If a child 1.5m tall walks away from a light at 2m/s and the light is above ground at 4m. how fast is shadow?

Question

how fast is the shadow growing when the boy is 5m away?

Answer 1

square root of(2) x 2m/s

2.828 m/s

Answer 2

You draw a triangle with the side opposite 4m long and the side adjacent made up of two segments: d the distance from the light to the child and s the length of the shadow from the child to the angle a. The child is 1.5m high and is where segment d ends and s begins and terminates on the hypoteneus. We have two similar triangles.
tan(a)=4/[s+d]=1.5/s
So 4s=1.5d+1.5s and 2.5s=1.5d or
s=.6d and since d is changing at 2m/s then
ds/dt=.6*2m/s=1.2m/s

Answer 3

1. Your first step is to create a function that relates the length of the shadow (call it "L") to the child's distance from the light post (call that "x"). In other words, complete this equation:

L = (some formula with "x" in it).

(Hint: It's probably easier if you draw a diagram.)

2. Your next step is to find dL/dx (that is, differentiate that formula with respect to "x").

3. You now have "dL/dx", but what you really want is the _speed_ (the derivative with respect to _time_) at which "L" is changing. That is, you really want "dL/dt". To find that, use this formula:

dL/dt = (dL/dx)(dx/dt)

(Note that "dx/dt" is the rate at which the child is moving away from the post; namely 2m/s.)

4. Now you have a formula for dL/dt. So just plug "x=5m" into the formula and that's your answer.