Problem 8
Question
A balloon maintains the shape of a sphere as it is being inflated. Find the rate of change of the surface area with respect to the radius at the instant when the radius is 2 in.
Step-by-Step Solution
Verified Answer
The rate of change of the surface area with respect to the radius at a radius of 2 inches is \( 16 \pi \; \text{square inches per inch} \).
1Step 1: Understand the Formula for Surface Area of a Sphere
The surface area (A) of a sphere is given by the formula: \( A = 4 \pi r^2 \), where \( r \) is the radius.
2Step 2: Differentiate the Surface Area with Respect to Radius
To find the rate of change of the surface area with respect to the radius (\( \frac{dA}{dr} \)), differentiate the surface area formula with respect to \( r \): \( \frac{dA}{dr} = \frac{d}{dr}(4 \pi r^2) \).
3Step 3: Apply the Power Rule of Differentiation
Using the power rule, \( \frac{d}{dx}(x^n) = nx^{n-1} \), we get: \( \frac{dA}{dr} = 4 \pi \cdot 2r = 8 \pi r \).
4Step 4: Substitute the Given Radius
Substitute \( r = 2 \) inches into the differentiated formula: \( \frac{dA}{dr} = 8 \pi \cdot 2 = 16 \pi \).
5Step 5: Simplification
Simplify the equation to obtain the rate of change of the surface area when the radius is 2 inches: \( \frac{dA}{dr} = 16 \pi \; \text{square inches per inch} \).
Key Concepts
differentiationsurface area of a spherepower rule of differentiation
differentiation
Differentiation is a key concept in calculus that refers to the process of determining the derivative of a function. This derivative, often denoted as \(f'(x)\) or \(\frac{df}{dx}\), represents the rate of change of the function with respect to a variable. For instance, in the given exercise, we are looking for the rate of change of the surface area of a sphere with respect to its radius. Differentiation breaks down complex changes into infinitely small steps to understand how a function behaves at any given point. By using differentiation, we can determine not just the value of a function but how it evolves as the input changes.
Applications of differentiation are widespread, including in fields such as physics for velocity and acceleration, economics for cost functions, and biology for growth rates.
Applications of differentiation are widespread, including in fields such as physics for velocity and acceleration, economics for cost functions, and biology for growth rates.
surface area of a sphere
The surface area of a sphere is a fundamental concept in geometry. It measures the total area covering the outer layer of a spherical object. The formula for the surface area of a sphere is given by \(A = 4 \pi r^2\), where \(A\) is the surface area and \(r\) is the radius. This formula helps us understand how the surface area expands as the radius of the sphere increases. For example, if the radius of a balloon inflates, knowing this formula allows us to calculate how much more area the balloon's surface will cover.
Understanding the surface area is crucial for practical applications like designing spherical tanks, balloons, and understanding phenomenon such as planet surface measurements.
Understanding the surface area is crucial for practical applications like designing spherical tanks, balloons, and understanding phenomenon such as planet surface measurements.
power rule of differentiation
The power rule of differentiation is one of the most straightforward and commonly used differentiation rules. It states that if you have a function in the form \(x^n\), its derivative is given by \(nx^{n-1}\). This rule simplifies the process of finding the rate of change of polynomial functions.
In our example, the surface area of the sphere is expressed as \(4 \pi r^2\). Using the power rule, we differentiate \(r^2\) to \(2r\), and then multiply by the constant \(4 \pi\), resulting in \(8 \pi r\). This application of the power rule helps us easily find the rate of change of the surface area with respect to the radius. It is a fundamental tool in calculus and essential for solving a wide range of problems.
In our example, the surface area of the sphere is expressed as \(4 \pi r^2\). Using the power rule, we differentiate \(r^2\) to \(2r\), and then multiply by the constant \(4 \pi\), resulting in \(8 \pi r\). This application of the power rule helps us easily find the rate of change of the surface area with respect to the radius. It is a fundamental tool in calculus and essential for solving a wide range of problems.
Other exercises in this chapter
Problem 8
Find the derivative of the given function. $$ f(x)=\left(4 x^{2}+7\right)^{2}\left(2 x^{3}+1\right)^{4} $$
View solution Problem 8
Find the derivative of the given function. $$ g(x)=\sqrt[3]{\left(3 x^{2}+5 x-1\right)^{2}} $$
View solution Problem 8
A man \(6 \mathrm{ft}\) tall is walking toward a building at the rate of \(5 \mathrm{ft} / \mathrm{sec}\). If there is a light on the ground \(50 \mathrm{ft}\)
View solution Problem 8
In Exercises 1 through 10, find the first and second derivative of the function defined by the given equation. $$ g(r)=\sqrt{r}+\frac{1}{\sqrt{r}} $$
View solution