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The length of a rectangle is increasing at a rate of $ 8 cm/s $ and its width is increasing at a rate of $ 3 cm/s. $ When the length is $ 20 cm $ and the width is $ 10 cm, $ how fast is the area of the rectangle increasing?

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03:35

Alex Lee

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 9

Related Rates

Derivatives

Differentiation

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

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.

02:26

Expanding rectangle A rect…

05:02

A rectangle initially has …

01:35

The area of a rectangle is…

02:12

The width of a rectangle i…

03:38

A rectangle expands with t…

02:28

The length $x$ of a rectan…

And this question we're asked that, given the rate of change of the length and the width of a rectangle, we want to find the length and the width of the, We want to find the rate of change of the area at a certain length and a certain width. Let's remember the formula for the area of a rectangle. We know that the area is equal to the length times the width, but All three of these are functions of time, so it's more correct to say that as a function of time, the area is, the length is a function of time, multiplied by the width times a function of time. Okay people, now that we have this information, we can go ahead and straight up, apply implicit differentiation and differentiate both sides with respect to time. Now the derivative of the area with respect to time is the rate of change of the area. And we know that that's going to be equal to the right hand side, which we're going to have to use the product rule which states to keep the first function and take the derivative of the second function and add it with the second function times the derivative of the first function. Now we know that were given to fixed lengths and width. That's these two numbers here. So let's plug these in. We know that the length is 20 And the width is 10. Yeah. And we're going to multiply that by the rate of change of the length and the rate of change of the width, Which is going to be three times eight. And when we add and when we simplify this together, we're going to get 60 plus 80 which is equal to 1 40 centimetres squared her second. And that's the answer to this question.

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