# Related Rate Shadow & Spotlight Question

• Mar 10th 2009, 01:43 AM
Sajiko
Related Rate Shadow & Spotlight Question
I have a question that I am stuck on for the Related Rates problems. I did it and got a negative answer when it's suppose to be positive I did it again a little differently and still got a negative answer!! Here is the problem:

Joe is 1.8 meters tall is walking directly towards a spotlight located on the ground 20 meters away from the theatre. If Joe walks at a constant rate of 1/5 m/s find the rate at which the length of his shadow on the wall is changing when he is 6 meters from the lamp.

So the 2nd time I went through it I set it up as:

Given: dx/dt = 1/5 m/s
Find: dy/dt when x = 6

y/20 = 1.8/20-x
y = 36/(20-x)
dy/dt = [-36/(20-x)^2](dx/dt)
dy/dt = [-36/(20-6)^2](1/5)
dy/dt = -9/245

Worked from there. First time I got an answer of -1/5, second time an answer of -9/245. But I know if he is moving towards the spotlight the answer will have to be a positive rate. I think there is something fundamentally wrong with my procedure, if anyone could show me their through process on this I would greatly appreciate it!
• Mar 10th 2009, 02:56 AM
earboth
Quote:

Originally Posted by Sajiko
I have a question that I am stuck on for the Related Rates problems. I did it and got a negative answer when it's suppose to be positive I did it again a little differently and still got a negative answer!! Here is the problem:

Joe is 1.8 meters tall is walking directly towards a spotlight located on the ground 20 meters away from the theatre. If Joe walks at a constant rate of 1/5 m/s find the rate at which the length of his shadow on the wall is changing when he is 6 meters from the lamp.

So the 2nd time I went through it I set it up as:

Given: dx/dt = 1/5 m/s
Find: dy/dt when x = 6

y/20 = 1.8/20-x
y = 36/(20-x)
dy/dt = [-36/(20-x)^2](-1)(dx/dt) You have to use the chain rule and you forgot about the derivation of the inner function
dy/dt = [-36/(20-6)^2](1/5)
dy/dt = -9/245

Worked from there. First time I got an answer of -1/5, second time an answer of -9/245. But I know if he is moving towards the spotlight the answer will have to be a positive rate. I think there is something fundamentally wrong with my procedure, if anyone could show me their through process on this I would greatly appreciate it!

...
• Mar 10th 2009, 02:57 AM
Sajiko
Quote:

Originally Posted by earboth
...

AHhhhhhhhhhh!!!! I always make such stupid mistakes! Thank you so much!

But otherwise then does 9/245 look like the right answer?
• Mar 10th 2009, 03:15 AM
earboth
Quote:

Originally Posted by Sajiko
AHhhhhhhhhhh!!!! I always make such stupid mistakes! Thank you so much!

But otherwise then does 9/245 look like the right answer?

Correct (Clapping)

Don't forget that a speed of $\dfrac15\ \dfrac ms$ describes a more snail like movement.
• Mar 10th 2009, 03:45 AM
Sajiko
Thank you and yeah such a slow rate lol, but that is what the teacher gave us.

If you wouldn't mind could you take a look at the other topic I have in this forum? It's about 2 trough questions I have no approach to tackle them; I know how to solve a normal trough with isosceles triangles at each end when they give you a certain volume but not with time.

If you could that would be great!