Chapter 6.4, Problem 34E

a.

To determine

Graph the curves.

Answer to Problem 34E

Use Wolfram Alpha you can type:Equations $x\text{ and }y$ then

Graph of theequations shown below:

  

Explanation of Solution

Given:Use Wolfram Alpha you can type: Equations $x\text{ and }y$ then

Calculation:

Use Wolfram Alpha you can type: Equations $x\text{ and }y$ then

Graph of theequations shown below:

  

b.

To determine

Find the length of Cornu’s spiral from the origin to the point.

Answer to Problem 34E

The length of Cornu’s spiral from the origin to the point is $t$

Explanation of Solution

Given:Cornu’s spiral

Calculation:

As $t\to \infty ,$ both Fresnel functions C and S approach $0.5$ like before,

But since the upper limit of the integral is less than the lower limit that

Negates the integral. So the spiral on the left approaches point $\left(-0.5,-0.5\right)$

The form of the arc length using these equations would be

  $L={\displaystyle \underset{0}{\overset{t}{\int }}\sqrt{\left[c^{\prime }{\left(t\right)}^2\right]+\left[S^{\prime }{\left(t\right)}^2\right]}}dt$

So we need the derivatives.

From the Fundamental Theorem of Calculus we get:

  $\begin{array}{l}C\left(t\right)={\displaystyle \underset{0}{\overset{t}{\int }}\cos \frac{\pi u^2}{2}}du\\ C^{\prime }\left(t\right)=\cos \frac{\pi t^2}{2}\\ S\left(t\right)={\displaystyle \underset{0}{\overset{t}{\int }}\sin \frac{\pi u^2}{2}}du\\ S^{\prime }\left(t\right)=\sin \frac{\pi t^2}{2}\end{array}$

Plug them in and simplify.

  $\begin{array}{l}L={\displaystyle \underset{0}{\overset{t}{\int }}\sqrt{{\left[\cos {\frac{\pi t}{2}}^2\right]}^2+{\left[\sin {\frac{\pi t}{2}}^2\right]}^2}}dt\\ ={\displaystyle \underset{0}{\overset{t}{\int }}\sqrt{1}}dt\\ ={\displaystyle \underset{0}{\overset{t}{\int }}dt}\\ ={\left[t\right]}_0^t\\ =t-0\\ =t\end{array}$

The length fromthe origin to the point with parameter value t is the same as t.

Thus, the length of Cornu’s spiral from the origin to the point is $t$