Question

A man 6ft tall walks at a rate of 6ft/s away from a lamppost that is 23 ft high. At what rate is the length of his shadow changing when he is 65 ft away from the lamppost?

Answer 1

Suppose,
x = distance of the man
s = length of the shadow

Using the idea of similar triangles

6/s = 23/(x + s) 

Simplifying:
we get,
6(x + s) = 23s 

6x + 6s = 23s 

6x = 17s 

Differentiating with respect to time, 

6(dx/dt) = 17(ds/dt) 

Manipulating the above equation for ds/dt,

ds/dt = (17/6)(dx/dt) 

ds/dt = (17/6)(6) 

ds/dt = 17 ft/sec.