \frac{4}{25}\ \text{rad/s}\]
Questions
A coin expands when heated. Its radius is 5 mm and grows at 0.2 mm/s. Its area is A and its circumference is C.
How fast does the area A grow?
- \(2\pi\ \text{mm}^2/\text{s}\)
- \(10\pi\ \text{mm}^2/\text{s}\)
- \(\pi\ \text{mm}^2/\text{s}\)
- \(5\pi\ \text{mm}^2/\text{s}\)
Answer
\(2\pi\ \text{mm}^2/\text{s}\)
How fast does the circumference C grow?
- \(0.4\pi\ \text{mm/s}\)
- \(2\pi\ \text{mm/s}\)
- \(10\pi\ \text{mm/s}\)
Answer
\(0.4\pi\ \text{mm/s}\)
The side \(s\) of an equilateral triangle grows, and its area is \(A = \tfrac{\sqrt3}{4}s^2\). When \(s = 6\) cm, the area grows at \(6\sqrt3\) cm²/min. How fast does \(s\) grow?
- 2 cm/min
- 24 cm/min
- 4 cm/min
Answer
2 cm/min
Summary
You differentiate the relation, isolate the requested rate with symbols, and then substitute the values for the instant.
This works because the requested rate equals the given rate times a ratio, such as \(x/s\), whose value depends on the moment.
You use it when a problem gives every value at one instant and asks for one rate.
This is one of four skills in topic 4.5 of Which & Why: AP Calculus.