/*! 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 39 If the radius of a circle of are... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the radius of a circle of area \(A\) and circumference \(C\) is doubled, find the new area and circumference of the circle in terms of \(A\) and C. (Consult Chapter 0 if necessary.)

Short Answer

Expert verified
The new area is \( 4A \) and the new circumference is \( 2C \).

Step by step solution

01

Recall Key Circle Formulas

The area of a circle is given by the formula \( A = \pi r^2 \), and the circumference is \( C = 2\pi r \) where \( r \) is the radius of the circle.
02

Express Radius in Terms of Area and Circumference

From the area formula \( A = \pi r^2 \), solve for \( r \) to get \( r = \sqrt{\frac{A}{\pi}} \). Similarly, from the circumference formula \( C = 2\pi r \), solve for \( r \) to get \( r = \frac{C}{2\pi} \). These expressions will help to substitute back once the radius is doubled.
03

Double the Radius

If the radius \( r \) is doubled, the new radius \( r' \) is \( 2r \).
04

Calculate the New Area

Substitute \( r' = 2r \) into the area formula: The new area is \( A' = \pi (2r)^2 = 4\pi r^2 \). Since \( A = \pi r^2 \), the new area is \( A' = 4A \).
05

Calculate the New Circumference

Substitute \( r' = 2r \) into the circumference formula: The new circumference is \( C' = 2\pi (2r) = 4\pi r \). Since \( C = 2\pi r \), the new circumference is \( C' = 2C \).
06

Summarize the Results

The new area of the circle is \( 4A \) and the new circumference is \( 2C \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Area of a Circle
The area of a circle is a measure of the space enclosed within its boundary. Imagine the circle filled with tiny squares, the sum of those squares gives you the area. For any circle, the area can be calculated using the formula: \[A = \pi r^2\]where \( A \) represents the area, \( \pi \) (pi) is approximately 3.14159, and \( r \) is the radius of the circle. When the radius is doubled, in this context, the area of the circle changes significantly. Since the area is proportional to the square of the radius, doubling the radius increases the area by a factor of four. This means if the original area was \( A \), the new area will be \( 4A \). This scaling property is crucial in geometry and physics, especially when calculating areas in different shapes or changing dimensions.
Understanding this relationship helps us recognize how areas grow in relation to changes in size.
Circumference of a Circle
The circumference of a circle is the total length around the circle, akin to a perimeter of a polygon. For a circle, the circumference provides insights into the boundary's length, wrapped around the circle once. The formula for calculating the circumference is: \[C = 2\pi r\]where \( C \) stands for the circumference and \( r \) is the radius.
When the radius of a circle is doubled, the circumference is also directly affected. It is directly proportional to the radius, meaning if the radius doubles, so does the circumference. So, if the initial circumference is \( C \), when the radius becomes twice its size, the circumference becomes \( 2C \). This linear relation between the radius and circumference is foundational in geometry, showing how linear dimensions influence boundary lengths.
It also is important in practical applications, such as determining distances traveled by wheels or circular tracks.
Radius of a Circle
The radius of a circle is the distance from the center to any point on its edge. It represents half of the diameter, which is the longest distance across the circle. The radius is a fundamental part of the key circle formulas, as it helps define both the area and the circumference:
  • From the formula \( A = \pi r^2 \), we understand the significant impact of the radius on the area.

  • In the circumference formula, \( C = 2\pi r \), the radius directly scales the circle's boundary length.
The beauty of geometry lies in how these measurements interrelate. In our example, when the radius is initially solved through the formulas \( r = \sqrt{\frac{A}{\pi}} \) or \( r = \frac{C}{2\pi} \), it reveals its overarching influence.
When doubled, the new radius allows us to explore the expanded properties of larger circles, evidencing how a simple linear change has exponential impacts on a circle's area, and linear yet proportional effects on its circumference.
Geometry in Physics
Geometry plays a significant role in physics, especially when describing the physical world. Circles and circular motions are common in physical systems and engineering, manifested in gears, orbits, and rotations. When examining systems through geometry, understanding the relationship between radius, area, and circumference helps us to understand different physical phenomena. For instance,
  • The area can represent sections of wheels impacting friction or contact area.
  • Circumference might relate to the path in circular motion or revolution cycles.
In problems like the one we tackled, doubling the radius can symbolize scaling up systems to handle more load, speed, or volume. Understanding these geometric principles aids in conceptualizing and calculating physical possibilities efficiently. This connection between mathematics and physics highlights how foundational concepts such as circle formulas are, not just in pure mathematics but across numerous physical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of \(60 g,\) but lasting only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g} ?\)

Galileo used marbles rolling down inclined planes to deduce some basic properties of constant accelerated motion. In particular, he measured the distance a marble rolled during specific time periods. For example, suppose a marble starts from rest and begins rolling down an inclined plane with constant acceleration \(a\). After 1 s, you find that it moved a distance \(x\). (a) In terms of \(x,\) how far does it move in the next 1 s time period-that is, in the time between \(1 \mathrm{~s}\) and \(2 \mathrm{~s} ?\) (b) How far does it move in the next second of the motion? (c) How far does it move in the \(n\) th second of the motion?

(a) The pilot of a jet fighter will black out at an acceleration greater than approximately \(5 g\) if it lasts more than a few seconds. Express this acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) and \(\mathrm{ft} / \mathrm{s}^{2} .\) (b) The acceleration of the passenger during a car crash with an air bag is about \(60 g\) for a very short time. What is this acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) and \(\mathrm{ft} / \mathrm{s}^{2} ?\) (c) The acceleration of a falling body on our moon is \(1.62 \mathrm{~m} / \mathrm{s}^{2}\). How many \(g^{\prime} \mathrm{s}\) is this? (d) If the acceleration of a test plane is \(24.3 \mathrm{~m} / \mathrm{s}^{2},\) how many \(g\) 's is it?

A state trooper is traveling down the interstate at \(30 \mathrm{~m} / \mathrm{s}\). He sees a speeder traveling at \(50 \mathrm{~m} / \mathrm{s}\) approaching from behind. At the moment the speeder passes the trooper, the trooper hits the gas and gives chase at a constant acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\). Assuming that the speeder continues at \(50 \mathrm{~m} / \mathrm{s},\) (a) how long will it take the trooper to catch up to the speeder? (b) How far down the highway will the trooper travel before catching up to the speeder?

You are driving eastbound on the interstate at \(70 \mathrm{mi} / \mathrm{h}\). You observe that you are approaching a truck in your lane at a relative speed of \(20 \mathrm{mi} / \mathrm{h}\). (a) How fast is the truck moving relative to the highway? (b) If the truck were instead traveling at this speed in the westbound lane, what would be the relative velocity between you and the truck?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.