/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Helium gas is being pumped into ... [FREE SOLUTION] | 91Ó°ÊÓ

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Helium gas is being pumped into a spherical balloon at a rate of 5 cubic feet per minute. The pressure in the balloon remains constant. a. What is the volume of the balloon when its diameter is 20 inches? b. At what rate is the radius of the balloon changing when the diameter is 20 inches?

Short Answer

Expert verified
a) Volume is approximately 2.42 cubic feet. b) Radius is increasing at 0.573 feet/min.

Step by step solution

01

Convert Units

First, convert the diameter from inches to feet for consistency. The diameter of the balloon is given as 20 inches. Since 1 foot is 12 inches, we convert 20 inches to feet by dividing by 12: \[ \text{Diameter in feet} = \frac{20}{12} \approx 1.6667 \text{ feet} . \]
02

Calculate the Radius

The radius of the balloon in feet is half of the diameter. So, \[ r = \frac{1.6667}{2} \approx 0.8333 \text{ feet}. \]
03

Calculate the Volume of the Balloon

The volume \( V \) of a sphere is given by the formula \( V = \frac{4}{3}\pi r^3 \). Substitute the radius into the formula: \[ V = \frac{4}{3}\pi (0.8333)^3 \approx 2.423 \text{ cubic feet} . \]
04

Differentiate Volume with Respect to Time

We use the related rates formula to find \( \frac{dr}{dt} \), the rate of change of the radius: \[ V = \frac{4}{3}\pi r^3 \] differentiates with respect to \( t \) as \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} . \] Given \( \frac{dV}{dt} = 5 \text{ cubic ft/min} \), substitute the radius and solve for \( \frac{dr}{dt} \): \[ 5 = 4\pi (0.8333)^2 \frac{dr}{dt} . \]
05

Solve for Rate of Change of Radius

Simplify and solve for \( \frac{dr}{dt} \): \[ 5 = 4\pi (0.6944) \frac{dr}{dt} , \] \[ 5 = 8.722 \frac{dr}{dt} , \] \[ \frac{dr}{dt} \approx \frac{5}{8.722} \approx 0.573 \text{ feet per minute} . \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume of a Sphere
The volume of a sphere is a fundamental concept in geometry that helps us measure how much space a spherical object occupies. To find the volume, we use the formula\[V = \frac{4}{3}\pi r^3\]where \(V\) represents the volume, \(r\) is the radius of the sphere, and \(\pi\) is a constant approximately equal to 3.14159.
Understanding the sphere's radius is key, as it's the distance from the center of the sphere to any point on its surface. Doubling the radius doesn't just double the volume; it increases by quite a bit more. This is because the volume is a function of the cube of the radius.
  • Always ensure the radius is in the correct unit, so the volume calculation is accurate.
  • The formula is derived from integral calculus, but it has been simplified for easy use.
By applying this formula, you can determine the volume of any sphere, helping you understand the capacity of spherical objects, whether it's a balloon or a planet.
Differentiation
Differentiation allows us to understand how a function changes relative to changes in its input variables. For spheres, when dealing with volume changes over time, we differentiate the volume formula to find rates of change.
The differentiation of the volume formula \( V = \frac{4}{3}\pi r^3 \) with respect to time \( t \) gives us:
\[\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\].
Here, \( \frac{dV}{dt} \) represents the rate of change of the volume, and \( \frac{dr}{dt} \) represents the rate of change of the radius.
  • Each part tells us how a small change in \(r\) impacts \(V\).
  • This is a core part of calculus and gives a powerful method to solve real-world problems involving changing quantities.
Differentiation helps us peek into dynamic systems, providing insights into how changing one factor, like radius, influences another, like volume.
Conversion of Units
To ensure calculations are correct, it's important to convert all measurements to consistent units. In our problem, the diameter was originally given in inches and needed to be converted to feet.
Knowing the conversion factor—1 foot equals 12 inches—is essential.
  • To convert from inches to feet, divide by 12.
  • For feet to inches, multiply by 12.
Consistency in units prevents errors in calculations. It ensures that computations, especially involving geometrical shapes like spheres, are accurate and reliable.
For example, converting 20 inches to feet results in approximately 1.6667 feet. This change is crucial, as inconsistent use of units can lead to mistakes in measuring rates or volumes.
Rate of Change
The rate of change tells us how quickly a variable changes over time. In this exercise, we are interested in how fast the radius of a balloon is changing as helium is pumped into it.
Using the related rates formula, we found that:\[\frac{dr}{dt} = \frac{5}{8.722} \approx 0.573 \text{ feet per minute}\]
  • \( \frac{dr}{dt} \) indicates speed at which the balloon's radius expands.
  • This concept applies broadly in physics, finance, and any area involving change over time.
Understanding the rate of change helps predict future states of a system. It provides a mathematical way to analyze concepts like speed, growth, and inflation in various real-world situations.

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Most popular questions from this chapter

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