The width of a rectangle increases at a rate of 2 cms. The length is five times its width. Find the rate at which the area is increasing when its width is5 cm.

College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter3: Polynomial Functions
Section3.5: Mathematical Modeling And Variation
Problem 7ECP: The kinetic energy E of an object varies jointly with the object’s mass m and the square of the...
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29..answer no 7 do like example 3 solution 

An emply container is filled with liquid. The height of the liquid in the container is x cm
1.
and its volume is V cm, vwhere V =
0.05 cm per second, find the rate of change in volume when x = 2 cm.
2 nx ( 5
x ). If x increases at the rate of
The volume of liquid with depth x cm is given by V =
12) + 24 cm.
2.
The liquid is added to the container at a constant rate of 0.5 cm'/s
Find the rate of change of the depth of the liquid at the instant the depth is 5 cm.
Water is poured into a container at a rate of 12 cm/sec. The volume of water in the
3.
3.
(h + 8h ) where h is the height of water in the container.
container is given by V =
Find the rate of change at which the height of water is increasing when h = 8.
3p
Show that the area of an equilateral triangle with sides p cm is A =
4
If the side is changing at a rate of 0.05 cms, how fast is the area changing when the
side is 5 cm.
The area of a triangle is given by A =
x². If the area is decreasing at the rate of
5.
4 cm?/min, find the rate at which x is changing at the instant when the area is 200 cm2.
6. Two sides of a triangle are 4m and 5m in length and the angle between them is
increasing at a rate of 0.06 rad/second. Find the rate at which the area of the triangle is
TC
increasing when the angle 0 is
3
(Hint:
Area of triangle
ab sin C)
The width of a rectangle increases at a rate of 2 cms. The length is five times its
width. Find the rate at which the area is increasing when its width is 5 cm.
Transcribed Image Text:An emply container is filled with liquid. The height of the liquid in the container is x cm 1. and its volume is V cm, vwhere V = 0.05 cm per second, find the rate of change in volume when x = 2 cm. 2 nx ( 5 x ). If x increases at the rate of The volume of liquid with depth x cm is given by V = 12) + 24 cm. 2. The liquid is added to the container at a constant rate of 0.5 cm'/s Find the rate of change of the depth of the liquid at the instant the depth is 5 cm. Water is poured into a container at a rate of 12 cm/sec. The volume of water in the 3. 3. (h + 8h ) where h is the height of water in the container. container is given by V = Find the rate of change at which the height of water is increasing when h = 8. 3p Show that the area of an equilateral triangle with sides p cm is A = 4 If the side is changing at a rate of 0.05 cms, how fast is the area changing when the side is 5 cm. The area of a triangle is given by A = x². If the area is decreasing at the rate of 5. 4 cm?/min, find the rate at which x is changing at the instant when the area is 200 cm2. 6. Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06 rad/second. Find the rate at which the area of the triangle is TC increasing when the angle 0 is 3 (Hint: Area of triangle ab sin C) The width of a rectangle increases at a rate of 2 cms. The length is five times its width. Find the rate at which the area is increasing when its width is 5 cm.
Example 3:
The volume of water in a bowl is given by V= (36xx) where x cm is the depth
of the water tf water is poured in at a rate of 10 cm's at what rate is the level rising
when the depth is 8 cm ?
Solution:
dV
= 10. = ?, x = 8
dx
dt
n ( 36x
x')
V =
dV
* ( 72x - 3x )
dx
24x x
%3D
dV
dx
clx
dt
dt
AP
1
10 x
24m X
1.
10 x
24 (8)
R(8)
5
64
cms
Alternative Solution:
dV
= 10, = ?, x = 8
dx
dt
3*(36x - x)
V.
dV
%3D
dl
10 - (12(8) - 3(8y 1
(384 )
10 =
dx
dt
64
cms !
Transcribed Image Text:Example 3: The volume of water in a bowl is given by V= (36xx) where x cm is the depth of the water tf water is poured in at a rate of 10 cm's at what rate is the level rising when the depth is 8 cm ? Solution: dV = 10. = ?, x = 8 dx dt n ( 36x x') V = dV * ( 72x - 3x ) dx 24x x %3D dV dx clx dt dt AP 1 10 x 24m X 1. 10 x 24 (8) R(8) 5 64 cms Alternative Solution: dV = 10, = ?, x = 8 dx dt 3*(36x - x) V. dV %3D dl 10 - (12(8) - 3(8y 1 (384 ) 10 = dx dt 64 cms !
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