Changing dimensions in a rectangle The length l of a...

Changing dimensions in a rectangle The length $l$ of a rectangle is decreasing at the rate of $2 \mathrm{cm} /$ sec while the width $w$ is increasing at the rate of $2 \mathrm{cm} / \mathrm{sec}$. When $l=12 \mathrm{cm}$ and $w=5 \mathrm{cm},$ find the rates of change of (a) the area, (b) the perimeter, and (c) the lengths of the diagonals of the rectangle. Which of these quantities are decreasing, and which are increasing?

Key Concepts

Related Rates

Related rates problems involve finding the rate at which one quantity changes with respect to time by relating it to the rates of change of other quantities. This often requires differentiating an equation that connects the involved variables with respect to time.

Product Rule

The product rule is a fundamental rule in differentiation used when finding the derivative of a product of two functions. In applications like determining the rate of change of the area of a rectangle, where the area is the product of the length and width, the product rule is essential.

Chain Rule

The chain rule is used to differentiate composite functions, where one function is nested inside another. It is particularly useful in problems involving changing geometric quantities, such as finding the rate of change of the length of a diagonal that depends on both the length and width of a rectangle.

Pythagorean Theorem

The Pythagorean theorem relates the sides of a right triangle and is used to compute distances such as the diagonal of a rectangle. In dynamic problems, differentiating the relationship given by the theorem allows one to determine the rate at which the diagonal is changing.

Transcript

00:01 This question gives us a rectangle, and it tells us that the length and the width are changing.

00:08 The rate at which the length is changing, it says it's decreasing, so the l by dt, sorry, is negative because it's decreasing, and it's two centimeters per second.

00:21 And then the width is changing as well.

00:24 It says dw by dt is increasing at two centimeters per second.

00:30 And then it says at the point we're talking, the length is 12 centimeters and the width is 5 centimeters at the instant and time that we're talking.

00:41 And it says find the rates of chains of.

00:43 First of all, part a is the area.

00:45 So the area is length times width, which means da by dt is going to be l, dw by dt, plus wdl by dt.

00:59 So we're going to just take l is 12, dw by dt is 2.

01:04 Then we're going to take w, which is 5, and multiply it by the l by dt, which is minus 2.

01:09 And you're going to end up with 24 minus 10.

01:12 So that gives us 14 and it's centimeter squared per second.

01:18 The area is increasing at 14 centimeters squared per second.

01:23 And it asks us about the perimeter.

01:25 Okay, so for part b, the perimeter is going to be two l's and two ws, right, because it's just the length around the outside.

01:33 So d .p by dt is going to be 2 times dl by dt plus two times dw by dt.

01:43 All right.