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A heap of rubbish in the shape of a cube is being compacted into a smaller cube. given that the volume decreases at a rate of 4 cubic meters per minute, find the rate of change of an edge, in meters per minute, of the cube when the volume is exactly 125 cubic meters.

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meerkat18
Working Formula:

V = s^3

Given:

dV/dt = -4 cubic meters per minute
V = 125 cubic meters

Required: ds/dt (rate of change of edge per minute)  at V = 125 m^3

Solution:

Differentiate, equation below 
V = s^3
dV/dt = 3*s^2 (ds/dt)
-4 = 3*s^2 (ds/dt)       
ds/dt = -1.33/s^2  -----------> eq. (1)

V = s^3
(125)^0.33 = (s^3)^0.33
s = 5                   ------------> eq. (2)

Substitute eq. (2) to eq. (1), we get

ds/dt = -1.33/(5)^2 = -0.053 meters per minute

ANSWER: -0.053 meters per minute

Using implicit differentiation, it is found that the rate of change of an edge is of -0.0533 meters per minute.

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The volume of a cube of edge e is given by:

[tex]V = e^3[/tex]

In this problem, the volume is of 125 m³, thus, we solve the above equation to find the length of an edge, in metres.

[tex]V = e^3[/tex]

[tex]125 = e^3[/tex]

[tex]e = \sqrt[3]{125}[/tex]

[tex]e = 5[/tex]

Now, for the rate of change, we need to apply the implicit differentiation, thus:

[tex]V = e^3[/tex]

[tex]\frac{dV}{dt} = 3e^2\frac{de}{dt}[/tex]

[tex]\frac{dV}{dt} = 3(5)^2\frac{de}{dt}[/tex]

[tex]\frac{dV}{dt} = 75\frac{de}{dt}[/tex]

Volume decreases at a rate of 4 cubic meters per minute, thus:

[tex]\frac{dV}{dt} = -4[/tex]

The rate of change of an edge is [tex]\frac{de}{dt}[/tex]. Then:

[tex]-4 = 75\frac{de}{dt}[/tex]

[tex]\frac{de}{dt} = -\frac{4}{75}[/tex]

[tex]\frac{de}{dt} = -0.0533[/tex]

The rate of change is of -0.0533 cubic meters per minute.

A similar problem is given at https://brainly.com/question/24158553

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