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Chapter 12: Problem 6

The volume of a right circular cylinder with radius \(r\) and height \(h\) is \(V=\pi r^{2} h .\) Is the volume an increasing or decreasing function of the radius at a fixed height (assume \(r>0\) and \(h>0\) )?

Short Answer

Expert verified
Answer: The volume is an increasing function of the radius when the height is fixed.

Step by step solution

01

Write down the given expression for the volume of a right circular cylinder.

The volume of a right circular cylinder is given by \(V = \pi r^2 h\), where \(r\) is the radius of the cylinder and \(h\) is the height of the cylinder.
02

Take the derivative of the volume with respect to the radius.

To find out if the volume is an increasing or decreasing function of the radius, we need to find how the volume changes with respect to the radius. To do this, we will take the derivative of the volume expression with respect to the radius (\(r\)). Using the chain rule, we get: \(\frac{dV}{dr} = \frac{d(\pi r^2 h)}{dr} = 2\pi rh\).
03

Analyze the sign of the derivative.

We now need to analyze the sign of \(\frac{dV}{dr}\) (the derivative of the volume with respect to the radius) to determine if the volume is an increasing or decreasing function of the radius. Since both the constants \(\pi\) and \(h\) are positive (assuming \(h>0\)) and the radius \(r\) is also positive (assuming \(r>0\)), the product of these variables (\(2\pi rh\)) will always be positive. If \(\frac{dV}{dr} > 0\), then the volume is an increasing function of the radius. Since the derivative of the volume with respect to the radius (\(2\pi rh\)) is always positive, we can conclude that the volume is an increasing function of the radius when the height is fixed.

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