Problem 99 - A Graphical Approach to Precalculus with Limits

Question

Solve each application of openations and composition of functions. Volume of a Sphere The formula for the volume of a sphere is \(V=\frac{4}{3} \pi r^{3},\) where \(r\) represents the radius of the sphere. (a) Write a function \(D(r)\) that gives the volume gained when the radius of a sphere of \(r\) inches is increased by 3 inches. (b) Graph \(y=D(r)\) found in part (a), using \(x\) for \(r,\) in the window \([0,10]\) by \([0,1500]\) (c) Use your calculator to graphically find the amount of volume gained when a sphere with a 4 -inch radius is increased to a 7 -inch radius. (d) Verify your result in part (c) analytically.

Step-by-Step Solution

Answer

The volume gained is found by calculating \(D(r) = \frac{4}{3} \pi [(r+3)^3 - r^3]\). For a change from 4 to 7 inches, compute the difference. Graph the result in the set window and verify calculations.

Step 1: Define the Original Volume Function

The original volume function for a sphere is given by the formula \(V(r) = \frac{4}{3} \pi r^3\), where \(r\) is the radius of the sphere.

Step 2: Define the New Volume Function

When the radius is increased by 3 inches, the new radius becomes \(r+3\). Therefore, the new volume function increases to \(V(r+3) = \frac{4}{3} \pi (r+3)^3\).

Step 3: Define the Volume Gain Function

The volume gained, \(D(r)\), is the difference between the new volume and the original volume. Thus, \(D(r) = V(r+3) - V(r) = \frac{4}{3} \pi [(r+3)^3 - r^3]\).

Step 4: Compute for Volume Function for r = 4 and r = 7

For \(r = 4\), calculate the original and new volume. The original volume is \( \frac{4}{3}\pi (4)^3 \) and for \(r = 7\), it's \( \frac{4}{3}\pi (7)^3 \). Subtract these to find \(D(4) = V(7) - V(4)\).

Step 5: Verify the Result Analytically

Calculate \((r+3)^3 - r^3\) algebraically to check the volume gain from the 4-inch to 7-inch radius. Substituting \(r = 4\), compute again and verify against the calculator result.

Step 6: Graph the Function

Use graphing software or a calculator to graph the function \(y = D(r)\) from \(r = 0\) to \(r = 10\). Plot the function in the window \([0,10]\) by \([0,1500]\).

Step 7: Graphically Find Volume Gain

On the calculator, set \(r = 4\) and find \(r = 7\). Calculate \(D(4)\) to find the volume gain by evaluating the difference graphically.

Key Concepts

Function Composition

Function composition involves taking two functions and combining them to form a new function. When we deal with functions such as calculating the volume of a sphere, the idea of function composition becomes useful, especially in finding changes in volume.
In the given exercise, we're working with the original formula for a sphere's volume, denoted as \( V(r) = \frac{4}{3} \pi r^3 \). The task is to determine the gain in volume when the radius of the sphere is increased by 3 inches. To achieve this, we use function composition by defining two functions:

• The original volume function \( V(r) \).
• The new volume function \( V(r+3) \), which is derived by adjusting the input radius in the original formula.

To find the volume gain, the function \( D(r) \) is defined as the difference between the two functions: \( D(r) = V(r+3) - V(r) \). By composing these functions, you get the formula to calculate exactly how much more space a sphere takes up when its radius increases by a fixed amount.

Graphing Functions

Graphing functions is a visual way to understand how a function behaves over a specified interval. In this exercise, we are asked to graph the function \( y = D(r) \) over a range of \( r \) values, seeing how the function reacts to changes in the radius between 0 and 10 inches.
Using a graphing calculator or software, plotting \( y = D(r) \) allows us to visually analyze the volume increase as the radius of the sphere changes. When graphing:

• Consider the domain of \( r \). For this exercise, it’s from 0 to 10.
• Set the range for \( y \) appropriately, here it's 0 to 1500, to accommodate all possible values of the volume gain \( D(r) \).

This graphical representation not only aids in verifying employment of the formula correctly but also quickly shows us the pattern and rate of volume change. When \( r \) changes, the steepness of the curve can indicate how significantly the volume grows. It gives immediate, tangible insight into the underlying mathematical principles.

Analytical Verification

Analytical verification is crucial for confirming solutions obtained either graphically or through direct computation. It involves breaking down the calculations step-by-step to ensure all aspects check out analytically.
In our problem, we analytically verify the increase in volume from a 4-inch to a 7-inch radius. This involves computing the volume for both radii using \( V(4) = \frac{4}{3} \pi (4)^3 \) and \( V(7) = \frac{4}{3} \pi (7)^3 \). The resulting difference provides \( D(4) = V(7) - V(4) \) as analytically computed volume gain.By further dissecting \((r+3)^3 - r^3\), we confirm and substantiate the graphical findings. Calculating explicitly:

• Expand \((r+3)^3 \) to ensure the coefficients and powers align.
• Subtract the original \( r^3 \) from this expansion.

This analytical method is paramount as it reinforces the result’s accuracy, ensuring no calculator errors occurred. Analytical verification gives a solid foundation to trust the solution process fully.