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Two cars start moving from the same from the same point. One travels south at $ 60 mi/h $ and the other travels west at $ 25 mi/h. $ At what rate is the distance between the cars increasing two hours later?
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02:50
Wen Zheng
01:20
Amrita Bhasin
Calculus 1 / AB
Chapter 3
Differentiation Rules
Section 9
Related Rates
Derivatives
Differentiation
Jake D.
March 16, 2022
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Jake N.
February 26, 2019
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Baylor University
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Lectures
04:40
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.
44:57
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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Two cars start moving from…
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(a) One car leaves a given…
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a. One car leaves a given …
00:25
Two cars start from a poin…
04:48
Two cars leave an intersec…
02:42
Two cars start from the sa…
Alright, so here we have a fun scenario, we have at one central point, we have um to folks that are starting to travel, we've got um one person traveling south with a velocity of 60 MPH and then we have another person traveling west with a velocity of 25 mph. And if you were to um kind of imagine then with time they would each travel a certain distance. Well let me make that a little bit nicer. I will use a straight line. Okay, so after a certain time the second person will have traveled horizontally to uh in the west direction and um the other person that's going south will travel even further. I'll make it really big. Okay? Um and then you can kind of imagine this has been a right triangle, which it is. So basically I'm going to call the distance traveled by the second person as X two. And the distance traveled by the first person going south as X one. And the question is basically saying if we call this distance steve, well, how fast is that distance changing with time? So what's the rate of change of that kind of hypotenuse with time. Alright, so to solve this, we just need to know our basic physics that the distances rate times time. So X one is V one times time And X two then is V two times time. But we have values, so let's plug it in. V one is 60, so do 60 t. And V two is 25. So we'll do 25 t. Alright, so let's find our distance with time and then we can take the derivative. So distances, pythagorean theorem because it's the hypotenuse of a right triangle. So we're gonna basically do X one squared plus X two squared and then square root the whole thing. So we can plug in what we have, we have 6080 Squared plus 25 T squared. And why don't we clean that up a little bit? Let's see that will give us 3600 T squared Plus 625 T squared for a net. Um Let's see. 4225 T squared. Well we can certainly remove the tea from the square root because it's squared inside and it turns out 4225 is actually uh 65 squared. So this whole thing actually cleans up to simply be 65 T. Which is pretty cool. Makes our derivative a whole lot easier. So now if I want to take the derivative of that hypotenuse split time, it's just 65. That's so cool. The units will be distant. So miles her our So that hypotenuse is traveling or getting bigger um at 65 mph. So that's pretty cool. Alright so there we have solved it. That was fun. Alright, I'll see you next time
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