Related rates, question 18

Topic: Related rates
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §5 CLP-1 §3.2Key results

A glassmaker draws out a hot glass rod. The rod stays a solid cylinder of volume $100\pi$ cm³, with radius $r$ cm and length $L$ cm, while its radius shrinks at 0.1 cm per second. Two students find $\dfrac{dL}{dt}$ at the moment the radius is 2 cm, as shown. Which solution is right, and why?

\begin{aligned} &\textbf{Solution A} \\ \text{A1:}\quad & \pi r^{2} L = 100\pi \\[4pt] \text{A2:}\quad & r = 2: \quad 4\pi L = 100\pi, \ \text{ so } L = 25 \\[4pt] \text{A3:}\quad & \frac{dL}{dt} = \frac{d}{dt}(25) = 0 \text{ cm/s} \\[14pt] &\textbf{Solution B} \\ \text{B1:}\quad & \pi\Bigl(2rL\,\frac{dr}{dt} + r^{2}\,\frac{dL}{dt}\Bigr) = 0 \\[4pt] \text{B2:}\quad & \frac{dL}{dt} = -\frac{2L}{r}\,\frac{dr}{dt} \\[4pt] \text{B3:}\quad & r = 2,\ L = 25,\ \frac{dr}{dt} = 0.1: \\ & \frac{dL}{dt} = -\frac{2(25)}{2}(0.1) = -2.5 \text{ cm/s} \end{aligned}

A is right; B overlooks that at r = 2 the length is fixed at L = 25 cm, so it cannot be changing.
B is right; A substitutes r = 2 before differentiating, which freezes the radius and makes the length look constant.
Both are right: the rate depends on whether r is set to 2 before or after differentiating, and either order is allowed.
Neither is right: A substitutes r = 2 before differentiating, and B uses dr/dt = 0.1 for a radius that is shrinking.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §5 CLP-1 §3.2Key results

Worked solution

Strategy

Differentiate the volume relation with respect to time before putting in any values, then substitute the values of the instant, with the sign of each rate. Compare each line of A and B with that.

Working

$\pi r^{2} L = 100\pi$
A1 is right: the volume stays 100π while r and L both change.
$\pi\Bigl(2rL\,\frac{dr}{dt} + r^{2}\,\frac{dL}{dt}\Bigr) = 0 \quad\Longrightarrow\quad \frac{dL}{dt} = -\frac{2L}{r}\,\frac{dr}{dt}$
Differentiate first, with the product rule, as B1 and B2 do. A2 instead put r = 2 in first, which turned L into the constant 25 and forced A3 to give 0.
$r = 2, \qquad L = \frac{100}{4} = 25, \qquad \frac{dr}{dt} = -0.1$
Now put in the values of the instant. The radius shrinks, so its rate is negative; B3 used +0.1.
$\frac{dL}{dt} = -\frac{2(25)}{2}(-0.1) = 2.5$
The length grows at 2.5 cm/s, as a thinning rod of fixed volume must.

Answer

\text{Neither:}\quad \frac{dL}{dt} = 2.5 \text{ cm/s}

Check

A tenth of a second later r = 1.99 cm and L = 100/1.99² ≈ 25.25 cm, so the rod lengthens at about 2.5 cm/s, ruling out both 0 and −2.5.

Where this goes wrong

Putting the instant’s values in before differentiating. Once r = 2 is substituted, L becomes the number 25 and its rate is lost; substitute only after the derivative is taken, and give a shrinking quantity a negative rate.