How fast is the length of his shadow on the building decreasing when he is 4 m from the building

Question

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

How do you solve this word problem. I have drawn a picture to figure out the solution but I have failed to come up with anything.

Let x be the distance between the man and the spotlight, the distance between him and the building shall be $12-x$. $12-x=4$ means $x=8$

At this point I don't know how to proceed.

Please help

Answer 1

As was aptly noted in the comments section, you can easily find the function for the length of the man's shadow cast on the wall by using similar triangles:

$$\frac{2}{1.6t}=\frac{l(t)}{12}\implies l(t)=\frac{15}{t}\ m$$

Then, the function for the rate at which the length of the shadow on the wall is decreasing is going to be this:

$$ l'(t)=-\frac{15}{t^2}\ m/s $$

He is 4 meters away from the building at time: $$12-1.6t=4\implies t=5\ s$$

And this is how fast the length of his shadow on the wall is decreasing when he is 4 meters aways from the building:

$$l'(5)=-\frac{15}{5^2}=-0.6\ m/s$$

Answer 2

Man's distance $x = 12-4 =8 $ meters from spotlight, $4$ times the man's height $h=2,$ which is constant for similar triangles being corresponding sides ratio.

$$ \frac{h}{x} = \frac{s}{12};\quad s.x =const $$

Differentiating with respect to time,

$$ \frac{\dot s}{\dot x} = - \frac{s}{x} ;\, \frac{\dot s}{1.6}= - \frac{12/4}{8} \rightarrow \dot s= -0.6 $$

meters per second is the rate at which shadow is shortening.