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The perimeter of a rectangle is growing at the rate of $6 \mathrm{cm} / \mathrm{sec}$. Its area is growing at the rate of $10 \mathrm{cm}^{2} / \mathrm{sec} .$ At a certain time one side is $50 \mathrm{cm}$ long and the other side is $20 \mathrm{cm}$ long. Find the rate of change of each side. Are they growing or shrinking?

Longer side is increasing at $14 / 3 \mathrm{cm} / \mathrm{sec}$, shorter decreasing at $5 / 3 \mathrm{cm} . \mathrm{sec}$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

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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.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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The perimeter of a rectang…

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'The length of a rect…

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The length / of a rectangl…

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The length of a rectangle …

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Expanding rectangle A rect…

09:31

Changing dimensions in a r…

10:22

The length $l$ of a rectan…

04:18

05:26

Changing Dimensions in a R…

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08:48

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A rectangle initially has …

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The length $x$ of a rectan…

it's always beneficial to begin a related rates problem with a picture. So for this problem, we're going to start with a picture of a wreck, tank flanks and wit. And we're told that both the perimeter and the area of this rectangle are changing. So let's write down what we know. We know the perimeter is growing. That's a positive change at the rate of six centimeters per second. And we're told that the area is also growing again, a positive change at the rate of 10 square centimeters per second. And the question we're asking is the rate of change of each side. So I want to know how d l is changing with respect to t N d. W. Now we are told that at this time that we're gonna be looking at we're told that one side is 50 centimeters will call that length because it's the longer of the two, and the width is 20. Those were not going to go on a picture, however, because they're not constant. Those were just a snapshot at a specific instance. In time, because they are in flux. I'm leaving them as variables on our picture. Okay, now, we have two different pieces of information, one about perimeter and one about area. So let's try perimeter first. What equation would relate the perimeter to the sides of the rectangles? Perimeter equals two times the length, plus two times the width Right before we take the derivative. Let's make sure we have all of the rates we need. I have three variables. PLN W I know DP DT and the other two or what I'm trying to find. So three unknown. Three variables. Three rates we could take the derivative dp DT well to l derivative is two times d l d w a r d t and then to D W D t. Now let's substitute in what we know. I know that dp DT equals six. The other two are the two unknowns were trying to find. So they're going to stay here and I can simplify this a little. I could divide everything by three. Okay, so though that is my first piece of information, however, it's not enough to solve because still have two unknowns. So let's move on to area. Well, the area of a rectangle is length, times width. Again. I have three variables, But I know one of those rates and the other two or what we're trying to find. So three and three we can move on to take the derivative de a d. T. Now this is a product, so that's going to be the first times the derivative of second Plus the second times the derivative of the first. Just like before, we'll substitute in what we know. So I have 10 equals 50 d W d t plus 20 d l d t Simplify this a bit. I could divide everything by 10 and I end up with one equals five d W d t plus two d l d t So now I have two equations into unknowns. What? We can solve this. This is just some basic algebra. So let's take I'm going to do We could do substitution or elimination. I'm going to do substitution and I'm going to solve this for D W d T. You could do either one. I just picked one at random. Hey, and now we'll substitute So I end up with one minus five times. There's my substitution. Plus two d l d t. If you don't mind, I'm going to just scroll a little bit to get myself a little more room. Let's get rid of our parentheses. I have one minus 15, plus de l D T e did. It is a minus and it was supposed to be in equals. I apologize for that. That's a miss copy on my part. One equals 15. And let's get this because that messed me up there. So 15 of minus five D l d t plus two d l d t Let's subtract that 15 from both sides. That is the negative 14 on the right. I have negative three d l d t. And when I saw for this, that gives me 14 3rd and I am in centimeters per second. So that's our first rate. So that means the length is increasing. It's a positive numbers. That means it's increasing at a rate of 14 3rd centimeters per second. Now, what about our wits? Okay, well, we're gonna go back toe one of our equations. I think I'll probably go to that one right there since says dwt t equals three minus D l d t. Well, common denominator. That means I have nine thirds minus 14 3rd or negative five thirds centimeters per second because that's a negative. That means at this point my with is actually decreasing. So that is how but both my length and my with their changing one increasing and one decreasing.

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