Aman 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.
(a) when he is 10 feet from the base of the light, at what rate is the tip of his shadow moving?
(b) when he is 10 feet from the base of the light, at what rate is the length of his shadow changing?

A) When the man is 10 ft from the base of the light, the rate at which the tip of his shadow is moving is; dy/dt = 10 ft/s

B) When the man is 10 ft from the base of the light, at what rate is the length of his shadow changing is; d(BD)/dt = 4 ft/s

We are given;

Height of man = 6 ft

Height of light = 15 ft

Rate of walk by man; dx/dt = 6 ft/s

I have attached an image showing the triangle formed by the man and the light.

From the diagram;

AB is height of light = 15 ft

CD is height of man = 6 ft

A) From the attached diagram, if BE = y, then we can use ratio of similar triangles to get;

15/y = 6/(y - x)

Cross multiply to get;

15(y - x) = 6y

15y - 15x = 6y

15y - 6y = 15x

9y = 15x

Divide both sides by 3 to get;

3y = 5x

Differentiating both sides with respect to t gives;

3(dy/dt) = 5(dx/dt)

dy/dt = (5/3)(dx/dt)

Putting 6 ft/s for dx/dt yields;

dy/dt = (5/3)(6)

dy/dt = 10 ft/s

B) From the diagram, length of shadow is;

BD = y - x

Rate at which the shadow is changing will be gotten by differentiation of this length with respect to time.

Thus;

d(BD)/dt = dy/dt - dx/dt

Plugging in the relevant values gives;

d(BD)/dt = 10 - 6

d(BD)/dt = 4 ft/s

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(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light

height of man

From diagram

differentiate both sides

Tip of shadow is moving at the rate of

(b)rate at which length of his shadow  is changing

Length of shadow is

differentiating w.r.t time

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