Simple Related Rates Question (implicit differentiation)?

Question

A spotlight on the ground shines on a wall 10m away. A man 2 m tall walks from the spotlight towards the wall at a speed of 1.2m/s. How fast is his shadow on the wall decreasing when he is 3 m from the wall?

Answer 1

Let x be the distance of the man from spotlight on ground. Let his shadow on wall be s. Then we have, x/10 = 2/s, by similarity of triangles. or s = -x/20 if we see that as x increases s decreases.

This gives ds/dt = -(1/20).(dx/dt) = -.06 m/s independent of x. so the Shadow decreases at this rate even when the man is 3m from wall or x = 7 m

Answer 2

Let x be the distance of the man from spotlight on ground. Let his shadow on wall be s.
Then we have,
s/10 = 2/x
Differentiating with respect to time,
ds/dt = -20 (1/x^2) (dx/dt).
ds/dt is the rate of change of the length of the shadow; dx/dt is the speed of the man.
Negative sign shows that shadow decreases as the man approaches the wall.

It is given dx/dt = 1.2 m/s.

Therefore when x = 3m, ds/dt = -20 x (1/ 9) x 1.2 = - 2.66m/s.

Answer 3

I think "letslearn" has the right idea but has made a mistake somewhere.

The first part is correct: by similar triangles, x/2 = 10/s.
However, you cannot go around arbitrarily introducing minus signs.

s = 20/x
ds/dx = -20/x^2 <-- this shows that the shadow decreases as x increases

ds/dx * dx/dt = -20/x^2 dx/dt
ds/dt = -20/x^2 dx/dt
When the man is 3 m from the wall, x = 7
x = 7, dx/dt = 1.2 --> ds/dt = -0.490 m/s