Problem 36 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Suppose the amount of power generated by an energy generating system is a function of \(v\), the volume of water flowing through the system. The function is given by \(P=P(v)\). The volume of water in the sytem is determined by \(r\), the radius of an adjustable valve; \(v=v(r) .\) The radius varies with time: \(r=r(t) .\) (a) Express \(\frac{d P}{d r}\), the rate of change of the power with respect to a change in the valve's radius, in terms of the functions \(P(v)\) and \(v(r)\) and their derivatives. (b) Express \(\frac{d P}{d t}\), the rate of change of the power with respect to time, in terms of the functions \(P(v), v(r)\), and \(r(t)\) and their derivatives.

Step-by-Step Solution

Answer

The rate of change of the power with respect to a change in the valve's radius is \(\frac{dP}{dr} = \frac{dP}{dv} . \frac{dv}{dr}\) and the rate of change of the power with respect to time is \(\frac{dP}{dt} = \frac{dP}{dv} . \frac{dv}{dr} . \frac{dr}{dt}\).

Step 1: Differentiate Power function with respect to volume and volume function with respect to radius

Firstly, using the chain rule, \(\frac{dP}{dv} . \frac{dv}{dr}\). Here, \(\frac{dP}{dv}\) represents the derivative of the power function \(P(v)\) with respect to \(v\), and \(\frac{dv}{dr}\) represents the derivative of the volume function \(v(r)\) with respect to \(r\).

Step 2: Substituting Step 1 back into chain rule for part (a)

The result is \(\frac{dP}{dr} = \frac{dP}{dv} . \frac{dv}{dr}\), which is the answer to part (a). Here, \(\frac{dP}{dr}\) represents the rate of change of power with respect to the change in the valve's radius.

Step 3: Differentiate radius function with respect to time

Now, to extend this equation to differentiate for time, do: \(\frac{dr}{dt}\). This represents the derivative of the radius function \(r(t)\) with respect to \(t\).

Step 4: Substituting Step 1 and Step 3 back into chain rule for part (b)

The result is \(\frac{dP}{dt} = \frac{dP}{dv} . \frac{dv}{dr} . \frac{dr}{dt}\), which is the answer to part (b). Here, \(\frac{dP}{dt}\) represents the rate of change of power with respect to time in terms of the functions \(P(v)\), \(v(r)\), and \(r(t)\), and their derivatives.

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus, especially useful for differentiating composite functions.
Imagine you have a function that depends on another variable, which in turn depends on another different variable.
This is where the chain rule becomes essential.In our problem, power is a function of volume, which is a function of radius, which itself varies over time.

• To find the rate of change of power with respect to radius (\(\frac{dP}{dr}\) ) using the chain rule, we set up the relationship: \(\frac{dP}{dr} = \frac{dP}{dv} \cdot \frac{dv}{dr}\) .This means we multiply the derivative of power with respect to volume by the derivative of volume with respect to radius.
• To extend to the time variable, we use the chain rule again, resulting in \(\frac{dP}{dt} = \frac{dP}{dv} \cdot \frac{dv}{dr} \cdot \frac{dr}{dt}\) .

This helps us express how power changes over time, accounting for how volume changes with radius, and how radius changes with time.

Derivatives

Derivatives define how a function changes as its input changes.
In essence, they provide the rate at which one quantity changes concerning another.
In this problem, we encounter several derivatives:

• \(\frac{dP}{dv}\): Refers to how much power changes with respect to a change in volume. This derivative highlights the sensitivity of power to variations in water volume.
• \(\frac{dv}{dr}\): Indicates the rate at which the volume of water changes as the radius of the valve changes. It shows the dependency of water volume on the valve's opening.
• \(\frac{dr}{dt}\): Illustrates how the radius of the valve changes over time, indicative of the speed at which the valve adjusts.

Understanding these derivatives is crucial as they piece together the physical relationships between power, volume, radius, and time.

Related Rates

Related rates problems involve finding a rate at which one quantity changes by relating it to other quantities whose rates of change are known.
They often involve multiple variables that are interdependent. In the provided exercise, related rates help us understand how the power generated changes over time through relationships of other variables:

• The rate of change of the power (\(\frac{dP}{dt}\) ) depends on both how the volume changes concerning the radius and how the radius changes with time.
• Using the chain rule, we interconnect these rates which are: the rate at which power changes due to volume changes (\(\frac{dP}{dv}\) ), volume changes with radius (\(\frac{dv}{dr}\) ), and radius changes with time (\(\frac{dr}{dt}\) ).

Related rates provide a powerful lens for analyzing systems that involve multiple interconnected changes, making them especially useful in engineering and physics.