# Thread:

1. ## Relative Rates

I've added an attachment to this post to explain the problem.

Two carts, A and B, are connected by a rope 39 feet long that passes over a pulley P. The point Q is on the floor h = 12 ft directly beneath P and between the carts. Cart A is being pulled away at a speed of 2.5 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q. (round to 2 decimal places)

From the problem, I know the following things:

PQ = h = 12
AQ = 5
AP = 13 (by Pythagoras)
APB = 39
PB = 39-13 = 26
QB = $2\sqrt{133}$ (by Pythagoras)

(work to follow in next posting for organization)
Attached Thumbnails

2. My strategy is to work clockwise around the triangle to find $\frac{dQB}{dt}$ as follows:

First, find $\frac{dAP}{dt}$ since this must be equal to $\frac{dPB}{dt}$ since the length of these two segments is constant:

$\frac{dAP}{dt}(AP)^2 = \frac{dAQ}{dt}(AQ)^2$ (note that PQ is a fixed length so its derivative is zero)

$\frac{dAP}{dt} = \frac{AQ}{AP} \frac{dAQ}{dt} = \frac{5}{13} (2.5) = \frac{25}{26} ft/sec$

Again, if $\frac{dAP}{dt} = \frac {25}{26}$, then $\frac{dPB}{dt} = \frac{-25}{26}$ since the length of this segment is constant. THIS IS A KEY ASSUMPTION! IS IT CORRECT???? (sorry for the caps)

Then, I find my ultimate answer using Pythagoras:

$\frac{dQB}{dt}(2\sqrt{133})^2 = \frac{dPB}{dt} (PB)^2 - \frac{dPQ}{dt} (PQ)^2$

$(4\sqrt{133}) \frac{dQB}{dt} = \frac{-25}{26} (26^2) - 0$

$\frac{dQB}{dt} = \frac{-650}{4\sqrt{133}} = 0.30545$ feet / second

Unfortunately, this answer doesn't check out, so I'm doing something wrong, probably in my key assumption, but I don't see it. Can somebody help?

Thanks.

3. Hello, joatmon!

Two carts, A and B, are connected by a rope 39 feet long that passes over a pulley P.
The point Q is on the floor h = 12 ft directly beneath P and between the carts.
Cart A is being pulled away at a speed of 2.5 ft/s.
How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q.
(Round to 2 decimal places)

Code:
```              P
              *
             /:\
            / : \
           /  :  \
          /   :   \
         /    :12  \
        /     :     \
       /      :      \
    A * - - - * - - - * B
          x   Q   y```

Let $x = AQ,\;y = QB$
We are given: . $AP + PB \,=\,39\,\text{ and }\,\dfrac{dx}{dt} = 2.5$

In right triangle $PQA\!:\;AP \,=\,\sqrt{x^2+12^2}$

. . Then: . $PB \,=\,39 - \sqrt{x^2+144}$

In right triangle $PQB\!:\;y^2 \:=\:(39 - \sqrt{x^2+144})^2 - 12^2$

. . . . . . $y^2 \;=\;x^2 - 78(x^2+144)^{\frac{1}{2}} + 1521$ .[1]

Differentiate implicitly:

. . $2y\,\dfrac{dy}{dt} \;=\;\bigg[2x - 78\cdot\frac{1}{2}(x^2+144)^{-\frac{1}{2}}\cdot2x\bigg]\,\dfrac{dx}{dt}$

. . . . $\displaystyle \frac{dy}{dt} \;=\;\frac{1}{y}\bigg[x - \frac{39x}{\sqrt{x^2+144}}\bigg]\,\frac{dx}{dt}$ .[2]

Substitute $x = 5$ into [1]:

. . $y^2 \;=\;25 - 78(13) + 1521 \:=\:532 \quad\Rightarrow\quad y \:=\:2\sqrt{133}$

Substitute into [2]:

. . $\displaystyle \frac{dy}{dt} \;=\;\frac{1}{2\sqrt{133}}\left[5 - \frac{39(5)}{\sqrt{169}}\right](2.5) \;=\;\frac{1}{2\sqrt{133}}(-10)(2.5)$

$\displaystyle\text{Therefore: }\frac{dy}{dt} \;=\;-\frac{25}{2\sqrt{133}} \;\approx\;-1.08\text{ ft/sec}$

But check my work . . . please!
.

4. Originally Posted by joatmon
My strategy is to work clockwise around the triangle to find $\frac{dQB}{dt}$ as follows:

First, find $\frac{dAP}{dt}$ since this must be equal to $\frac{dPB}{dt}$ since the length of these two segments is constant:

$\frac{dAP}{dt}(AP)^2 = \frac{dAQ}{dt}(AQ)^2$ (note that PQ is a fixed length so its derivative is zero)

$\frac{dAP}{dt} = \frac{AQ}{AP} \frac{dAQ}{dt} = \frac{5}{13} (2.5) = \frac{25}{26} ft/sec$

Again, if $\frac{dAP}{dt} = \frac {25}{26}$, then $\frac{dPB}{dt} = \frac{-25}{26}$ since the length of this segment is constant. THIS IS A KEY ASSUMPTION! IS IT CORRECT???? (sorry for the caps)
Yes, that is correct. AP+ PB= 39 so (AP)'+ (PB)'= 0, (BP)'= -(AP)'.

Then, I find my ultimate answer using Pythagoras:

$\frac{dQB}{dt}(2\sqrt{133})^2 = \frac{dPB}{dt} (PB)^2 - \frac{dPQ}{dt} (PQ)^2$
I don't understand why you have the squares there nor where you got $2\sqrt{33}$. What formula did you differentiate to get this?

$(4\sqrt{133}) \frac{dQB}{dt} = \frac{-25}{26} (26^2) - 0$

$\frac{dQB}{dt} = \frac{-650}{4\sqrt{133}} = 0.30545$ feet / second

Unfortunately, this answer doesn't check out, so I'm doing something wrong, probably in my key assumption, but I don't see it. Can somebody help?

Thanks.

5. Thanks to both of you. Soroban, your work helped me immensely. We were together in calculating the segment lengths. Where I went wrong was in applying the differential equation. I didn't equate x and y correctly, so my differential was all screwed up. Thanks for setting me straight.

Halls, the $2\sqrt{133}$ comes from the Pythagoream theorem. We determined that the PB hypotenuse was 26 and the height of 12 was given. Thus, the segment at the bottom that needed to be measured in order to differentiate was calculated like this:

$\sqrt{26^2 - 12^2} = 2\sqrt{133}$

Thanks again!

6. I originally thought the sum of the bases of the triangle (x and y in your diagram) would have to be constant by thinking that in real life if the base was not constant and say it was getting shorter that would require there to be some slack in the rope. But it seems that x+y is not constant here. I guess that's because our assumption is that this problem can be modeled with a triangle no matter what.
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