Question

A man , 2 m tall, walks at the rate of $1\frac{2}{3}$ m/s towards a street light which is $5\frac{1}{3}$ m above the ground. At what rate is the tip of his shadow moving and at what rate is the length of the shadow changing when he is $3\frac{1}{3}$ m from the base of the light?

Answer 1

Let AB be the street light post and CD be the height of man i.e., CD = 2 m.

Let BC = x m, CE = y ma and $\frac{dx}{dt}=\frac{-5}{3}m/s$
From $△ABEand△DCE$, we see that,
$△ABE$ ~ $△DCE$
$\therefore \frac{AB}{DC}=\frac{BE}{CE}\Rightarrow \frac{\frac{16}{3}}{2}=\frac{x+y}{y}\Rightarrow \frac{16}{6}=\frac{x+y}{y}\Rightarrow 16y=6x+6y\Rightarrow 10y=6x\Rightarrow y=\frac{3}{5}x$
On differentiating both sides w.r.t. t, we get
$\frac{dy}{dt}=\frac{3}{5}.\frac{dx}{dt}=\frac{3}{5}.(-1\frac{2}{3})=\frac{3}{5}.\left(\frac{-5}{3}\right)=-1m/s$
Let z = x + y
Now, differentiating both sides w.r.t.t, we get
$\frac{dz}{dt}=\frac{dx}{dt}+\frac{dy}{dt}=-(\frac{5}{3}+1)=-\frac{8}{3}=-2\frac{2}{3}$
Hence, the tip of shadow is moving at the rate of $2\frac{2}{3}$ m/s towards the light source and length of the shadow is decreasing at the rate of 1 m/s.