Like

Report

Two carts, $\mathrm{A}$ and $\mathrm{B},$ are connected by a rope $39 \mathrm{ft}$ long that passes over a pulley $P$ (see the figure). The point $Q$ is on the floor $12 \mathrm{ft}$ directly beneath $P$ and between the carts. Cart A is being pulled away from $Q$ at a speed of $2 \mathrm{ft} / \mathrm{s}$. How fast is cart B moving toward $Q$ at the instant when cart A is $5 \mathrm{ft}$ from $Q ?$

$\frac{20}{\sqrt{523}} \mathrm{ft} / \mathrm{sec}$

You must be signed in to discuss.

Sharieleen A.

October 23, 2020

This will help alot with my midterm

Catherine A.

October 27, 2020

Thought I needed a tutor to help with Calculus: Early Transcendentals, but this helps a lot more.

Campbell University

Oregon State University

Harvey Mudd College

University of Nottingham

So you have two carts A. And B. And they are connected by a rope that's 39 ft long and it passes over Pulley P. So we're gonna put the pulley up here recalling it, letter P. And these are connected by a rope. And that rope is 39 ft long. The Point Q. is on the floor, 12 ft directly Beneath P. So Q. Is down here and the distance down Is going to be 12 ft. And we are pulling cart A. In this direction away from Q At two ft/s. And we want to know how fast is cart be moving toward Q. So when we pull on this cart, this pulley is going to pull this rope and this be is going to move toward Q. And we want to know how fast that is moving at the instant when cart A. Is five ft from point Q. So what we've got going here is we know that the distance from A. To P plus the distance from P. To be has to be the 39 ft of rope. Now, if you think in terms of the fact that there are two right triangles in this picture, one has a five and a. 12 here. And on the other side We've got a 12 and some unknown value. So what we can do is we can call this A. We can call this b. And if we were to use the Pythagorean theorem on each of those triangles, then a squared plus 12 squared has to equal the hypotenuse squared. So therefore this hypotenuse would be the square root of A squared plus 12 squared. And if we did the same thing on the other triangle, we'd have b squared plus 12 squared equals the hypotenuse squared. So therefore this hypotenuse would end up being the square root of B squared plus 12 squared. And we know that again from a two P is now that square root of a squared plus 12 squared. And from B two P is b squared plus 12 squared. And It has to end up adding up to 39 from there. We are talking in terms of related rates and when we're talking in terms of rates, we're talking in terms of derivatives. So we're going to need to take the derivative of this equation. So I'm going to rewrite it as being a squared plus 144 to the half power plus B squared plus 144 to the half power. So I'm now going to take the derivative. And when I take the derivative, I will have to employ the chain rule uh in two different instances. So my derivative will be one half A squared plus 144 to the negative half power time's D. Or to A. And then times D A D. T plus I'll take the derivative of this next term. So I'll have one half B squared plus 144 raised to the negative half power Multiplied by two B. DBT. And when I take the derivative of what's on the other side of the equal sign, the derivative of 39 is going to be zero. So I'm gonna want to clean that up a little bit. So when I clean that up I can reduce the half with this to A. And just be left with a. This expression here because it has a negative power is the same thing as saying over the square root of a squared Plus 144 time's D. A. D. T. Yeah. Plus this too. And this half would cancel. So you would be left with B. Because of this negative power. We could say over the square root of B squared plus 144 DB DT equals sarah. So at that point we've got to talk about the components. So we know that A. Was five. So everywhere we see an A we can put a five. So now we'll have five over the square root of five squared plus 1 44. D. A. D. T plus B. Over the square root of B squared plus 144 multiplied by DBT Has to Equal zero. Now let's think about the value of B. We want to be able to fill in B. So we know that if this is five And this is 12 Then we have a 5, 12, 13 triangle. So this high pot news here would end up being 13 at the moment when the cart cart a. Is five ft away from Q. And we know that when we add up This side plus this side we have to end up with 39. So we could then say that 13 plus the square root of B squared plus 12 squared has to equal 39. And that's going to help us solve for r value of B. That we will eventually bring back down into our Equation. So we'll end up with 13 plus the square root of B squared plus 144 equals 39. If we subtract 13 from both sides, we have the square root of B squared plus 144 equals 26. If we square both sides then we will get B squared Plus 144 equals 676. And if we subtract 144 from both sides, we know that B squared will equal the square root of 530. Or sorry B squared will not be the square root of B squared will be 532. So therefore be will equal the square root of 532. So therefore we can now go back and substitute our value of B into our equation down below. And we could say five Over the square root of 25 plus 144 D. A. D. T plus The Square Root of 532. Over the square root. Now we know that B squared is 532. So we could say 532 plus 144. DBT equals zero. Now let's think about the fact that we are pulling a cart a away from point Q. At two ft per second. So we're pulling this at two ft/s. So that is going to be how A. Is changing. So now in place of the A. D. T. We can put to So as we place that in, we're going to clean up as well. So we're gonna get five over the square root of 169, multiplied by two plus The square root of 532 can be reduced. That can be reduced or simplified into radical four times radical 133. And when I simplify the denominator I'll get the square root of 676. DBT equal to zero. And I can keep going here. And I could multiply these and I could say 10 over the square root of 169 is 13 Plus two. Radical 1 33. and the square root of 676 is 26. DBT. I can simplify the two and the 26. So now I would have 10/13 plus the square root of 1 33/13. DBT Is equal to zero. My goal is to solve for DV DT So I'll subtract 23rd from both sides. So I'm at the square root of 133 over 13. DBT Is equal to negative 10/13. If I were to multiply both sides of the equation by a 13, my denominators will drop out so I'll have the square root of 133. DBT is equal to negative 10, dividing both sides by the square root of 1 33. D B D T equals negative 10 over the square root of 1 33. Now that would be an acceptable answer. But oftentimes we rationalize instead of leaving the radical in the denominator. So you would end up with negative 10. Radical 133 Over 133. Now, if I brought in my calculator to just get my head around it a little bit, so I'm going to bring in my calculator. If I were to Say negative 10 multiplied by The square root of 133 divided by 133, you're going to get an answer of approximately negative point 8671. And our units of measure on this is going to be in feet per second and that concludes your problem