Problem 6 - Calculus Early Transcendentals

Question

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 \(h>0\) )?

Step-by-Step Solution

Answer

Answer: The volume of a right circular cylinder is an increasing function of the radius when the height is fixed.

Step 1: Write down the formula for the volume of a right circular cylinder

The formula for the volume of a right circular cylinder is \(V=\pi r^{2}h\), where \(r\) is the radius and \(h\) is the height.

Step 2: Differentiate the volume formula with respect to the radius

To find the derivative of the volume with respect to the radius, we will use the power rule. So, we get: \(\frac{dV}{dr} = \frac{d(\pi r^2 h)}{dr} = \pi h \frac{d(r^2)}{dr}\). Now, differentiate \(r^2\) with respect to \(r\): \(\frac{d(r^2)}{dr} = 2r\). So, the derivative of the volume with respect to the radius is: \(\frac{dV}{dr} = \pi h(2r) = 2\pi rh\).

Step 3: Determine if the derivative is positive or negative

We need to look at the sign of the derivative \(\frac{dV}{dr} = 2\pi rh\) to know if the function is increasing or decreasing. Since \(r>0\) and \(h>0\) are both given, the expression \(2\pi rh\) is positive.

Step 4: State the conclusion

Since the derivative \(\frac{dV}{dr}>0\), the volume of a right circular cylinder is an increasing function of the radius when the height is fixed.