Chapter 8.4, Problem 20E

a.

To determine

To Find: A curve through the point $\left(0,1\right)$ .

Answer to Problem 20E

A curve through the point $\left(0,1\right)$ is

  $x=\frac{-1}{y}+1$ .

Explanation of Solution

Given information: A curve through the point $\left(0,1\right)$ whose length integral is $L={\displaystyle \underset{1}{\overset{2}{\int }}\sqrt{1+\frac{1}{y^4}}dy}$ .

Formula used: if a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$ $a<b,c>d,$ then the length of the curve is

  $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx$

If y is a smooth function of x on $\left[a,b\right]$

  $L={\displaystyle \underset{c}{\overset{d}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}dy$

If x is a smooth function of y on $\left[c,d\right]$

Calculation:

A curve through the point $\left(0,1\right)$ whose length integral is $L={\displaystyle \underset{1}{\overset{2}{\int }}\sqrt{1+\frac{1}{y^4}}dy}$

if a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$ $a<b,c>d,$ then the length of the curve is

  $L={\displaystyle \underset{c}{\overset{d}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}dy$

If $x$ is a smooth function of y on $\left[c,d\right]$

Here

  $\begin{array}{l}{\left(\frac{dx}{dy}\right)}^2=\frac{1}{y^4}\\ \frac{dx}{dy}=\frac{1}{y^2}\\ {\displaystyle \int dx}={\displaystyle \int \frac{1}{y^2}dy}\\ x=\frac{-1}{y}+c\\ \text{curve through the point }\left(0,1\right)\\ c=1\\ x=\frac{-1}{y}+1\end{array}$

b.

To determine

To Find: Numbers of such functions are there.

Answer to Problem 20E

There is only one function.

Explanation of Solution

Given information: A curve through the point $\left(0,1\right)$ whose length integral is $L={\displaystyle \underset{1}{\overset{2}{\int }}\sqrt{1+\frac{1}{y^4}}dy}$ .

Formula used: if a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$ $a<b,c>d,$ then the length of the curve is

  $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx$

If y is a smooth function of x on $\left[a,b\right]$

  $L={\displaystyle \underset{c}{\overset{d}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}dy$

If x is a smooth function of y on $\left[c,d\right]$

Calculation:

A curve through the point $\left(0,1\right)$ whose length integral is $L={\displaystyle \underset{1}{\overset{2}{\int }}\sqrt{1+\frac{1}{y^4}}dy}$

if a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$ $a<b,c>d,$ then the length of the curve is

  $L={\displaystyle \underset{c}{\overset{d}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}dy$

If $x$ is a smooth function of y on $\left[c,d\right]$

Here

  $\begin{array}{l}{\left(\frac{dx}{dy}\right)}^2=\frac{1}{y^4}\\ \frac{dx}{dy}=\frac{1}{y^2}\\ {\displaystyle \int dx}={\displaystyle \int \frac{1}{y^2}dy}\\ x=\frac{-1}{y}+c\\ \text{curve through the point }\left(0,1\right)\\ c=1\\ x=\frac{-1}{y}+1\end{array}$

So, there is only one function.