Related Rates — Question 6

Question 6 - Solution

Let: $$x(t) = \text{horizontal distance from the spotlight to the person (m)}$$ $$s(t) = \text{height of the shadow on the wall (m)}$$

From similar triangles: $$\frac{s(t)}{10} = \frac{2}{x(t)} \Rightarrow s(t) = \frac{20}{x(t)}$$

Differentiate both sides: $$\frac{ds}{dt} = -\frac{20}{x^2} \cdot \frac{dx}{dt}$$

Given: $$x = 5 \text{ m}, \quad \frac{dx}{dt} = -1.2 \text{ m/s}$$

$$\frac{ds}{dt} = -\frac{20}{25}(-1.2) = 0.96 \text{ m/s}$$

(a) Rate of change of shadow height: $$\boxed{\frac{ds}{dt} = 0.96 \, \text{m/s}}$$

(b) Speed of the tip of the shadow:

The tip moves upward at the same rate: $$\boxed{0.96 \, \text{m/s upward}}$$