Suppose a 6ft. man is 10 ft. away from a 24 ft. tall lamp post. If the person is moving away from the lamp post at at rate of 2 ft/sec, at what rate is the length of his shadow changing?
Why is it "let the man be x ft from the lamppost. "
Why not 10 feet away?
& Why is it "also notice that the fact that he is 24 feet from the post does not enter the picture at all"
When the lamp post is 24 feet tall?
I put the wrong number in when I made that last comment. It should have said " .... he is 10 feet from ..."
It has no effect on the solution.
We cannot use the 10 feet in the "general" case, the 10 feet is one specific instant.
in rate of change problems you <bnever use the specific case data until you have differentiated the equation.
let the man be x ft from the lamppost.
At that time, let the length of his shadow be y ft
by ratios:
24/(x+y) = 6/y
24y = 6x+6y
18y=6x
3y=x
3dy/dt = dx/dt
but dx/dt = 2
dy/dt = 2/3 ft/s
so his shadow is increasing in length at 2/3 ft/sec,
notice that the end of his shadow would be MOVING at 2 +2/3 or 8/3 ft/se
also notice that the fact that he is 24 feet from the post does not enter the picture at all.
To find the rate at which the length of the man's shadow is changing, we can use similar triangles.
Let's call the length of the man's shadow "s" and the distance between the man and the lamp post "x". We want to determine ds/dt, the rate at which s is changing with respect to time.
We know that the height of the lamp post is 24 ft, and the distance between the man and the lamp post is 10 ft. So we have the following proportion:
(Height of lamp post) / (Length of shadow) = (Distance from lamp post) / (Distance from man)
Using this proportion, we can express s in terms of x:
24 ft / s = 10 ft / x
Cross-multiplying, we get:
24x = 10s
Rearranging the equation to solve for s:
s = (24x) / 10
Now, we want to determine ds/dt, the rate at which s is changing with respect to time. To do this, we can implicitly differentiate both sides of the equation with respect to time (t):
d/dt(s) = (d/dt(24x)) / 10
ds/dt = (24 * dx/dt) / 10
We know that dx/dt, the rate at which the person is moving away from the lamp post, is given as 2 ft/sec. So substituting this value:
ds/dt = (24 * 2) / 10
ds/dt = 4.8 ft/sec
Therefore, the length of the man's shadow is changing at a rate of 4.8 ft/sec.