Ladder problem with related rates

[Ingenium]

1. The ladder is 24 ft long and is leaning against a house. The ladder is moving away from the house at a rate of 3 ft/s. I'm supposed to find the rate the slope of the ladder is decreasing when it is 14 ft away from the house.

3. I'm thinking its got something to do with the second derivative of the ladder, but i can think of how to do it in this context.

any help you guys could give would be helpful.
also: yay, first post

 

[grief]

So, let x be the horizontal distance of the ladder, y be the vertical distance. We know x^2+y^2 = 24^2

The slope is z = y\x.
I think we are given that dx\dt = 3.

This should be enough information to find dz/dt.

 

[jakncoke]

agree with what grief said, one more hint. Chain rule!

 

[Ingenium]

where would the chain rule come in? i see where i use quotient rule but that's all I'm seeing.

 

[jakncoke]

since z=y/x

d/dt(z) = d/dt(y/x)
== dz/dt = 1/x dy/dt - y/x^2 dx/dt

we know dx/dt = 3ft/s
we know dy/dt = dy/dx*dx/dt

u know x^2+y^2=24^2

 

[Ingenium]

ok i got it now. thanks a lot guys for the help