Problem 65 - Calculus for Biology and Medicine

Question

Assume that the side length \(x\) and the volume \(V=x^{3}\) of a cube are differentiable functions of \(t .\) Express \(d V / d t\) in terms of \(d x / d t\)

Step-by-Step Solution

Answer

\(\frac{dV}{dt} = 3x^2 \frac{dx}{dt}\).

Step 1: Identify Given Information and Requirement

We are given that the side length of a cube, \(x\), and the volume \(V = x^3\), are functions of time \(t\). We need to find the rate of change of volume \(dV/dt\) in terms of the rate of change of the side length \(dx/dt\).

Step 2: Apply Chain Rule for Derivatives

The volume \(V\) is given by \(V = x^3\). To find \(dV/dt\), we use the chain rule for derivatives: \[\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}.\]

Step 3: Differentiate Volume with Respect to Side Length

First, differentiate \(V = x^3\) with respect to \(x\). Using the power rule, \(\frac{dV}{dx} = 3x^2\).

Step 4: Substitute Back into Chain Rule Equation

Substitute \(\frac{dV}{dx} = 3x^2\) into the chain rule equation \(\frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt}\).

Step 5: Final Expression for dV/dt

Now we have the expression for the rate of change of volume: \[\frac{dV}{dt} = 3x^2 \frac{dx}{dt}.\] This shows how the volume changes with respect to time in terms of the change in side length.

Key Concepts

Understanding the Chain Rule

In calculus, the chain rule is a fundamental tool for differentiating composite functions. Imagine you have a function nested within another function. The chain rule helps you find the derivative, or the rate of change, of such composite functions efficiently. For instance, if you have two functions, say \(y = f(u)\) and \(u = g(x)\), then \(y\) is indirectly dependent on \(x\) through \(u\). The chain rule formula for this situation is expressed as:

• \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)

Applying this rule allows you to break down the differentiation process into simpler steps. Here, \(\frac{dy}{du}\) is the derivative of \(y\) with respect to \(u\), and \(\frac{du}{dx}\) is the derivative of \(u\) with respect to \(x\). This approach effectively links changes in one variable to changes in another.
In our exercise with the cube, the chain rule connects the rate of change of the side length to the volume of the cube.

Rate of Change in Functions

The concept of the rate of change is pivotal in understanding how different quantities vary with respect to time or other variables. Simply put, it is the speed at which a particular variable changes. This is often expressed as a derivative. For example, in physics or everyday life, rate of change could describe how fast a car is moving or how quickly a plant grows.
In mathematical terms, if \(y = f(x)\), the rate of change of \(y\) concerning \(x\) is given by the derivative \(\frac{dy}{dx}\). This value tells you the slope of the function at any point, offering insights into whether the function is increasing or decreasing at that point.
In the context of our initial exercise, the rate at which the side length of the cube changes, denoted by \(\frac{dx}{dt}\), directly impacts how the cube's volume \(V = x^3\) changes over time, represented by \(\frac{dV}{dt}\). This link is crucial in understanding dynamic systems in calculus.

Exploring Derivatives

Derivatives are the cornerstone of calculus. They measure how a function's output value changes as its input value changes. The derivative of a function can give profound insights into the behavior of that function.

• If \(f'(x) > 0\), the function \(f(x)\) is increasing at \(x\).
• If \(f'(x) < 0\), the function \(f(x)\) is decreasing at \(x\).
• If \(f'(x) = 0\), the function \(f(x)\) has a stationary point, which could be a maximum, minimum, or a point of inflection.

In the case of our cube problem, we differentiate the volume function \(V = x^3\) with respect to \(x\) to find \(\frac{dV}{dx} = 3x^2\). This derivative tells us how the volume changes with a small change in the side length of the cube. When we multiply this by \(\frac{dx}{dt}\), it effectively gives us the rate of volume change with respect to time. Understanding how derivatives function and what they represent is pivotal in solving real-world problems effectively.