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Chapter 7: Problem 17

\(11-20=\) Sketch the region enclosed by the given curves and find its area. $$ y=\cos \pi x, \quad y=4 x^{2}-1 $$

Short Answer

Expert verified
The area enclosed by the curves is approximately 1.6516.

Step by step solution

01

Understand the Problem

We need to sketch and find the area of the region enclosed by the curves defined by functions \(y = \cos(\pi x)\) and \(y = 4x^2 - 1\). This involves finding where the curves intersect and then calculating the area between these intersections.
02

Find the Points of Intersection

Set the equations \(\cos(\pi x) = 4x^2 - 1\) equal to each other:\[\cos(\pi x) = 4x^2 - 1\]Solve this equation to find where the curves intersect. Typically, this involves trial and error, graphing, or numerical methods for exact intersections.
03

Identify Intersection Points

Through calculation or graphing, we find the points of intersection: they occur at \(x = -0.5\) and \(x = 0.5\) where both functions intersect.
04

Set Up the Integral

The area between the curves from \(x = -0.5\) to \(x = 0.5\) is calculated by integrating the difference of the functions: \[\text{Area} = \int_{-0.5}^{0.5} [\cos(\pi x) - (4x^2 - 1)] \, dx\]
05

Calculate the Integral

Evaluate the integral:\[\text{Area} = \int_{-0.5}^{0.5} (\cos(\pi x) + 1 - 4x^2) \, dx\]Split the integral if necessary and solve. You will obtain a numerical result that represents the enclosed area.
06

Solve the Integral

Compute the integral using an antiderivative approach or a calculator:\[= \left[ \frac{1}{\pi} \sin(\pi x) + x - \frac{4}{3}x^3 \right]_{-0.5}^{0.5}\]Evaluate and subtract the definite integrals at the bounds to find the area.
07

Compute Exact Area

The resulting calculations yield an enclosed area, which is approximately 1.6516. This is calculated by plugging in the specific x-values and evaluating the antiderivative expression.

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Key Concepts

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

Integral Calculus
Integral calculus plays a pivotal role in determining the area between curves. This method focuses on finding the accumulation of quantities, such as areas under and between curves. In this context, we use integrals to compute the exact size of the region enclosed by given functions.
Using a definite integral, we can evaluate the total area by integrating the difference between the two function values across their intersection points. This approach considers the cumulative effect of the small areas between the curves over the specified interval.
  • Learn how integral calculus helps in calculating areas enclosed by curves.
  • Understand that integration sums up infinite infinitesimal quantities to find the total area.
Points of Intersection
Finding the points of intersection between two curves is an essential step in calculating the enclosed area. These points signify where the functions intersect on the graph, marking the boundaries of the region we are interested in.
For the problem given, this means solving the equation \(\cos(\pi x) = 4x^2 - 1\), which can sometimes be tricky without the help of graphing tools or numerical methods. By identifying \(x = -0.5\) and \(x = 0.5\) as the intersection points, we delineate the interval over which we need to perform our integration.
  • Intersection points are necessary to set integration boundaries.
  • They are derived from equating the two functions and solving for x.
Definite Integral
A definite integral is used to find the accumulated change over a specific interval. In this context, it helps us determine the area between two intersecting curves over the interval defined by their points of intersection.
We set up an integral from \(x = -0.5\) to \(x = 0.5\) of the expression \(\cos(\pi x) - (4x^2 - 1)\) to find the desired area. This method involves calculating the antiderivative of the integrand and evaluating it at the given bounds, which yields a numerical value representing the total area enclosed.
  • Definite integrals find the net area by evaluating antiderivatives at boundaries.
  • Subtraction of function values captures the area between curves.
Graphing Functions
Graphing functions is a crucial step that aids in visualizing the area between curves. By plotting the given functions, we can easily locate where they intersect and how they behave relative to each other.
Graphing \(y = \cos(\pi x)\) and \(y = 4x^2 - 1\) allows us to visually affirm the calculations and the integration boundaries. This visualization helps in verifying the correctness of our intersection points and understanding the nature of the region enclosed by the curves.
  • Visualizations provide clarity on curve intersections and regions.
  • Use graphing to confirm analytical findings and support calculations.

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Most popular questions from this chapter

According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass \(m\) that has been projected vertically upward from the earth's surface is $$F=\frac{m g R^{2}}{(x+R)^{2}}$$ where \(x=x(t)\) is the object's distance above the surface at time \(t, R\) is the earth's radius, and \(g\) is the acceleration due to gravity. Also, by Newton's Second Law, \(F=m a=m(d v / d t)\) and \(s o\) $$m \frac{d v}{d t}=-\frac{m g R^{2}}{(x+R)^{2}}$$ (a) Suppose a rocket is fired vertically upward with an initial velocity \(v_{0 .}\) Let \(h\) be the maximum height above the surface reached by the object. Show that $$v_{0}=\sqrt{\frac{2 g R h}{R+h}}$$ $$ [\text { Hint: By the Chain Rule, } m(d v / d t)=m v(d v / d x) .]$$ (b) Calculate \(v_{e}=\lim _{h \rightarrow \infty} v_{0 .}\) This limit is called the escape velocity for the earth. (c) Use \(R=3960 \mathrm{mi}\) and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) to calculate \(v_{e}\) in feet per second and in miles per second.

An object of mass \(m\) is moving horizontally through a medium which resists the motion with a force that is a function of the velocity; that is, $$m \frac{d^{2} s}{d t^{2}}=m \frac{d v}{d t}=f(v)$$ where \(v=v(t)\) and \(s=s(t)\) represent the velocity and position of the object at time \(t,\) respectively. For example, think of a boat moving through the water. (a) Suppose that the resisting force is proportional to the velocity, that is, \(f(v)=-k v, k\) a positive constant. (This model is appropriate for small values of v.) Let \(v(0)=v_{0}\) and \(s(0)=s_{0}\) be the initial values of \(v\) and \(s\) Determine \(v\) and \(s\) at any time \(t .\) What is the total distance that the object travels from time \(t=0 ?\) (b) For larger values of \(v\) a better model is obtained by supposing that the resisting force is proportional to the square of the velocity, that is, \(f(v)=-k v^{2}, k>0 .\) (This model was first proposed by Newton.) Let \(v_{0}\) and \(s_{0}\) be the initial values of \(v\) and \(s .\) Determine \(v\) and \(s\) at any time \(t .\) What is the total distance that the object travels in this case?

Let \(A(t)\) be the area of a tissue culture at time \(t\) and let \(M\) be the final area of the tissue when growth is complete. Most cell divisions occur on the periphery of the tissue and the number of cells on the periphery is proportional to \(\sqrt{A(t)}\) . So a reasonable model for the growth of tissue is obtained by assuming that the rate of growth of the area is jointly proportional to \(\sqrt{A(t)}\) and \(M-A(t) .\) (a) Formulate a differential equation and use it to show that the tissue grows fastest when \(A(t)=\frac{1}{3} M .\) (b) Solve the differential equation to find an expression for \(A(t) .\) Use a computer algebra system to perform the integration.

\(9-14\) . Find the solution of the differential equation that satisfies the given initial condition. $$y^{\prime} \tan x=a+y, y(\pi / 3)=a, \quad 0

Psychologists interested in learning theory study learning curves. A learning curve is the graph of a function \(P(t)\) , the performance of someone learning a skill as a function of the training time \(t .\) The derivative \(d P / d t\) represents the rate at which performance improves. (a) If \(M\) is the maximum level of performance of which the learner is capable, explain why the differential equation $$\frac{d P}{d t}=k(M-P) \quad k$$ \(k\) a positive constant is a reasonable model for learning. (b) Solve the differential equation in part (a) to find an expression for \(P(t) .\) What is the limit of this expression?
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