Question

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

I am assuming the building and the wall are the same thing.

Answer 1

`dy/dt=0.6` m/s

Explanation:

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At the specified moment in the problem, the man is standing at point `D` with his head at point `E`.

At that moment, his shadow on the wall is `y=BC`.

The two right triangles `DeltaABC` and `DeltaADE` are similar triangles. As such, their corresponding sides have equal ratios:

`(AD)/(AB)=(DE)/(BC)`

`8/12=2/y, :. y=3` meters

If we consider the distance of the man from the building as `x` then the distance from the spotlight to the man is `12-x`.

`(12-x)/12=2/y`

`1-1/12x=2*1/y`

Let's take derivatives of both sides:

`-1/12dx=-2*1/y^2dy`

Let's divide both sides by `dt`:

`-1/12*dx/dt=-2/y^2*dy/dt`

At the specified moment:

`dx/dt=1.6` m/s

`y=3`

Let's plug them in:

`-1/12(1.6)=-2/9*dy/dt`

`dy/dt=(1.6/12)/(2/9)=1.6/12*9/2=0.6` m/s

This is the rate at which the length of his shadow is decreasing at the specified moment.