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?

Question

myname · Answer

Let x be the distance between the man and the lamppost.
Let y be the length of the shadow.
You can draw a sketch and you will see 2 similar triangles.

So, from the relation of similar triangles you can get
     23/(x+y)=6/y
Now, you can just solve for y in terms of x
     6x+6y=23y
     6x=17y
     y=6x/17 
Here, we already have the rate of change of the distance between the man and the lamppost given(6ft/sec), which is in algebrical terms refered as :dx/dt= 6ft/sec
 Then lets implicitly differentiate the function y=6x/17
dy/dt=(6/17)*6 ft/sec
dy/dt= 36/17 (ft/se) 
Therefore the rate of change of the man’s shadow when he is 65 fts away from the lamppost is 36/17(ft/sec).
I hope this helps.