a man is walking away from a lamppost he walks at three feet per second away from a twenty-four foot tall lamppost; if he himself is six feet tall, what is the rate of change of the length of his shadow?
length of
shadow guy distance to
some guy lamp!
looks like it's Rite Triangle to the rescue try to lay everything out in a shape that you can recognize
height of lamp guy b1 = lenght of
shadow b2 = distance
to guy

the trick is that there's a similar triangle
θ = angle to
guy's head also = θ
 also = b2
so what can we do with similar triangles? we can use the similar triangles to relate b1 to b2

since b'
2 is given, we can solve that equation for b'
1, which is the desired result
b'
2 = given b'
1 = answer
from rules of similar triangles
b1 h1 = b2 h2
multiply by h1 to help solve for b1 again, we're trying to solve this ekwation for b1 because its rate-of-change is the desired result
h1 b1 h1 = h1 b2 h2
cancel on one side; multiply on the other
h1 b1 1 h1 = h1 b2 h2
we'll try to get the h's into their own fraction because, we'll see that h1 and h2 are both just constants, which is to say, they are numbers that don't change
h1· b1 1 = h1 1 b2 h2

b1 = h1      b2 1      h2

b1 = h1      b2 1 h2

b1 = h1 h2 b2
h2 is the difference between the height of the guy and the height of the lamp
lamp height h1= guy's height h2= lamp - guy
since both h1 and h2 are constants, so is their ratio over the course of this short walk, the guy's height will not change any appreciable amount, or even an amount that can be detected with known instruments, and is therefore outside the scope of the accuracy of this question
b1 = h1 h2 b2
this derivative is rather simple
b′
1 = h1 h2 b′
2
b'
1 is the answer when computed with these values it is the rate of change of the length of the shadow
b′
1 = h1 h2 b′
2 b′
1 = ans h1 = lamp - guy = 24ft - 6ft h2 = guy = 6ft b′
2 = 3ft/s
replace those values in the equation
b′
1 ans = h1 18ft h2 6ft b'
2 3ft/s
do some mathrobatics
ans = 18ft 6ft 3ft/s
reduce; cancel values and units
ans = 318 ft 16 ft 3ft/s
getting closer ...
ans = 3 1 3ft/s

ans = 3 3 ft/s
the rate of change of the length of the shadow...
9ft/s
what if you want to know that rate of change of the tip of the shadow from the lamp?

well, that is simply the sum of both b′
1 and b′
2

but how do we prove that?

we know that the entire base is equal to the sum of the two base-parts
base = ... b1 + ... b2

we can set up this simple equation
base = b1 + b2
this derivative is also quite easy
base = b1 + b2
since the derivative of a sum is the sum of the derivatives
b′
ase = b′
1 + b′
2
the rate of the tip is the answer with these values
b′
ase = b′
1 + b′
2 b′
ase = rate of tip b′1 = 9ft/s b′2 = 3ft/s
replace those values
b′
ase ans = b′
1 9ft/s + b′
2 3ft/s
use a computron brand comptometer to add these values
ans = 9ft/s + 3ft/s
looks like a good answer
12ft/s