# Math Help - Related Rate-Ladder against the wall

1. ## Related Rate-Ladder against the wall

Here is the problem I'm working on:

A ladder 20ft long leans against a house. If the foot of the ladder is moving away from the house at the rate of 2ft/s, find how fast the slope of the ladder is decreasing when the foot of the ladder is 12ft from the house.

Here is what I have so far:

Given:
• $\frac{dx}{dt}=2$
• x=12
• z=20
Find:
• $\frac{d\theta}{dt}$ when x = 12
My Work:
• By the theorem of Pythagoras:
• y=16
• $sin\theta = \frac{y}{z} = \frac{16}{20} = \frac{4}{5}$
• $cos\theta=\frac{x}{20}$
• $x=20cos\theta$
• $\frac{dx}{dt}$ = $20(-sin\theta)*\frac{d\theta}{dt}$
• $2 = 20*\frac{-4}{5}*\frac{d\theta}{dt}$
• $\frac{d\theta}{dt} = \frac{-1}{8} rad/s$

• The book, however, came up with $\frac{25}{72} per second$
Any tips on where I went wrong?

2. Well, the most important thing I can see is that I don't think you're solving for the right variable. It's asking for how fast the slope is changing at that point. You attempted to calculate (I haven't checked the accuracy) how fast the angle the ladder makes with the floor is changing. Those are not the same.

What is the slope in general? How could you find its derivative in time?

3. Well, the general definition of slope is $\frac{\Delta y}{\Delta x}$

@ 6 seconds $\frac{\Delta y}{\Delta x} = \frac{9}{12} = \frac{3}{4} ??$

I'm not quite sure how I would go about finding its derivative...another hint?

4. What about $tan \theta = \frac{y}{x}$

Taking the derivative, I get:

$sec^2 \theta*\frac{d\theta}{dt} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

...am I on the right track?

5. What you're calculating (incorrectly there) is the actual slope at the time when you're given data. However, that's not what the problem is asking for, either. You're asked to find how fast the slope is changing. Let's say you define the slope as

$m(t)=\dfrac{y(t)-0}{0-x(t)}=-\dfrac{y(t)}{x(t)}.$

$\dfrac{dm}{dt}.$

I would agree that this is a bit confusing: you're trying to take the derivative of a slope! No, it's not really a second derivative, because here you're taking the time derivative, whereas the slope is a sort of spatial derivative in this case.

How can you find this time derivative?

6. I'm still stuck...can't seem to wrap my head around this one

7. Do you understand how I got m(t)?

8. Not really...I was thinking $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$

9. Exactly. But now let's suppose the point $(x_{1},y_{1})$ is the point where the ladder touches the floor, and let's suppose the point $(x_{2},y_{2})$ is the point where the ladder touches the wall. What are $x_{2}$ and $y_{1}?$

10. ok..i see that they are both equal to zero

I can't just plug in numbers at this point bc I would get $\frac{-4}{3}$

So I am a bit confused as to what y(t) and -x(t) are and how I would take the derivative of them?

11. Right. So, since there's only one x coordinate to keep track of, and only one y coordinate to keep track of, we'll drop the subscripts and get the m I have in post # 5. You follow that now? (Keep in mind also that both x and y change in time).

12. I'm still stuck. I don't really understand what y(t) and -x(t) are?

13. y(t) is the height of the point at which the ladder touches the wall, and x(t) is the x-coordinate of the point at which the ladder touches the floor. You have to have the negative sign in the slope there, because the y and x do not correspond to the same point. Make sense?

Or think of it this way: y is positive up (along the wall), and x is positive to the right (along the floor). The slope of that ladder has to be negative, because it's going down as x increases. But y and x are both positive. Hence, m = - y / x. Make sense?

14. Yeah, that part makes sense, I just dont understand how to differentiate -y/x

15. Well, if you have a quotient there, what rule do you suppose you're supposed to use?

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