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Autious
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Homework Statement
A cone has the base radius of 8 cm and height of 10 cm. The height of the cone is changing at at a rate of 1cm/hour whilst radius of the base is changing with it keeping the volume constant.
At what rate is the surface area of the entire cone changing at this exact moment?
Homework Equations
None were given in the assignment. Such info is assumed to have been memorized.
The Attempt at a Solution
We know the following.
height
[tex]h=10cm[/tex]
radius
[tex]r=8cm[/tex]
change in height
[tex] dh/dt = 1cm/hour [/tex]
Starting out here, I figured that the formula for the volume of the cone was relevant as it gives a relationship between the height, radius and the volume.
[tex]V= (\pi r^2 * h )/3[/tex]
Given that the volume is constant, I can calculate it...
[tex] V =(\pi 640)/3 [/tex]
...and replace it in the equation
[tex] (\pi 640)/3 = (\pi r^2 * h )/3 [/tex]
...using some algebra I can get the following equation.
[tex] r = \sqrt{640/h} [/tex]
Which means I have a way to express r as a function of h.
Here comes the part that makes me think that I'm way off.
The formula for the full surface of the cone now is
[tex] A = \pi r s + \pi r^2[/tex]
Where s is
[tex] s = \sqrt{ r^2 + h^2 } [/tex]
Finally gives me a function of the cones total surface as a function of the cones height that is
[tex] A = \pi sqrt{640/h} \sqrt{640/h + h^2} + \pi 640/h [/tex]
Which I can then derivate to calculate the rate of change in the surface of the area.
At this point I start getting the feeling that I've taken the wrong turn somewhere as getting to the answer seems harder than it's "supposed to be".
Can anyone see an easier solution to the problem? Am I even on the path of a solution that would be correct?
Any insight into the issue is appriciated.