Calculating rates with derivatives

Question

A light shines toward the ground from the top of a building wall that is 40 ft high. A kid in a tree house that is 25ft to the right of the building drops an apple from the same height (40 ft). How fast is the shadow of the apple moving along the ground 1⁄2 sec later? Assume the apple falls at a distance 𝑠 = 16𝑡^2 in t seconds.

Answer 1

Let w(t) = distance from apple's shadow to base of the wall at time t

We're asked to find w'(1/2)

By carefully drawing and labeling a diagram, we can derive the formula 16t²/25 = 40/w(t)

Solving for w gives you w(t) = 125/2t²

Therefore w'(t) = -2*125/2t³, so w'(1/2) = -125*(1/2)^-3 = -125*8 = -1000 ft/sec

Answer 2

Well, we have to make an estimate of the light's intensity which says that it will start casting a shadow at some distance away after the apple is dropped and will cast an angle with the level ground. After 1/2 second the apple dropped 4 feet in 1/2 second. The angle Θ of the light's angle with the level ground is then 40'/Tan(Θ). Θ can be calculated from the 4' drop and horizontal distance of 25' given to the tree house, Θ = arcTan(4/25) = 25.91°. And it distance away is 40/Tan(25.91°) = 82.34 feet.°

Now we have to assume that the light's intensity might start a shadow at some distance further away, so I've chosen the shadow starts at an angle of Θ=10° for the shadow to be observed.

The shadow's distance is then 40/Tan(10°) =226.85' at the time zero.

The velocity is calculated as (d1 - d2)/time, with the time being 1/2 second, and (d1-d2)/(time)

is (226.85' - 82.34')/(1/2) = 289 feet per second.

= 289 feet per second velocity that the shadow is traveling after 1/2 second