Related Rates — Question 4

Question 4 - Solution

Let: - $x(t)$ be the distance from the person to the light pole - $s(t)$ be the length of the shadow - The tip of the shadow is at a distance $x + s$ from the pole

Using similar triangles: $$\frac{6}{s} = \frac{15}{x + s} \Rightarrow 6(x + s) = 15s \Rightarrow 6x + 6s = 15s \Rightarrow 6x = 9s \Rightarrow s = \frac{2}{3}x$$

Differentiate with respect to $t$: $$\frac{ds}{dt} = \frac{2}{3} \frac{dx}{dt}$$

Given: $$\frac{dx}{dt} = 3 \text{ ft/s} \Rightarrow \frac{ds}{dt} = \frac{2}{3}(3) = 2 \text{ ft/s}$$

Now, the tip of the shadow moves at: $$\frac{d}{dt}(x + s) = \frac{dx}{dt} + \frac{ds}{dt} = 3 + 2 = \boxed{5 \text{ ft/s}}$$

Answers:

• (a) The tip of the shadow is moving at $\boxed{5 \text{ ft/s}}$

• (b) The shadow is lengthening at $\boxed{2 \text{ ft/s}}$