Finding the Rate of Change of Shadow: A Man 6 ft Tall

[Amadeus]

Homework Statement

A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground. When he is 10 feet from the base of the light,

(A) at what rate is the tip of his shadow moving?
(B) at what rate is the length of his shadow changing?

Homework Equations

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The Attempt at a Solution

This is what I have so far...
dr/dt = 5 ft/sec
dx/dt = unknown
h = 10

Since the man is walking away, his shadow is getting bigger (right?).

for (A) I guess I'm suppose to find dx/dt since I already know what dy/dt is. But my question is which equation will work between the y and the x?

for (B) same question but my friend told me that it was a proportion; as in x/y = 6/15 but I'm not sure if that's right - even if it is, I'm still lost...

Help?

 

[Dick]

Could you maybe say what r,h,x and y mean to you? I can guess what some might mean, but it's still confusing.

 

[Dick]

Ok, I'm going to guess. r=dist from man to flagpole. x=dist from man to end of shadow. y=dist from end of shadow to flagpole. So y=x+r. h is the same thing as r, right? Then your friend is right 6/x=15/y. They are similar triangles. So 6*y=15*x and y=x+r. Can you differentiate that and reach a conclusion?

 

[Amadeus]

Sorry about the confusion, but thanks for your help! :)

I tried to differentiate it but I got the wrong answer (15 for part B)...

My differentiation looks something like this:
6y*dy/dt = 15x*dx/dt

Maybe I screwed up my differentiation...

 

[Dick]

Ok, so 6y=15x. How does differentiating that give you something like 6y*dy/dt?

 

[rocomath]

for (B) same question but my friend told me that it was a proportion; as in x/y = 6/15 but I'm not sure if that's right - even if it is, I'm still lost...

similar triangles