Problem 91 - Algebra and Trigonometry

Question

\(79-92\) Solve the equation for the indicated variable. $$ V=\frac{4}{3} \pi r^{3} ; \quad \text { for } r $$

Step-by-Step Solution

Answer

\( r = \sqrt[3]{\frac{3V}{4\pi}} \)

Step 1: Identify the Expression

The given equation is the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \). We need to solve for the variable \( r \), which represents the radius of the sphere.

Step 2: Isolate the Cubic Term

First, multiply both sides of the equation by \( \frac{3}{4} \) to isolate \( \pi r^3 \) on one side:\[\frac{3}{4}V = \pi r^3\]

Step 3: Divide by Pi

Divide both sides by \( \pi \) to completely isolate \( r^3 \):\[\frac{3V}{4\pi} = r^3\]

Step 4: Solve for the Radius

Take the cube root of both sides to solve for \( r \):\[r = \sqrt[3]{\frac{3V}{4\pi}}\]

Step 5: Final Expression for Radius

Now, the expression for the radius \( r \) of the sphere in terms of the volume \( V \) is established as:\[r = \sqrt[3]{\frac{3V}{4\pi}}\]

Key Concepts

Volume of a Sphere

The volume of a sphere is a concept fundamental to understanding shapes in three-dimensional space. The volume gives us an idea of how much space a sphere occupies. For a sphere, the formula to calculate volume is given by \( V = \frac{4}{3} \pi r^3 \). Here,

• \( V \) represents the volume of the sphere.
• \( \pi \) is a mathematical constant approximately equal to 3.14159.
• \( r \) is the radius of the sphere, which is the distance from the center of the sphere to any point on its surface.

Understanding this formula is crucial as it allows us to determine the space inside a spherical object. The formula tells us that the volume is directly related to the cube of the radius, meaning that even small changes in radius lead to larger changes in volume.

Isolate the Variable

To isolate a variable means to rearrange an equation so that the variable stands alone on one side. This process is essential in solving equations and finding unknown values. In this exercise, our goal was to isolate the variable \( r \), the radius of the sphere.When isolating, we start by reversing the operations applied to \( r \). Since the original volume equation was \( V = \frac{4}{3} \pi r^3 \), we perform operations to "undo" these.

• First, we multiplied both sides by \( \frac{3}{4} \) to remove the fraction in front of \( \pi r^3 \).
• Then, we divided by \( \pi \) to deal with the multiplication of \( \pi \).
• Finally, we solved for \( r \) by taking the cube root, as \( r \) was originally cubed.

These steps of manipulating the equation ensure that \( r \) becomes the subject of the formula. Isolating the variable helps us express \( r \) purely in terms of other given values, which is crucial in many areas of mathematics and applied sciences.

Algebraic Manipulation

Algebraic manipulation is the heart of solving equations and is a fundamental skill in mathematics. It involves applying algebraic operations to transform an equation to make solving it easier. In this case, we used algebraic manipulation to determine the value of \( r \), the radius, from the equation for the volume of a sphere.Key steps in algebraic manipulation for this exercise included:

• Understanding how to handle fractions and multiples, such as multiplying both sides by \( \frac{3}{4} \) to clear the fraction on the right hand side.
• Dividing by constants like \( \pi \) to simplify the terms involving \( r \).
• Knowing when to apply roots, such as taking the cube root to counteract a cubic term.

These techniques allow us to work through complex equations systematically and simplify expressions. By breaking down the equation in manageable steps, algebraic manipulation helps you find solutions that might look complicated at first glance. Mastering this skill is useful not only in math classes but also in various problem-solving situations in sciences and engineering.