Decrease rate of the shadow

Question

If a man of height $6ft$ moves with $5ft/sec$ velocity towards a lamp hanging at $15ft$ height, then at what rate his shadow will decrease?

My Work: Let the initial distance between the man and the lamp post be $x$ and the length of shadow be $y$.

Then, $\frac{x}{9} = \frac{y}{6} \implies y = \frac{2x}{3}$
Now how do I handle the $5ft/sec$ velocity and the decrease rate of the shadow?
I think the rest part is so easy. But I am missing somewhere.
Note: This is a problem from BdPhO 2016 but completely math related. So I posted it here.

Answer 1

Anatoly has given a simple solution,
but in case you are expected to use calculus, here's another simple solution.

Starting from $y = \dfrac23 x,\;\; \dfrac{dy}{dt} = \dfrac23 \dfrac{dx}{dt} = \dfrac 23\times -5 = -\dfrac{10}{3}$ ft/s,

the negative sign showing decreasing length of shadow.

Answer 2

Opps. I just found the answer.
In $1sec$ if he move $5ft$ then the length of shadow is
$y' = \frac{2(x-5)}{3} = \frac{2x-10}{3}$
So in one second $$y'-y = \frac{2x}{3} - \frac{2x-10}{3} = \frac{10}{3}$$
So, the decrease rate of the shadow is $\frac{10}{3}ft/sec$.

Correct me if I am wrong.

Answer 3

Alternative approach: you already correctly showed that $y=2/3 \,x$. Now note that, assuming $x $ and $y $ expressed in feet, the time needed for the man to reach the base of the lamp is $x/5$ seconds. In the same time, the length of the shadow decreases from $y=2/3 \, x \, $ to $0$. So the decreasing rate is $$\frac {2/3 \, x}{x/5}=\frac {10}{3} \, \text { feet per second.} $$