Problem 21 - Calculus for Biology and Medicine

Question

Assume that the radius \(r\) and the surface area \(S=4 \pi r^{2}\) of a sphere are differentiable functions of \(t\). Express \(d S / d t\) in terms of \(d r / d t\).

Step-by-Step Solution

Answer

\(\frac{dS}{dt} = 8\pi r \cdot \frac{dr}{dt}\).

Step 1: Understand the Surface Area Formula

The surface area of a sphere is given by the formula \(S = 4\pi r^2\). This equation relates the surface area \(S\) to the radius \(r\) of the sphere.

Step 2: Differentiate the Surface Area with Respect to Time

We want to find \(\frac{dS}{dt}\), the rate of change of the surface area with respect to time. Since \(S\) is expressed as a function of \(r\), and \(r\) itself is a function of \(t\), we apply the chain rule to differentiate \(S = 4\pi r^2\) with respect to \(t\).

Step 3: Apply the Chain Rule

The chain rule states that \(\frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt}\). First, find \(\frac{dS}{dr}\). We differentiate \(S = 4\pi r^2\) with respect to \(r\), which gives \(\frac{dS}{dr} = 8\pi r\).

Step 4: Combine Results

Using the chain rule, substitute the result from Step 3 into the equation. We obtain \(\frac{dS}{dt} = 8\pi r \cdot \frac{dr}{dt}\). This expresses the rate of change of surface area \(S\) concerning time \(t\) in terms of the rate of change of \(r\) with respect to \(t\), \(\frac{dr}{dt}\).

Key Concepts

Chain Rule

The chain rule is a vital concept in calculus for differentiating compound functions. When you have a function expressed in terms of another function, and you need their rate of change with respect to a third variable, the chain rule is your go-to tool. The rule can be concisely expressed with the formula:

• If you have a function, say, \( y = f(g(x)) \), differentiating \( y \) with respect to \( x \) involves multiplying the derivative of the outer function by the derivative of the inner function.
• Symbolically, we write this as \( \frac{dy}{dx} = \frac{dy}{dg} \cdot \frac{dg}{dx} \).

In the context of our problem, the surface area \( S \) of a sphere is a function of the radius \( r \), which in turn, is a function of time \( t \). Thus, to find the rate of change of the surface area with respect to time, we used the chain rule:

• Identify \( \frac{dS}{dr} \) where \( S = 4\pi r^2 \).
• Then express \( \frac{dS}{dt} \) as \( \frac{dS}{dr} \times \frac{dr}{dt} \).

This gives us an elegant way to compute the desired derivative, combining both direct differentiation and interdependence between variables.

Sphere Surface Area

Understanding the surface area of a sphere is crucial in geometry and calculus. The formula \( S = 4\pi r^2 \) expresses how the surface area \( S \) relates to the radius \( r \) of a sphere.

• This formula originates from geometric principles and provides a direct link between the size dimensions of a sphere and its external cover.
• The term \( 4\pi r^2 \) denotes that the surface area depends on the square of the radius, reflected by the exponent \( 2 \).

Mathematically, when you need to differentiate the surface area with respect to time, you're finding how quickly this surface is expanding or contracting as the radius changes. Applying calculus, specifically the dynamics of the radius,

• Allows insight into real-world applications such as expanding gases or growing cells.

Thus, grasping this geometrical link aids in perceiving broader physical phenomena.

Calculus Applications

Calculus plays a pivotal role in innumerable practical applications, many of which involve changes over time. Whether tracking growth, decay, or other dynamic systems, understanding how to apply concepts like differentiation is key.

• The chain rule example provided shows how concrete these ideas can be, by relating changes in a sphere's surface area due to its changing radius.
• This example transcends the realm of pure theory, offering a clear demonstration of calculus in real-life scenarios.

In physics, biology, and engineering, applications of such problems are abundant:

• From computing the evaporation of water droplets to the expansion of planets, the concepts hold firm.
• Moreover, it highlights the interconnectedness of different parameters and how calculus enables us to articulate these relationships quantitatively.

By following through these connections, students grasp not just abstract mathematical concepts but also their empirical relevance.