Chapter 4: Problem 21
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 \(d r / d t\).
Short Answer
Expert verified
\(\frac{dS}{dt} = 8\pi r \cdot \frac{dr}{dt}\).
Step by step solution
01
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.
02
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 \(t\).
03
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 \(\frac{dS}{dr} = 8\pi r\).
04
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 \(t\), \(\frac{dr}{dt}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( \frac{dy}{dx} = \frac{dy}{dg} \cdot \frac{dg}{dx} \).
- Identify \( \frac{dS}{dr} \) where \( S = 4\pi r^2 \).
- Then express \( \frac{dS}{dt} \) as \( \frac{dS}{dr} \times \frac{dr}{dt} \).
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 \( 2 \).
- Allows insight into real-world applications such as expanding gases or growing cells.
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.
- 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.