Question

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the building at a speed of 1.4 m/s, which is taken as the given dx/dt, how fast is the length of his shadow on the building decreasing when he is 4 m from the building? Step 1

Answer 1

Answer:

$\frac{dy}{dt}=-4.2m/s$

Explanation:

Now, in order to solve this problem we must start by representing it with a drawing. (See attached picture). Notice in the drawing that we are dealing with similar triangles, so we need to build our equation based on the similar equations. So let's build our equation:

$\frac{y}{2}=\frac{12}{12-x}$

Now, we can solve the equation for y, so we get:

$y=\frac{24}{12-x}$

now we can differenciate this equation, we can start by turning the fraction into a negative power, like this:

$y=24(12-x)^{-1}$

Nex we can differenciate the function by using the power and the chain rules, so we get:

$\frac{dy}{dt}=-24(12-x)^{-2}(-1)$

or:

$\frac{dy}{dt}=\frac{24}{(12-x)^{2}} \frac{dx}{dt}$

so now we can substitute the values we know:

$\frac{dy}{dt}=\frac{24}{(12-4)^{2}} (-1.4)$

when solving we get:

$\frac{dy}{dt}=-4.2m/s$

so the shadow decreases at a rate of -4.2m/s.