Question

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight towards the building at a speed of 1.6 m/s, how fast is the length of shadow on the buildingdecreasing when he is 4 m from the building?

Short answer

The height of the shadow decreases at a rate of 0.6 m/s.

Step-by-step solution

Step 1:Find the rate of change of distance in two directions

Let x represents the distance traveled by the man. Therefore;

\(\frac{{{\rm{d}}x}}{{{\rm{d}}t}} = 1.6\)

The figure below represents the schematic of man and spotlight.

Using the properties of similar triangles.

\(\begin{aligned}\frac{y}{{12}} &= \frac{2}{x}\\y &= \frac{{24}}{x}\end{aligned}\)

Find the rate of change of length of the shadow

Differentiate the equation \(y = \frac{{24}}{x}\) with respect to t.

\(\begin{aligned}\frac{{{\rm{d}}y}}{{{\rm{d}}t}} &= \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{24}}{x}} \right)\\ &= - \frac{{24}}{{{x^2}}}\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}}} \right)\end{aligned}\)

Substitute 8 for x and 1.6 for \(\frac{{{\rm{d}}x}}{{{\rm{d}}t}}\) in the equation\(\frac{{{\rm{d}}y}}{{{\rm{d}}t}} = - \frac{{24}}{{{x^2}}}\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}t}}} \right)\).

\(\begin{aligned} \frac{{{\rm{d}}y}}{{{\rm{d}}t}} &= - \frac{{24}}{{{{\left( 8 \right)}^2}}}\left( {1.6} \right)\\ &= - \frac{3}{8}\left( {1.6} \right)\\ = - 0.6\;{\rm{m/s}}\end{aligned} \)

So, the height of the shadow decreases at a rate of 0.6 m/s.