Question

Related Rates! A ladder 16 feet long is learning against the wall of a building. The base of the ladder is moving away from the wall at a rate of 1 foot per second.
Find the rate at which the angle theta between the ladder and the wall of the building is changing when the base of the ladder is 8 feet from the wall.

Answer 1

I've already drawn a picture and come up with the relationship: sin(theta) = x/16. Is that correct?

Answer 2

you re right

Answer 3

what is that @ carabell

Answer 4

Then here's my issue: If you try to differentiate that--you know, to get the rate of change--you don't end up using the x.
sin(theta) = x/16 --> cos(theta) = 1/16 dx/dt. dx/dt is just 1. And the x=8 doesn't come into play. Am I making sense? I'm just concerned.

Answer 5

Oh, oops. There's probably a dtheta/dt in there as well...

Answer 6

What do you get if you differentiate in respect to time? Or do you have to do that at all?

Answer 7

rate = d(theta)/dt = I/(16 x Cos theta)

Answer 8

Is is possible for you explain how you got that?

Answer 9

cos(theta) d(theta)/d (time) = 1/16 dx/dt
so from question, we know that dx/dt = 1 foot per sec
hence
cos(theta) d(theta)/d (time) = 1/16 (1)
d(theta)/dt = I/(16 x Cos theta)

Answer 10

Oh! And that x is multiplication?

Answer 11

yes

Answer 12

What I don't understand is why do they, in the question, specify x=8 if you're not going to use it?

Answer 13

may be to confuse..... you just need to get the idea....

Answer 14

Okay. Thank you for your help.

Answer 15

if you ll use d(8+x)/dt , still then you ll get dx/dt because 8 is constant

Answer 16

pleasure.....