Problem 13 - University Calculus: Early Transcendentals

Question

Volume The radius \(r\) and height \(h\) of a right circular cylinder are related to the cylinder's volume \(V\) by the formula \(V=\pi r^{2} h\). a. How is \(d V / d t\) related to \(d h / d t\) if \(r\) is constant? b. How is \(d V / d t\) related to \(d r / d t\) if \(h\) is constant? c. How is \(d V / d t\) related to \(d r / d t\) and \(d h / d t\) if neither \(r\) nor \(h\) is constant?

Step-by-Step Solution

Answer

a) \(\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}\), b) \(\frac{dV}{dt} = 2\pi rh \frac{dr}{dt}\), c) \(\frac{dV}{dt} = \pi (2r h \frac{dr}{dt} + r^2 \frac{dh}{dt})\).

Step 1: Understand the Formula

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

Step 2: Differentiate for Part a

If the radius \(r\) is constant, only \(h\) changes over time. Differentiate the volume formula with respect to time \(t\): \[\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}.\] This shows that \(\frac{dV}{dt}\) is dependent on \(\frac{dh}{dt}\) and the constant \(r\).

Step 3: Differentiate for Part b

If the height \(h\) is constant, only \(r\) changes over time. Differentiate with respect to time \(t\): \[\frac{dV}{dt} = 2\pi rh \frac{dr}{dt}.\]This demonstrates that \(\frac{dV}{dt}\) depends on \(\frac{dr}{dt}\) and the constants \(h\) and \(r\).

Step 4: Differentiate for Part c

When neither \(r\) nor \(h\) are constant, use the product rule for differentiation: \[\frac{dV}{dt} = \pi (2r h \frac{dr}{dt} + r^2 \frac{dh}{dt}).\]This equation shows how \(\frac{dV}{dt}\) is related to both \(\frac{dr}{dt}\) and \(\frac{dh}{dt}\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us determine how a function changes as its variables change. It's like finding the speed of a moving car at a particular moment. In problems involving rates of change, like the cylinder volume problem, we often need to find how one quantity affects another over time.

When we differentiate the volume of a cylinder formula, we test how the volume changes due to changes in its radius or height. This rate of change is linked to how fast these dimensions change over time, which we denote using differentiation.

• For example, if the radius is constant but the height changes, we focus on how the height's change rate (\(\frac{dh}{dt}\)) affects the volume.
• Similarly, if the height is constant, it's the change in radius (\(\frac{dr}{dt}\)) that matters.

Understanding differential equations in these contexts helps predict how dimensions affect volumes, which is crucial for engineering and physics applications.

Cylinder Volume

The volume of a cylinder is a measure of the space it occupies in three dimensions. For a right circular cylinder, like the ones in the exercise, the formula for calculating volume is given by \(V = \pi r^2 h\). This formula is intuitive:

• \(\pi r^2\) represents the area of the base of the cylinder, which is a circle.
• Multiplying this area by the height (\(h\)) gives us the total volume.

The formula highlights how both radius and height directly influence the cylinder's volume. In practical terms, increasing either the radius or height will increase the volume. The exercise shows how vital understanding these relationships is when both parameters change over time. Calculating how changes in dimensions affect volume is key in designing objects with specific volume constraints.

Product Rule

The product rule is an essential part of differentiation, especially for functions that are products of two or more variable expressions. When neither the height nor the radius of a cylinder is constant, we need to use the product rule to find the rate of change in volume over time.

The formula we use here is the product rule for derivatives, which states:

• The derivative of a product is the derivative of the first function times the second function plus the first function times the derivative of the second function.

In terms of the cylinder, if we're dealing with both radius and height as variable quantities, the product rule helps us differentiate the volume. We apply \(\frac{dV}{dt} = \pi (2r h \frac{dr}{dt} + r^2 \frac{dh}{dt})\).

This ensures comprehensive analysis of how simultaneous changes in dimensions affect the cylinder's volume. Mastering the product rule is crucial for analyzing complex functions involving multiple changing elements.