# Homework Help: Related rates of a ladder sliding

1. Nov 15, 2016

### Arnoldjavs3

1. The problem statement, all variables and given/known data
A 5 m ladder is sliding down the wall, and h is the height of the ladder's top at time t, and x is the distance from the wall to the ladder's bottom at time t.
Given that h(0) = 4 at t = 0 seconds and dh/dt = 1.2m/s, and the ladder is 5m long find x(2) and dx/dt at t=2 seconds

2. Relevant equations
x^2 + h^2 = 5^2
25 - 16 = x^2
x = 3

3. The attempt at a solution
$$2x*\frac{dx}{dt} + 2h*\frac{dh}{dt} = 0$$
since we know that dh/dt = 1.2 and x = 3:
$$2(3)*\frac{dx}{dt} + 2(4)*(1.2)= 0$$
$$\frac{-9.6}{6} = \frac{dx}{dt}$$
$$\frac{dx}{dt} = -1.6$$

I believe dx/dt is -1.6 at t = 0 seconds, but how do I use this information to find x(2) and dx/dt at 2 seconds? Am I wrong in approaching this problem?

2. Nov 15, 2016

### Staff: Mentor

In your problem statement, dh/dt appears to be constant at 1.2 m/s (really, this is -1.2 m/s). You're given that h(0) = 4, so what will h be at t = 2 sec.? From that you can find x(2).

3. Nov 15, 2016

### Arnoldjavs3

h(2) = 1.6m and x(2) = 4.73?

Is my initial approach wrong though? And why is it that I got $$\frac{dx}{dt} = -1.6$$ which is the same coefficient of h(2)? I get that from a logical perspective this problem is very easy but I still want to do the math behind it.

4. Nov 15, 2016

### Staff: Mentor

I don't get these values. dh/dt = -1.2 m/sec, and h(0) = 4. How did you get h(2) = 1.6 m?

Last edited: Nov 16, 2016
5. Nov 16, 2016

### Arnoldjavs3

Realized my mistakes were because I wasnt' careful enough
Okay so h(2) = 1.6m as 4-2(1.2) = 1.6
then we can use that to find x(2):
√(25-1.6^2) = 4.7377

And then i input these values in to:

$$2(4.73)*\frac{dx}{dt}=2(1.6)(1.2)$$
$$\frac{dx}{dt} = \frac{3.84}{2(4.737)}$$

and i get 0.405 for dx/dt.

Edit: changed this post as the previous was wrong

Last edited: Nov 16, 2016