Build and differentiate the rate relationship
A ladder that slides
A ladder 13 feet long leans against a wall. The orange segment on the ground is x, the distance from the wall to the foot. The orange segment on the wall is y, the height of the top.
Drag the foot away from the wall. The top moves down the wall, because the ladder cannot change its length.
The ladder is the longest side of a right triangle, so \(x^2 + y^2 = 169\) at every position. At \(x = 5\), \(y = 12\), because \(25 + 144 = 169\).
As time t passes, x and y change, so they are functions of t. A rate tells how fast a quantity changes in each unit of time. The relation holds at every moment, so it also links the rate of x to the rate of y.
Any two linked quantities
The ladder has a fixed length L, and x and y change with time.
Drag the foot. The relation \(x^2 + y^2 = L^2\) is true at every position.
A related rates problem links the rates of several quantities that one relation ties together. Both sides of the relation are the same number at every moment, so they change at the same rate.
Notation
Write \(x\) and \(y\) as functions of the time \(t\). Their rates are \(dx/dt\) and \(dy/dt\).
\[x = x(t), \qquad y = y(t)\]The green arrow shows dx/dt, which is positive when the foot moves away from the wall. The red arrow shows dy/dt, which is negative because the top moves down.
Write the relation that holds at every moment. The right side is a constant, because the ladder has a fixed length.
\[\cy{x}^2 + \cy{y}^2 = 169\]The relation must hold during an interval of time. Both \(x\) and \(y\) must be differentiable.
Differentiate both sides with respect to \(t\). Because \(x\) depends on \(t\), the chain rule puts the factor \(dx/dt\) after the derivative of \(x^2\). The same happens for \(y\).
\[2\cy{x}\,\frac{dx}{dt} + 2\cy{y}\,\frac{dy}{dt} = 0\]The right side is 0, because 169 is a constant.
Divide by 2. In words, \(x\) times the rate of \(x\) plus \(y\) times the rate of \(y\) is 0.
\[\cy{x}\,\frac{dx}{dt} + \cy{y}\,\frac{dy}{dt} = 0\]Because \(x\) and \(y\) are positive, the two rates have opposite signs.
When it fails
Suppose you put \(x = 5\) into the relation before you differentiate. The relation becomes \(25 + y^2 = 169\), whose derivative is \(2y\,dy/dt = 0\), so \(dy/dt = 0\).
The red mark holds the top at \(y = 12\), but the top moves down when you drag the foot. The equation \(25 + y^2 = 169\) describes one position, so it cannot show how \(x\) changes. Differentiate first, and substitute numbers after.
Example 1: A ripple
A stone makes a circular ripple. The radius \(r\) and the area \(A\) inside it both grow with time. Write an equation that links \(dA/dt\) and \(dr/dt\).
The area inside a circle depends on its radius.
\[\cb{A} = \pi \cy{r}^2\]Both \(A\) and \(r\) depend on \(t\), so differentiate both sides with respect to \(t\). The chain rule puts the factor \(dr/dt\) after the derivative of \(r^2\).
\[\frac{dA}{dt} = 2\pi \cy{r}\,\frac{dr}{dt}\]The picture explains the factor \(2\pi r\). When \(r\) grows by a small amount \(\Delta r\), the green ring is the area that is added. The ring is as long as the edge and \(\Delta r\) wide, so its area is about \(2\pi r\,\Delta r\).
At \(r = 4\), the edge is \(2\pi \cdot 4 = 8\pi\) long.
Example 2: A balloon
A spherical balloon is inflated. Its volume \(V\) and its radius \(r\) both change with time. Write an equation that links \(dV/dt\) and \(dr/dt\).
The volume of a sphere depends on its radius.
\[\cb{V} = \frac{4}{3}\pi \cy{r}^3\]A tempting wrong move: Differentiate the right side with respect to \(r\) only.
\[\frac{dV}{dt} = 4\pi r^2\]The green ring shows a thin shell added around the balloon. Its volume is the surface area \(4\pi r^2\), the red outline, times its thickness, the growth of \(r\). This equation leaves out the thickness, the factor \(dr/dt\).
Differentiate with respect to \(t\). The derivative of \(r^3\) is \(3r^2\,dr/dt\).
\[\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2\,\frac{dr}{dt} = 4\pi \cy{r}^2\,\frac{dr}{dt}\]
Practice
A rectangle has length \(l\) and width \(w\). Both change with time. Write an equation for \(dA/dt\), where \(A\) is the area.
In the picture, the area gains two green strips. When l grows, the new strip is w high. When w grows, the new strip is l wide.
Solution
The area is the length times the width.
\[A = l\,w\]Both \(l\) and \(w\) depend on \(t\), so differentiate with the product rule.
\[\frac{dA}{dt} = \frac{dl}{dt}\,w + l\,\frac{dw}{dt}\]The first term is the green strip on the right, \(w\) high. The second term is the green strip on top, \(l\) wide.
Questions
A can has radius \(r\) and height \(h\). Both change with time, and its volume is \(V\).
Which equation links \(V\), \(r\) and \(h\) at every moment?
- \(V = \pi r^2 h\)
- \(V = 2\pi r h\)
- \(V = \pi r^2 + h\)
Answer
\(V = \pi r^2 h\)
Both \(r\) and \(h\) change. Which equation is correct for \(dV/dt\)?
- \(\dfrac{dV}{dt} = 2\pi r h\,\dfrac{dr}{dt} + \pi r^2\,\dfrac{dh}{dt}\)
- \(\dfrac{dV}{dt} = 2\pi r h\,\dfrac{dr}{dt}\)
- \(\dfrac{dV}{dt} = \pi r^2\,\dfrac{dh}{dt}\)
Answer
\(\dfrac{dV}{dt} = 2\pi r h\,\dfrac{dr}{dt} + \pi r^2\,\dfrac{dh}{dt}\)
A sphere has surface area \(S = 4\pi r^2\), and \(r\) changes with time. Find \(dS/dt\).
- \(8\pi r\,\dfrac{dr}{dt}\)
- \(8\pi r\)
- \(4\pi r^2\,\dfrac{dr}{dt}\)
Answer
\(8\pi r\,\dfrac{dr}{dt}\)
Summary
You write an equation that is true at every moment, and you differentiate both sides with respect to \(t\).
This works because quantities that match at every moment change at matching rates, and the chain rule adds a rate factor to each.
You use it when a problem links several changing quantities, and you substitute numbers after you differentiate.
This is one of four skills in topic 4.5 of Which & Why: AP Calculus.