Physics, 07.01.2021 17:20 gabbytumey

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 1.1 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building

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Answer from: elizabethcswind5062

$\frac{dh}{dt}=-0.4125 m/s$

Explanation:

The very first thing we can do here is to draw a sketch of the situation the problem is presenting. (See attached picture)

As you may see in the picture, we can suppose the spotlight is directly located on the ground so the light of the spotlight together with the person and the wall make two similar triangles.

In this case we need to think about the dimensions of the triangle that will change over time. We will call them:

x= distance between the spotlight and the man.

h= height of the shadow.

So we can build a relation between the height and the length of each triangle:

$\frac{h}{12}=\frac{2}{x}$

the distance between the spotlight and the wall is constant, so I can directly write the 12 in my equation and the height of the man is constant as well, so I can write the 2 directly into the equation.

Next, we can solve the equation for h (since we are interested in figuring out how fast the height of the shadow is decreasing) so we get:

$h=\frac{24}{x}$

So next, we need to take the derivative of the equation, since the derivative of a position function will give us the velocity at which that position is changing.

First we rewrite the equation like this:

$h=24x^{-1}$

and take the derivative:

$dh=-24x^{-2}dx$

we can rewrite the derivative like this:

$\frac{dh}{dt}=-\frac{24}{x^{2}} \frac{dx}{dt}$

where:

dh/dt is the velocity at which the height of the shadow is decreasing (that's why our equation has a negative sign in front of it).

dx/dt is the velocity at which the value of x is increasing. In other words, how fast the man is moving away from the spotlight.

So we can go ahead and substitute:

we wish to find the velocity at which the shadow's height is decreasing when the distance between the man and the building is 4m, so in this case, x=12-4=8m

$\frac{dh}{dt}=-\frac{24}{8^{2}}(1.1 m/s)$

so we get:

$\frac{dh}{dt}=-0.4125 m/s$

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight tow

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