Question

Let \(V\) be the volume of a sphere of radius \(r\) that is changing with respect to time. If \(d r / d t\) is constant, is \(d V / d t\) constant? Explain your reasoning.

Short answer

No, \(\frac{d V}{d t}\) is not constant, it depends on the varying radius of the sphere

Step-by-step solution

Find the Derivative of V with respect to time

The formula for the volume of a sphere is \(V = \frac{4}{3}\pi r^3\). Differentiate both sides of the equation with respect to time \(t\), apply the Chain rule on the right side to get \(\frac{d V}{d t} = 4\pi r^2 \frac{d r}{d t}\)

Substitute the constant derivative of r

Given that \(\frac{d r}{d t}\) is constant, label it as \(k\). Replacing \(\frac{d r}{d t}\) in the equation from Step 1 gives \(\frac{d V}{d t} = 4\pi r^2 k\)

Analyze the resulting Derivative

Looking at the expression for \(\frac{d V}{d t}\), it is clear that \(4\pi k\) is indeed a constant, but \(r^2\) depends on the radius \(r\), which is changing with time. As the radius changes, \(r^2\) and thus \(\frac{d V}{d t}\) will change. So \(\frac{d V}{d t}\) is not constant, it depends on the value of \(r\).

Key concepts

Chain Rule

The Chain Rule is a fundamental concept in differential calculus. It helps us find the derivative of a composite function. Imagine a composite function as a process with multiple steps, where one function is nested inside another. The Chain Rule allows us to break down this process, so we can differentiate it step by step.

Here's how it works: if you have two functions, say, an outer function \(f(g(x))\), the Chain Rule states that the derivative of this composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Mathematically, this is expressed as:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \, g'(x) \)

This rule is especially handy when working with the rate of change in time, and it's used extensively in problems involving related rates.

In our original exercise, when finding how the volume \(V\) of a sphere changes with time \(t\), the Chain Rule was used to manage the change of the radius \(r\) over time. By differentiating both sides of the equation \(V = \frac{4}{3}\pi r^3\), we were able to establish the relation between the rate of change of the volume and the radius.

Volume of a Sphere

The volume of a sphere is a measure of the space contained within it. Calculating the sphere's volume is pretty straightforward using the standard formula:

• \( V = \frac{4}{3} \pi r^3 \)

Here, \(V\) represents the volume, and \(r\) denotes the radius of the sphere.

This formula gives us the volume in cubic units, and it serves as a foundational concept for understanding various phenomena in physics and engineering. For instance, when considering planets or bubbles, the volume formula helps calculate how much material or space they encompass.

In the differential calculus realm, knowing how to express the volume of a sphere is crucial for solving problems involving changes over time, like in the original exercise. The formula is differentiated with respect to time, using the Chain Rule, to explore how the volume alters as the radius dynamically grows or shrinks.

Related Rates

Related rates problems deal with finding the rate at which one quantity changes with respect to another over time. These problems often involve multiple changing quantities linked through an equation. Understanding how they relate to each other allows us to find unknown rates.

To tackle related rates problems, follow these steps:

• Identify all known and unknown variables and rates.
• Write down an equation that relates the different variables.
• Differentiate the equation with respect to time \(t\) using the Chain Rule.
• Plug in the known rates and solve for the unknown rate.

In the context of the original exercise, we were given that the radius \(r\) of a sphere changes at a constant rate. We had to find how the volume \(V\) changed as the radius grew. By using the related rates approach, we found that \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \).

This formula showed that the change in volume is not constant but varies with the square of the radius, illustrating how related rates problems can unveil dynamic relationships between different variables.