How Fast Does the Ladder Angle Change as It Slides Away?

[d.smith292]

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.0 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall?



So, I'm having a difficult time figuring out related rates. I have grasped derivatives just fine, but developing the diagram and labeling seem to be my problems.
I’ve set up my graph like this:

http://www.flickr.com/photos/23826260@N04/8520534293/ by d.smith292, on Flickr

θ is my changing rate in this problem. I've set my equation up like this :

x = 8ft
x' = 1ft/sec

cosθ = x/10

When I differentiate the problem I get

-sinθ = 1/10(x')

When I input this into my calculator I get some crazy long decimal answer, like -.1001674212 Where am I going wrong. I've probably got the entire thing wrong. I'm having trouble with this section. I appreciate the help.

 

[pasmith]

When I differentiate the problem I get

-sinθ = 1/10(x')

On the left hand side you should have
<br /> \frac{\mathrm{d}}{\mathrm{d}t} \cos\theta<br /> = (- \sin\theta) \frac{\mathrm{d}\theta}{\mathrm{d}t} <br />

 

[Mark44]

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.0 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall?



So, I'm having a difficult time figuring out related rates. I have grasped derivatives just fine, but developing the diagram and labeling seem to be my problems.
I’ve set up my graph like this:

http://www.flickr.com/photos/23826260@N04/8520534293/ by d.smith292, on Flickr

θ is my changing rate in this problem. I've set my equation up like this :

x = 8ft

No, x is variable that depends on (is a function of) time. At a particular moment x is 8, but at other times it has different values.

x' = 1ft/sec

cosθ = x/10

When I differentiate the problem I get

-sinθ = 1/10(x')

Not quite.
You are differentiating both sides with respect to t, so you should get
-sin(θ) dθ/dt = (1/10) dx/dt

When I input this into my calculator I get some crazy long decimal answer, like -.1001674212 Where am I going wrong. I've probably got the entire thing wrong. I'm having trouble with this section. I appreciate the help.

 

[d.smith292]

OK, so I see that I wasn't taking the derivative of θ with respect to time (t). Here is where I'm at now and stuck again.

Cos θ = x/10 → (Now I take the derivative with respect to time)

(-sin θ)dθ/dt = (1/10)dx/dt

Now since the rate of change for x = 1 ft/sec I get

(-sin θ)dθ/dt = 1/10

From here I don't know if θ should equal 8/10 or 6/10. Either way I continue to get some crazy answer.

And am I even on the right track here?

 

[eumyang]

OK, so I see that I wasn't taking the derivative of θ with respect to time (t). Here is where I'm at now and stuck again.

Cos θ = x/10 → (Now I take the derivative with respect to time)

(-sin θ)dθ/dt = (1/10)dx/dt

Now since the rate of change for x = 1 ft/sec I get

(-sin θ)dθ/dt = 1/10

From here I don't know if θ should equal 8/10 or 6/10.

It's neither. It should be
\cos \theta = \frac{8}{10}.

Solve for θ, and then plug that, and dx/dt = 1, into
-\sin \theta \frac{d\theta}{dt} = \frac{1}{10} \frac{dx}{dt}

and solve for dθ/dt.

 

[d.smith292]

So, I had everything right up until I solved for theta. I've rewritten the problem with detailed steps from the beginning so that I can remember this for future reference. Thank you.