Problem 80 - Calculus Volume 1

Question

For the following exercises, find the change in volume \(d V\) or in surface area \(d A\) $$ d A \text { if the radius of a sphere changes from } r \text { by } d r $$

Step-by-Step Solution

Answer

The change in surface area is \( dA = 8 \pi r \cdot dr \).

Step 1: Formula for the Surface Area of the Sphere

The surface area of a sphere is given by the formula \( A = 4 \pi r^2 \), where \( r \) is the radius of the sphere.

Step 2: Differentiate the Surface Area Formula with Respect to r

To find how the surface area changes with a small change in the radius \( r \), differentiate the surface area formula with respect to \( r \). Using the power rule, we have \( \frac{dA}{dr} = \frac{d}{dr}(4 \pi r^2) = 8 \pi r \).

Step 3: Express the Change in Surface Area, d A

The change in surface area \( dA \) due to a small change in radius \( dr \) is given by the derivative from the previous step multiplied by \( dr \):\[ dA = 8 \pi r \cdot dr \].

Key Concepts

Surface Area of a Sphere

The surface area of a sphere is a fundamental concept in geometry and calculus. To find the surface area, we use the formula:

• \( A = 4 \pi r^2 \)

Here, \( A \) represents the surface area, \( \pi \) is a constant approximately equal to 3.14159, and \( r \) is the radius of the sphere.
This equation tells us that the surface area of the sphere increases as the square of the radius increases. Consequently, if you double the radius, the surface area becomes four times larger.
This formula plays a crucial role when studying the physical properties of spherical objects, such as planets, and in understanding how these properties change with size.

Differentiation Techniques in Calculus

Differentiation is a key concept in calculus used to determine how a function changes. When applied to the surface area of a sphere, we focus on how small changes in the radius affect the surface area. Let's consider the steps:

• Differentiating the surface area formula \( A = 4 \pi r^2 \) with respect to \( r \), we use the power rule, resulting in \( \frac{dA}{dr} = 8 \pi r \).
• The power rule states that the derivative of \( r^n \) is \( nr^{n-1} \). Thus, differentiating \( 4 \pi r^2 \) gives \( 8 \pi r \).

The derivative \( \frac{dA}{dr} = 8 \pi r \) indicates the rate of change of the surface area with respect to the radius.
It's a tool that helps us understand the sensitivity of one quantity if another one is slightly changed.

Applied Calculus Problems

In applied calculus, we often deal with real-world problems where we need to calculate small changes. For instance, in determining the change in the surface area of a sphere when its radius changes slightly, we employ differentiation.
From the differentiation step, we get:

• \( dA = 8 \pi r \cdot dr \)

Here, \( dA \) represents the small change in the surface area, and \( dr \) is the small change in the radius.
This result helps us find out how much the surface area of a sphere increases or decreases due to minuscule variations in the radius.
It's particularly useful in fields such as physics and engineering, where precise calculations of area changes are necessary for design and analysis.