Calculus: Related Rates Problem?

Question

After a week or more of doing these, you'd think I'd get it. :(

The problem is:
A man 6ft tall walks at a rate of 5 ft/sec toward a light that is 20 ft above the ground. When he is 10 ft from the base of the light, (a)what rate is the tip of his shadow moving and (b) at what rate is the length of the shadow chaging?

Any explanation of just how to do it would be great. :) Thanks so much in advance.

I forgot to mention that there is a picture that comes with the problem. I do believe that it has something to do with similar triangles or something. That's all I know though... I can't think of how to use the rate that is given.

Answer 1

Sorry for the long answer but it seems you could use some explanation.

Similar triangles are correct. The two sides of interest in the similar triangles are

1) The height of the man (H=6 ft), the length of his shadow (S)
2) The height of the light (L = 20 ft) and the distance from the light to the tip of the shadow (T).

Because the triangles are similar:

H/S = L/T

One other parameter is the distance from the light to the man (D = 10 ft). Of course T = D + S meaning the total distance from the light to the shadow is the sum of his distance from the light and the length of the shadow. Combine these two equations:

H/S = L/(D + S)

Now we are getting somewhere. The only unknowns are the shadow length (S) and the distance from the light (D). Rearrange the equation to give S as a function of everything else. Do this because S is what we really need to figure:

H/S = L/(D + S)
H(D + S) = L*S
L*S - H*S = H*D
S = H*D/(L - H)

First figure how fast the shadow is changing length. How fast means change with time, or in calculus terms, dS/dt. Take the derivative of the final equation. Only S and D are variables so:

dS/dt = (H/(L - H)) * dD/dt

We know dD/dt is how fast he is moving or -5 ft/sec. I use negative because D is getting shorter over time. So the shadow is changing length by:

dS/dt = (6/(20 - 6)) * -5 ft/sec = -15/7 ft/sec.
The shadow is shortening by 2.14 ft/sec.

The speed of the tip of the shadow is dT/dt. The very first equation was:

H/S = L/T
T = SL/H

Taking the derivative: dT/dt = (L/H) dS/dt = 20/6 * -15/7 = -50/7 ft/sec.

The tip of the shadow is travelling at 7.14 feet/sec towards the light.

You might notice that the speed of the tip of the shadow is just the sum of the shortening and the walking speed. This is what you would expect.

Answer 2

Try using related triangles.

Draw a picture using the aspects that are changing such as the distance of the man from the light with aspects that aren't changing, the height of the pole and the man and see what kind of picture emerges.

Answer 3

well i don't really understand how the whole shadow thing works but umm

You have to have an equation to figure out the length of his shadow (it sounds like you're going to have to use pythagorean theorem or something)

Also they give you one rate say it's dD/dt (D--distance from light and t=time) so you already know that is dD/dt=5ft/s

gosh this is hard i'm sorry i wish i could help more but it's just the shadow thing throwing me off.