Problem 32 - Algebra 2

Question

Geometry The formula for the volume of a sphere is \(V=\frac{4}{3} \pi r^{3} .\) Find the radius to the nearest hundredth of a sphere with each volume. $$ 0.002 \mathrm{mm}^{3} $$

Step-by-Step Solution

Answer

The radius calculated need to be rounded to the nearest hundredth

Step 1 : Write down the given volume and the volume formula

The given volume of the sphere, \(V = 0.002 mm^3\). The volume formula of a sphere is \(V = \frac{4}{3} \pi r^3\) where V represents the volume, \(\pi\) is the mathematical constant (approximately equal to 3.14), and r is the radius.

Step 2: Substitute the given volume value in the formula

Insert the given volume (V) into the volume formula and it becomes: \(0.002 = \frac{4}{3} \pi r^3\)

Step 3: Solve for r

Solve this equation for r. To do this, multiply both sides by \( \frac{3}{4\pi}\), resulting in \(r^3 = \frac{0.002 \times 3}{4 \pi}\)

Step 4: Calculate \(r^3\) and find r

Calculate the value of \(r^3\), and then find r by taking the cube root of both sides. This results in \(r = \sqrt[3]{\frac{0.002 \times 3}{4 \pi}}\)

Step 5: Solve and Round

Use a calculator to solve for r and then round it to the nearest hundredth.

Key Concepts

Volume of Sphere

In geometry, understanding the volume of a sphere is an important concept. The volume measures how much space is inside a sphere. This is useful in various real-life situations such as engineering, physics, and even in daily life activities. The formula for calculating the volume of a sphere is given by \[ V = \frac{4}{3} \pi r^3 \]where

• \(V\) is the volume of the sphere.
• \(\pi\) is a constant approximately equal to 3.14159.
• \(r\) is the radius of the sphere.

To find the volume, you need to know the radius, which is the distance from the center of the sphere to any point on its surface. This formula emphasizes the cubic relationship of the radius to the volume, illustrating how even small changes in the radius can significantly affect the volume of a sphere.

Calculating Radius

Calculating the radius when you know the volume can seem tricky, but it follows straightforward algebraic principles. When you're given the volume \[ V = \frac{4}{3} \pi r^3 \]and need to solve for \(r\), it involves a few steps.

• First, rearrange the formula to isolate \(r^3\). You can do this by multiplying both sides of the equation by the reciprocal of \(\frac{4}{3}\pi\), which is \(\frac{3}{4\pi}\).
• This will give you: \[ r^3 = \frac{V \times 3}{4 \pi} \]

• Once you have \(r^3\), the next step is to take the cube root of both sides to solve for \(r\).

This process involves understanding and manipulating the equation so that radius \(r\) can be effectively calculated from the known sphere volume.

Mathematical Formulas

Mathematical formulas serve as tools to solve various problems systematically and provide exact solutions. Formulas are essential in geometry, especially for dealing with volumes, areas, and other properties of shapes. In the context of spheres, the formula \[ V = \frac{4}{3} \pi r^3 \]allows you to calculate the volume if the radius is known and vice versa. Mastery of these formulas aids in

• accurately describing properties of geometric shapes,
• understanding proportionality and relationships within shapes,
• solving complex real-world problems efficiently.

Comprehending and memorizing these formulas enables students to confidently tackle geometry problems. It’s crucial to understand how to rearrange these formulas algebraically to find unknown quantities, such as solving for the radius when the volume is given.

Cube Roots

Cube roots are mathematical operations used to reverse the process of cubing a number. In the formula for a sphere’s volume, after rearranging to find \(r^3\), you need to take the cube root to find \(r\).

• The cube root of a number \(x\), denoted as \(\sqrt[3]{x}\), is the number \(a\) which, when multiplied by itself twice (i.e., cubed), gives \(x\).
• For instance, if \(x = 27\), then \(\sqrt[3]{27} = 3\) because \(3 \times 3 \times 3 = 27\).

Using cube roots in problems like volume calculations helps to determine exact measurements effectively. Calculators often have dedicated functions for computing cube roots, making it easier to solve these problems accurately. Understanding cube roots deepens comprehension of mathematical relationships, ensuring accurate solutions in geometry exercises.