Related Rates: The ladder problem

[cphill29]

Homework Statement

An 18 meter ladder is sliding down a vertical wall at a rate of 2.5 m/s. Find the speed of the lower end of the ladder when it is 12 meters from the wall.

Homework Equations

Pythagorean Theorem

The Attempt at a Solution

Let h = height of the wall
L = length of the ladder
b = distance from the wall to the bottom of the ladder

dh/dt = 2.5 m/s
db/dt = ? when b = 12

18^2 = 12^2 + h^2
h = rad(180)

L^2 = h^2 + b^2
18^2 = 2h dh/dt + 2b db/dt
0 = 2h (2.5) + 2(12) db/dt
0 = 2[rad(180)] (2.5) + 2(12)(db/dt)

-2.79 m/s = db/dt

Could anyone verify that this is right? Please correct me if I'm wrong.

 

[lineintegral1]

You need to solve for h at the given instant in time (you can do this easily given the Pythagorean relationship; this should involve no differentials). Also, note that the length of the ladder does not change at all as time progresses. Therefore, \frac{dL}{dt} will be zero.

Also, there is an error in your setup. Is \frac{dh}{dt} increasing or decreasing?

 

[cphill29]

I suppose dh/dt would be decreasing, making my final answer positive?

 

[lineintegral1]

Yup! Draw a picture if it helps, but this should be evident from daily experiences. What the positive answer is saying, of course, is that the length of the bottom leg of the triangle is increasing.

 

[cphill29]

Right, thank you for your help! I need to be more aware of positives and negatives.