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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 $Q ?$
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04:21
Wen Zheng
01:19
Amrita Bhasin
Calculus 1 / AB
Chapter 3
Differentiation Rules
Section 9
Related Rates
Derivatives
Differentiation
Catherine A.
October 27, 2020
Thought I needed a tutor to help with Calculus: Early Transcendentals, but this helps a lot more.
Sharieleen A.
October 23, 2020
This will help alot with my midterm
Missouri State University
Harvey Mudd College
Baylor University
Lectures
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In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.
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In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.
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Alright, here's a cool problem. We have two cards held together by a 39 ft long robe. And the left cards moving to the left at two ft per second. And we want to find out At the position that the card is five ft away from the center basically. What is the rate that cart B. Is moving toward Q. Which is basically the center position here. So a pretty cool problem. Alright. So let's start off by just looking at card A. To kind of analyze what's happening here. So card A notice that as it goes to the left, we form a right triangle. You know the height is 12 ft. And in general his position is X. A. And so if we were to do a a pythagorean theorem, we'd say X. A. Square X. A squared plus 12 squared equals let's call this length length A. That's a link that's currently on the side of card A equals length A. Squared. I can take the derivative both sides. That will give Me two X. A. Um dx DT alcohol, D X A D. T Plus zero equals twice the length of the string. D. L. A. D. T. Because that continue to a time. Okay, we can divide through by two. And that gives us then our rate of change of length of the rope is equal to X. A. Over L. A. And then um D. X. A. D. T. Well, we can find that specific value based on the information we've been given. You know, X.A. is five ft at the point of interest. Um L. A. Is um by its when when X is five, we have a 5 12, 13. A special triangle. You could do pythagorean theorem, but it's not that hard to see that our length is actually um Going to be the 13 as our hypotenuse 13 ft. And then we are told we are traveling at two ft/s. So therefore we are getting uh 10/13. That is our way to change of length of the rope with time. And I'm focusing on the rope because the rope is the one thing that connects both part A and part B. So now we can go over and take a look at what's happening with part B. Because ultimately we want to know how fast Cardi B is moving towards Q. Okay, so same idea, if I were to draw a right triangle, that kind of looks more this way. Um Then I have um This length of the hypotenuse is now 39 -6. A. We still have a height of 12 ft and I'll call the distance away from Q. XB. Um Okay, so At this moment we know L. A. is 13. So if we subtract 39 -13, that gives us 26. So the current length of the string is 26. So I can find XB at the moment by pythagorean theorem. So XP a square root of 26 squared minus 12 squared. And that turns out to be um 532 in the square root. So it doesn't clean up. Nice. So there it is A X. B. Is route 5 32. And that would be in feet okay. Like before we can set up our pythagorean theorem, X p squared plus 12 ft. That should have been squared before anyway, and squared now equals L. B squared. So we take a derivative, we get this change color to back, just get a different color going. That will be twice X. B. D X B. D. T. Remember. That's what we want. We want this dx B. D. T. That will be our final answer. The constant derivative is zero. Now we'll get twice L. B. Time's D L. B. D. T. Okay, so now we can divide through by twos like we did before and really we want the XB D. T. And that will be equal to um lb over XP D L B G T. Let's see if we have room to squish in values here. Okay, so L. B we know is equal to 26 ft over XP XP is currently square root of 5 32 ft. So get feet over feet and the lbgt um it's going to be the same rate of changes so they could get shrinking. So I get minus 10, 13 feet per second. So um we can clean that up just a tad Um it turns out, you know, 26 divided by 13 is too, so I end up with -20 but way up top here. So the final answer for the rate of change of carby with time, um We're going to do absolute values since we're just doing how fast. So that will give me 20 Over square root of 5:32 ft her second. And we did it. Yay. Alright. Hopefully that helped have an amazing day and see you next time.
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