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Solve the problems in related rates.A rectangular image 4.00 in. high on a computer screen is widening at the rate of 0.25 in./s. Find the rate at which the diagonal is increasing when the width is 6.50 in.

Calculus 1 / AB

Chapter 24

Applications of the Derivative

Section 4

Related Rates

Derivatives

Harvey Mudd College

Baylor University

University of Nottingham

Idaho State University

Lectures

03:09

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31:55

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02:26

Expanding rectangle A rect…

05:02

A rectangle initially has …

00:33

Solve each problem algebra…

09:31

Changing dimensions in a r…

08:48

03:33

Let $l$ be the length of a…

04:18

03:51

Solve each problem. See Ex…

05:26

Changing Dimensions in a R…

01:24

For the following exercise…

here we have a rectangle with dimensions two by four. I'm just gonna label to as w for width and l for length for the four. We're also told that each side is going to begin increasing in length at a rate of one centimeter per second. So D W duty is one centimeter per second and the L D. T is one centimeter per second. What we want to know is how quickly the area is increasing after 20 seconds. Okay, so let's start with our area formula. We know area is length, times wet. I'm gonna take the derivative of both sides with respect to t. So de a t t is equal to We have a multiplication problem here. So we're gonna use our product rule first times the derivative on the second, plus the second times, the derivative of the first. All right, so let's plug in what we know. So de a t t is what we're trying to find out. L we Joan. Exactly. No, l Yet we know DW DT is one, and we know d l d t is one. So we gotta find l and w. So if we think about this a second. We know side links are increasing at one centimeter per second. Ok, well, if the length starts at four after 20 seconds, it will have increased by 20 centimeters. Which means L is 24. W started at two after 20 seconds. If it increased one centimeter per second, it would have increased 20 centimeters. So two plus 20 is 22. Right? So we're just gonna go ahead and add these together. De A T T is equal to 46 centimeters squared for a second.

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