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The perimeter of a rectangle is growing at the rate of $12 \mathrm{m} / \mathrm{sec}$. Its diagonal is growing at the rate of $4 \mathrm{m} / \mathrm{sec} .$ At a certain time one side is $4 \mathrm{m}$ long and the other side is $3 \mathrm{m}$ long. Find the rate of change of each side. Are they growing or shrinking?

Both increase, shorter side at $4 \mathrm{m} / \mathrm{sec}$, larger at $2 \mathrm{m} / \mathrm{sec}$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

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.

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.

05:54

The perimeter of a rectang…

07:38

'The length of a rect…

01:27

A rectangle has a length o…

01:20

05:26

Changing Dimensions in a R…

04:18

Changing dimensions in a r…

08:48

10:22

The length $l$ of a rectan…

07:18

09:31

02:12

The width of a rectangle i…

06:25

02:28

The length $x$ of a rectan…

03:33

Let $l$ be the length of a…

02:26

Expanding rectangle A rect…

It's always a good idea to start a related rape problem using a picture if you can, so we're going to start with a picture of a rectangle. I have a length l and A with W. I also know that I'm going to be referencing the diagonal, so we'll call that deep. So what do we know? We know that the perimeter is growing well. The perimeter is growing. That's positive 12 m per second. We also know that diagonal is growing, so that's D D D T. And it's growing again positive at a rate of 4 m per second. And we're going to take a snapshot in time at a certain time. I have one side, this 4 m, the other side three. So length is usually the longer ones will say. The Ellis four and W is three, and I just said centimeters is supposed to be meters, meters meters now. This is a snapshot in time. I'm not putting these numbers on the picture because they're not Constance. If I had a constant number, I put it on the picture to represent that, But these air changing. So they are variables right now. What is it. I'm trying to find I wanna find how each side is changing. So I want to see how the length changes in the wits. Those are my two unknowns that I'm looking for. So now that we've got our picture, we've looked at all of the information we know. Let's start to solve this. Well, I know that I'm gonna be doing something with the perimeter because it gave me that information. So perimeter of a rectangle is two lengths plus two times the width. This has three variables. I wanna make sure that all three variables are represented by a rate Somewhere in what I wrote, I have dp DT and the other two are the unknowns I'm looking for. So all the variables air covered Now we can take the derivative I have. Dp DT Times two d l d t plus two d w d t. When I plug in What I know the right hand side doesn't change, but the left hand side is now 12, and I can simplify this just a little bit by dividing everything by two. Okay, so there's my first equation. I can't go any farther because it's one equation but two unknowns. So I need to go to the other information I was given, which is about the diagonal. So I can say d squared equals W squared. Plus elsewhere Pythagorean theorem again. I have three variables. D w n l. Okay, I have d d d t given and again the other two or what I'm trying to find. So I have all three of these rates covered, which means I can take the derivative two d d d d t equals to w d W D T plus two l d l D T. Every term here has a two, so I could cancel those out. Now, let's substitute in now. Before we do, I could see that we have a d now using our equation that we came up with. It relates all of the's. If Ellis four and W. Is three, I can plug those in and I can see that D is going to be five. So I have five D D d. T is four by W. Is three. It's an unknown, and l is four, and this is an unknown, so I'm going to simplify this just a little bit. Get 20 and then the right hand side doesn't change. Okay, so two equations, two unknowns. This is algebra Weaken. Solve this, using either elimination or substitution. Let's try substitution. I'm just gonna pick one. Let's say D L d t So that becomes six minus de wtt well substituted into our other equation. 20 equals three D W d t plus four times the piece we just found and I'm going to put this up so I've got some more room to work. We'll get rid of our parentheses. That gives us 24 minus four D W d t. Okay, I'm going to subtract 24 from both sides. That gives me a negative four combined. The other two terms That gives me a negative D W D T or D W D. T equals four and my units, I'm working with our meters per second. This is a positive number, which means that my width is increasing at this rate at the given point that we're looking at. Okay, so there's my with what about length? Well, let's use our equation and solve the l d T ISS six minus four or to D l. D. T is 2 m per second again, it is positive. So it is also increasing

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