Chapter 5, Problem 68RE

a.

To determine

To find: on what intervals is C increasing.

Answer to Problem 68RE

C(x) is increasing when:

  $4n-1\le x^2\le 4n+1.$

Explanation of Solution

Given information: Given function $C(x)={\displaystyle \underset{0}{\overset{x}{\int }}\cos (\frac{1}{2}\pi t^2})dt$ and $S(x)={\displaystyle \underset{0}{\overset{x}{\int }}\sin (\frac{1}{2}\pi t^2})dt$ in.

his theory of the diffraction of light waves .

Formula Used:

2^nd Fundamental Theorem of calculus:

  $\frac{d}{dt}{\displaystyle \underset{a}{\overset{x}{\int }}f(t)dt=f(x)}$ .

Calculation:

  $C(x)={\displaystyle \underset{0}{\overset{x}{\int }}\cos (\frac{1}{2}\pi t^2})dt$ .

Differentiate

  $C\text{'}(x)=\cos (\frac{1}{2}\pi x^2)$ .

Since $\cos \theta$ is positive when it lies in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ .

Since it is periodic with periodicity $2\pi$ , and then $\cos \theta$ is positive when:

  $2n\pi -\frac{\pi }{2}\le \theta \le 2n\pi +\frac{\pi }{2}$ .

Here $\theta =\frac{\pi x^2}{2}$ .

C(x) is increasing when:

  $\begin{array}{c}2n\pi -\frac{\pi }{2}\le \frac{\pi x^2}{2}\le 2n\pi +\frac{\pi }{2}\\ 4n-1\le x^2\le 4n+1.\end{array}$

b.

To determine

To find: on what intervals is C concave upward.

Answer to Problem 68RE

n is a non −negative integer and

when x >0: $\sqrt{2(2n+1)}<x<2\sqrt{n+1}$ .

when x <0: $-\sqrt{2(2n+1)}<x<-2\sqrt{n}$ .

Explanation of Solution

Given information: Given function $C(x)={\displaystyle \underset{0}{\overset{x}{\int }}\cos (\frac{1}{2}\pi t^2})dt$ and $S(x)={\displaystyle \underset{0}{\overset{x}{\int }}\sin (\frac{1}{2}\pi t^2})dt$ in.

his theory of the diffraction of light waves.

Calculation:

  $C(x)={\displaystyle \underset{0}{\overset{x}{\int }}\cos (\frac{1}{2}\pi t^2})dt$ .

Differentiate

  $\begin{array}{l}C\text{'}(x)=\cos (\frac{1}{2}\pi x^2)\\ $ C\text{'}\text{'}(x)=-\sin (\frac{1}{2}\pi x^2).{(\frac{1}{2}\pi x^2)}^{\text{'}}$C\text{'}\text{'}(x)=-\pi x\sin (\frac{1}{2}\pi x^2)\end{array}$

For x > 0, $C\text{'}\text{'}(x)=-\pi x\sin (\frac{1}{2}\pi x^2)$ is positive when the sin part is negative that is when:

  $\begin{array}{l}2n\pi +\pi <\frac{\pi x^2}{2}<2n\pi +2\pi \\ 2(2n+1)<x^2<4(n+1)\\ \sqrt{2(2n+1)}<x<\sqrt{4(n+1)}\end{array}$

For x < 0, $C\text{'}\text{'}(x)=-\pi x\sin (\frac{1}{2}\pi x^2)$ is positive when the sin part is positive, that is when:

  $\begin{array}{l}2n\pi +0<\frac{\pi x^2}{2}<2n\pi +\pi \\ 4n<x^2<2(2n+1)\\ \sqrt{4n}<x<\sqrt{2(2n+1)}\\ -2\sqrt{n}>x>-\sqrt{2(2n+1)}\end{array}$

Put the negative sign at the end, and also reverse the inequality signs.

c.

To determine

To solve: the following equation correct to two decimal places using the graph:

  ${\displaystyle \underset{0}{\overset{x}{\int }}\cos (\frac{1}{2}\pi t^2})dt=0.7$

Answer to Problem 68RE

x = 0.76 and x = 1.22 are the two solutions.

Explanation of Solution

Given information:

Calculation:

The solution of ${\displaystyle \underset{0}{\overset{x}{\int }}\cos (\frac{1}{2}\pi t^2})dt=0.7$ is the x -coordinate of the point of intersection between the graphs of y = ${\displaystyle \underset{0}{\overset{x}{\int }}\cos (\frac{1}{2}\pi t^2})dt$ and y =0.7.

In the graph below:

The red curve is C(x) and y =0.7 is the horizontal straight line.

  

From the graph below, x = 0.76 and x = 1.22 are the two solutions.

d.

To determine

To plot: the graph of C and S on the same screen and how are these graphs related.

Answer to Problem 68RE

Both curves converge to $\frac{1}{2}\text{as}x\to \infty$ and converge to $-\frac{1}{2}\text{as}x\to -\infty$ .

Explanation of Solution

Given information: Given function $C(x)={\displaystyle \underset{0}{\overset{x}{\int }}\cos (\frac{1}{2}\pi t^2})dt$ and $S(x)={\displaystyle \underset{0}{\overset{x}{\int }}\sin (\frac{1}{2}\pi t^2})dt$ in.

his theory of the diffraction of light waves

Calculation:

In the graph below:

  $S(x)={\displaystyle \underset{0}{\overset{x}{\int }}\sin (\frac{1}{2}\pi t^2})dt$ is the blue curve.

  $C(x)={\displaystyle \underset{0}{\overset{x}{\int }}\cos (\frac{1}{2}\pi t^2})dt$ is the red curve.

  

Note that both curves converges to $\frac{1}{2}\text{as}x\to \infty$ and converge to $-\frac{1}{2}\text{as}x\to -\infty$ .