Question

A spotlight at a building corridor is fastened to a wall 8m above the floor. A lady 1.75m tall moves away from the wall at a speed of \(\displaystyle{0.75}\frac{{m}}{{s}}\).
a.)at what rate is the lenght of her shadow increasing?
b.)at what speed is the tip of her shadow moving?

Answer 1

Step 1
Please have a look at the picture below to understand what's going on:

For the sake of clarity,
s = length of the shadow = AD
and x = distance of the lady from the wall = BD
\(\displaystyle\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={0.75}\frac{{m}}{{s}}\)
Step 2
Triangle ADE is similar to triangle ABC (AAA criterion of similarity)
Hence, \(\displaystyle{D}\frac{{E}}{{B}}{C}={A}\frac{{D}}{{A}}{B}\) (Corresponding sides of similar triangles are proportional)
Hence, \(\displaystyle\frac{{1.75}}{{8}}=\frac{{s}}{{{s}+{x}}}\)
Hence, \(\displaystyle{1.75}{\left({s}+{x}\right)}={8}{s}\)
Or, \(\displaystyle{1.75}{x}={\left({8}-{1.75}\right)}{s}={6.25}{s}\)
Step 3
Part (a)
Differentiate both sides w.r.t time t to get:
\(\displaystyle\frac{{{1.75}{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}=\frac{{{6.25}{d}{s}}}{{{\left.{d}{t}\right.}}}\)
Hence, the rate at which length of her shadow is increasing = \(\displaystyle\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}=\frac{{{\left(\frac{{1.75}}{{6.25}}\right)}{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={\left(\frac{{1.75}}{{6.25}}\right)}{x}{0.75}={0.21}\frac{{m}}{{s}}\)
Step 4
Part (b)
the rate at which the tip of her shadow is moving = rate t which she is moving + rate at which the length of the shadow is increasing = \(\displaystyle\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}+\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}={0.75}+{0.21}={0.96}\frac{{m}}{{s}}\)
Step 5
Final answers:
Part (a) \(\displaystyle{0.21}\frac{{m}}{{s}}\)
Part (b) \(\displaystyle{0.96}\frac{{m}}{{s}}\)