Textbook Question
9–10. Velocity graphs The figures show velocity functions for motion along a line. Assume the motion begins with an initial position of s(0)=0. Determine the following.
d. A piecewise function for s(t)
Ch. 6 - Applications of Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 6, Problem 6.6.23d
Determine whether the following statements are true and give an explanation or counterexample.
d. Let f(x)=12x^2.. The area of the surface generated when the graph of f on [−4, 4] is revolved about the y-axis is twice the area of the surface generated when the graph of f on [0, 4] is revolved about the y-axis.
Verified step by step guidance1
Identify the function and the interval: We have \( f(x) = 12x^2 \) and two intervals: \([-4,4]\) and \([0,4]\). We are revolving the graph about the y-axis to find the surface area generated.
Recall the formula for the surface area of a curve revolved about the y-axis: \[ S = \int_a^b 2\pi x \sqrt{1 + \left(\frac{df}{dx}\right)^2} \, dx \]. Here, \(x\) is the radius from the y-axis, and \(f(x)\) is the height.
Compute the derivative \( \frac{df}{dx} \) for \( f(x) = 12x^2 \): \[ \frac{df}{dx} = 24x \]. Substitute this into the surface area integral.
Set up the two surface area integrals:
- For \([-4,4]\): \[ S_1 = \int_{-4}^4 2\pi x \sqrt{1 + (24x)^2} \, dx \]
- For \([0,4]\): \[ S_2 = \int_0^4 2\pi x \sqrt{1 + (24x)^2} \, dx \].
Analyze the integrand's behavior over \([-4,4]\): Since \(x\) appears as a factor and \(x\) is negative on \([-4,0)\), the integrand is negative there, so the integral over \([-4,4]\) is not simply twice the integral over \([0,4]\). This suggests the statement is false because the surface area integral is not symmetric about zero due to the \(x\) term.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
1mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Surface Area of Revolution
The surface area generated by revolving a curve around an axis is found using an integral formula involving the function, its derivative, and the radius from the axis of revolution. For revolution about the y-axis, the radius is the x-value, and the formula accounts for the curve's length and distance from the axis.
Recommended video:
09:07
Example 1: Minimizing Surface Area
Symmetry of Functions and Intervals
When a function is even (symmetric about the y-axis), the graph on [−a, a] is symmetric. This symmetry affects integrals over symmetric intervals, often allowing simplification by doubling the integral from [0, a]. However, for surface areas revolving around an axis, the radius term may break this symmetry.
Recommended video:
06:21Properties of Functions
Effect of Radius in Surface Area Integrals
In surface area calculations revolving around the y-axis, the radius is the x-coordinate, which is negative on [−a, 0] and positive on [0, a]. Since radius appears as an absolute value or squared term, the contribution from negative x-values may differ, impacting whether the surface area over [−a, a] is exactly twice that over [0, a].
Recommended video:
09:07Example 1: Minimizing Surface Area
Related Practice
Textbook Question
Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.
d. What is the displacement of the object over the interval [0, 8]?
Textbook Question
Where do they meet? Kelly started at noon (t=0) riding a bike from Niwot to Berthoud, a distance of 20 km, with velocity v(t) = 15 / (t + 1)² (decreasing because of fatigue). Sandy started at noon (t=0) riding a bike in the opposite direction from Berthoud to Niwot with velocity u(t) = 20 / (t + 1)² (also decreasing because of fatigue). Assume distance is measured in kilometers and time is measured in hours.
d. More generally, if the riders’ speeds are v(t)=A(t+1)² and u(t)=B(t+1)² and the distance between the towns is D, what conditions on A, B, and D must be met to ensure that the riders will pass each other?
1
views
Textbook Question
Compressing and stretching a spring Suppose a force of 30 N is required to stretch and hold a spring 0.2 m from its equilibrium position.
d. How much additional work is required to stretch the spring 0.2m if it has already been stretched 0.2m from its equilibrium position?
Textbook Question
Acceleration A drag racer accelerates at a(t)=88 ft/s². Assume v(0)=0, s(0)=0, and t is measured in seconds.
d. How long does it take the racer to travel 300 ft?
Textbook Question
Displacement and distance from velocity Consider the graph shown in the figure, which gives the velocity of an object moving along a line. Assume time is measured in hours and distance is measured in miles. The areas of three regions bounded by the velocity curve and the t-axis are also given.
d. What is the displacement of the object over the interval [0,5]?