Let / and w represent the length and width of the rectangle in cm, and let A represent the area of the rectangle in cm². Writing an equation for A in terms of/ and w gives the following result. A = 96cm Step 2 We need to determine how fast the area of the rectangle is increasing. In other words, we are looking for a rate of change of the area. In this problem, the volume, the length, and the width are all functions of the time t, where t is measured in seconds. The rate of increase of the area with respect to time is the derivative dA. The rate of increase of the length and width with respect to time are the derivatives dt dA d(/w) dt dt l ur dA dt dt Therefore, we need to differentiate each side of our area equation with respect to t using the product rule. Doing so gives the following result where dA is measured in cm²/s. dt = = dw 1) dr + ([ dt dw dt dl dt and respectively.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Question
The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 6 cm/s. When the length is 12 cm and the width is 9 cm, how fast is the area of the rectartime lefteasing
(in cm²/s)?
0:34:49
Step 1
Let / and w represent the length and width of the rectangle in cm, and let A represent the area of the rectangle in cm2. Writing an equation for A in terms of/ and w gives the following result.
A = 96cm
Step 2
We need to determine how fast the area of the rectangle is increasing. In other words, we are looking for a rate of change of the area. In this problem, the volume, the length, and the width are all
functions of the time t, where t is measured in seconds.
The rate of increase of the area with respect to time is the derivative dA. The rate of increase of the length and width with respect to time are the derivatives and respectively.
dt
dA
dt
dA
dt
lw
Therefore, we need to differentiate each side of our area equation with respect to t using the product rule. Doing so gives the following result where is measured in cm²/s.
dt
dA
=
=
d(/w)
dt
dw
dt
+
dl
dt
dl
dt
dw
dt
Transcribed Image Text:The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 6 cm/s. When the length is 12 cm and the width is 9 cm, how fast is the area of the rectartime lefteasing (in cm²/s)? 0:34:49 Step 1 Let / and w represent the length and width of the rectangle in cm, and let A represent the area of the rectangle in cm2. Writing an equation for A in terms of/ and w gives the following result. A = 96cm Step 2 We need to determine how fast the area of the rectangle is increasing. In other words, we are looking for a rate of change of the area. In this problem, the volume, the length, and the width are all functions of the time t, where t is measured in seconds. The rate of increase of the area with respect to time is the derivative dA. The rate of increase of the length and width with respect to time are the derivatives and respectively. dt dA dt dA dt lw Therefore, we need to differentiate each side of our area equation with respect to t using the product rule. Doing so gives the following result where is measured in cm²/s. dt dA = = d(/w) dt dw dt + dl dt dl dt dw dt
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