# Ladder problem, no equation to use for dy/dt

1. Oct 12, 2009

### Slimsta

1. The problem statement, all variables and given/known data
http://img132.imageshack.us/img132/8048/prob2r.jpg [Broken]

2. Relevant equations
dy/dt = dx/dt * dy/dx

3. The attempt at a solution
would dy/dx = 0? because the ladder is 11 m, the derivative of 11 is 0..
and then would dy/dt be the rate of the top of the ladder sliding down?

Last edited by a moderator: May 4, 2017
2. Oct 12, 2009

### lanedance

you need to find y in terms of x first, thinking about triangles

3. Oct 12, 2009

### alecst

You know that $$\frac{dy}{dt}$$ is equal to $$\frac{dy}{dx}*\frac{dx}{dt}$$

You know $$\frac{dx}{dt}$$ is 3. So all that's left to do is solve for $$\frac{dy}{dx}$$

By the pythagorean theorem, $$a^2+b^2=c^2$$. In this case, on the y-axis you have your a, on the x-axis you have your b and the hypotenuse you have c.

Therefore, $$y^2+x^2=c^2$$, so $$y=\sqrt{c^2-x^2}$$. Differentiate with respect to x.

$$\frac{dy}{dx}=-x/\sqrt(121-x^2)$$

Substitute 8.316 for x, and solve for $$\frac{dy}{dx}$$

4. Oct 12, 2009

### Slimsta

for $$\frac{dy}{dx}=-x/\sqrt(121-x^2)$$ of $$y=\sqrt{c^2-x^2}$$
wouldnt it be first (c-x)/(sqrt(c^2-x^2)) then plug in numbers?
here is what i got so far:
http://img94.imageshack.us/img94/8545/prob2o.jpg [Broken]

Last edited by a moderator: May 4, 2017
5. Oct 12, 2009

### lanedance

might be even easier to use implict differntiation wrt x on the following equation
$$x^2+y^2=c^2$$

note c is constant and y = y(x)

6. Oct 13, 2009

### blake knight

Theres a problem in this problem: What is the starting position of the ladder? Is it initially fully vertical? If it is, what makes it slide down?

7. Oct 13, 2009

### Slimsta

so you are saying that in the picture that i uploaded, the only thing that is wrong is where i wrote $$x^2+y^2=11^2$$ instead of $$x^2+y^2=c^2$$?
everything else is good?

8. Oct 13, 2009

### lanedance

no c = 11 as you said but i was suggesting to try implicit differntiation on the initial equation, though teh way you have looked at it is fine

9. Oct 14, 2009

### lanedance

this is a related rates problem, the speed & ladder position are given, so those questions are irrelevant

10. Oct 14, 2009

### blake knight

I maybe wrong with my assumption lanedance but I think there is a difference in speed as the top of the ladder approaches 7.2' if it starts from 10' or if it starts from 7.3', visualize the scenario.

11. Oct 14, 2009

### lanedance

the way I read it, the problem states the bottom of the ladder slides at a constant rate, so the initial position has no impact

12. Oct 14, 2009

### Slimsta

well, i did everything you mentioned already..
and i get dy/dt = 3.464
so what am i doing wrong?

13. Oct 14, 2009

### lanedance

$$x^2 + y^2 = 11^2$$
y = 7
$$x = \sqrt{11^2-7^2} = \sqrt{72} = \sqrt{2.2^2 3^2} = 6.\sqrt{2}$$
x' = 3
differentiate wrt t
$$2xx' + 2yy' = 0$$
rearrange for y'
$$y' = -xx'/y = -(6\sqrt{2})(3)/7$$

think you just missed the negative

14. Oct 14, 2009

### Slimsta

well it was negative at first for both the rate change and for dy/dt but then the system say it should be positive...
yeah that works.. i thought they both supposed to be the same answer.
thanks for help! that makes sense now!