EXERCISE 6.4

  1. Use the Theorem of Pythagoras to find the length of the line segment $y=2 x$ from $(1,2)$ to $(2,4)$, and confirm that the value is consistent with the length computed using (a) Formula (4) (b) Formula (5).

 

Solution:   Given points are $(1,2)$ and $(2,4)$.

$$
\begin{aligned}
& \text { Length of the line segment using Pythagoras } \\ \\
& =\sqrt{(2-1)^{2}+(4-2)^{2}}=\sqrt{1+2^{2}}=\sqrt{1+4}=\sqrt{5}
\end{aligned}
$$

(a) Formula (4) is given by

$$
L=\int_{a}^{b} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x——-(1)
$$

Equation of the line is

$$
y=2 x——–(2)
$$

differentiating w.r.t $x$

$$
\frac{d y}{d x}=2——-(3)
$$

Given points : $$\left(x_{1}, y_{1}\right)=(1,2)$$

$$
\left(x_{2}, y_{2}\right)=(2,4)
$$

limits of integration will be $x_{1}=a=1, x_{2}=b=2.$

Substituting (3) into (1) and integrating over the interval $[1,2]$.

$$
\begin{aligned}
& L=\int_{1}^{2} \sqrt{1+(2)^{2}} d x=\int_{1}^{2} \sqrt{1+4} d x=\int_{1}^{2} \sqrt{5} d x=\sqrt{5}[x]_{1}^{2} \\ \\
& L=\sqrt{5}(2-1)=\sqrt{5}
\end{aligned}
$$

(b)

Formula (5) is given below

$$
L=\int_{c}^{d} \sqrt{1+\left(\frac{d x}{d y}\right)^{2}} d y———(4)
$$

From (3) part (a), we have

$$
\frac{d y}{d x}=2 \Rightarrow \frac{d x}{d y}=\frac{1}{2}——-(5)
$$

limits of integration in this case will be

$$
y_{1}=c=2, \quad y_{2}=d=4
$$

Substituting (5) into (4) and integrating over the interval $[2,4]$

$$
L=\int_{2}^{4} \sqrt{1+\left(\frac{1}{2}\right)^{2}} d y=\int_{2}^{4} \sqrt{1+\frac{1}{4}} d y=\int_{2}^{4} \sqrt{\frac{4+1}{4}} d y
$$

$$

L=\int_{2}^{4} \sqrt{\frac{5}{4}} d y=\int_{2}^{4} \frac{\sqrt{5}}{2} d y=\frac{\sqrt{5}}{2}[y]_{2}^{4}=\frac{\sqrt{5}}{2}[4-2]=\frac{\sqrt{5}}{2}[2]=\sqrt{5}

$$

2.  use the theorem…

Solution: Given points are $$\left(x_{1}, y_{1}\right)=(0,0)$$

$$
\left(x_{2}, y_{2}\right)=(1,5)
$$

length of the line segment using pythagoras theorem

$$
=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}
$$

$$
=\sqrt{(1-0)^{2}+(5-0)^{2}}=\sqrt{1+25}=\sqrt{26}
$$

(3)

(a) Formula (4) is given by

$$
L=\int_{a}^{b} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x———–(1)
$$

Given equation of line segment

$$
y=5 x
$$

differentiating w.r.t $x$

$$
\frac{d y}{d x}=5———-(2)
$$

limits of integration will be $x_{1}=a=0, x_{2}=b=1$ Substituting (2) into (1) and integrating over the interval $[0,1]$

$$
\begin{aligned}
& L=\int_{0}^{1} \sqrt{1+(5)^{2}} d x=\int_{0}^{1} \sqrt{1+25} d x=\sqrt{26} \left[x \right]_{0}^{1}=\sqrt{26}(1-0) \\ \\
& L=\sqrt{26}
\end{aligned}
$$

(b) From (2) part (a)

$$
\frac{d y}{d x}=5 \Rightarrow \frac{d x}{d y}=\frac{1}{5}———–(3)
$$

Formula (5) is given by

$$
L=\int_{c}^{d} \sqrt{1+\left(\frac{d x}{d y}\right)^{2}} d y \quad—(4)
$$

substituting (3) into (4) and integrating over the interval $[0, 5]$

$$
\begin{aligned}

&L=\int_{0}^{5} \sqrt{1+\left(\frac{1}{5}\right)^{2}} d y \\ \\ & L=\int_{0}^{5} \sqrt{1+\frac{1}{25}} d y \\ \\& L=\int_{0}^{5} \sqrt{\frac{25+1}{25}} d y \\ \\
&L=\int_{0}^{5} \frac{\sqrt{26}}{5} d y=\frac{\sqrt{26}}{5}\left [y \right]_{0}^{5}=\frac{\sqrt{26}}{5}(5-0)=\sqrt{26}
\end{aligned}
$$

3-8. Find the exact arc length of the curve over the interval.

3.  $y=3 x^{3 / 2}-1$ from $x=0$ to $x=1$.

differentiating w.r.t. $x$

$$
\frac{d y}{d x}=3\left(\frac{3}{2}\right) x^{1 / 2}=\frac{9}{2} x^{1 / 2}
$$

Arc length formula:

$$
\begin{aligned}
& L=\int_{a}^{b} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \\ \\
& L=\int_{0}^{1} \sqrt{1+\left(\frac{9}{2} x^{1 / 2}\right)^{2}} d x \\ \\
& L=\int_{0}^{1} \sqrt{1+\frac{81}{4} x} d x=\int_{0}^{1}\left(1+\frac{81}{4} x\right)^{1 / 2} d x
\end{aligned}
$$

 

The derivative of $1+\frac{81}{4} x$ is $\frac{81}{4}$

Now multiplying and dividing (5) by $\frac{81}{4}$

$$
\begin{aligned}
\therefore L & =\frac{4}{81} \int_{0}^{1} \frac{81}{4}\left(1+\frac{81}{4} x\right)^{1 / 2} d x=\frac{4}{81}\left[\frac{\left(1+\frac{81}{4} x\right]^{3 / 2}}{3 / 2}\right]_{0}^{1} \\ \\
L & =\frac{4}{81} \cdot \frac{2}{3}\left[\left(1+\frac{81}{4} x\right)\right]_{0}^{1} \\ \\
L & =\frac{8}{243}\left[\left(1+\frac{81}{4}(1) \right)^{3 / 2}-(1+0)^{3 / 2}\right]\right. \\ \\
L & =\frac{8}{243}\left[\left(\frac{4+81}{4}\right)^{3 / 2}-1\right]=\frac{8}{243}\left[\frac{(85)^{3 / 2}}{(4)^{3 / 2}}-1\right] \\ \\
L & =\frac{8}{243}\left[\frac{(85)^{3 / 2}}{\left(2^{2}\right)^{3 / 2}}-1\right]=\frac{8}{243}\left[\frac{(85)^{3 / 2}}{2^{3}}-1\right]=\frac{8}{243}\left[\frac{(85)^{3 / 2}}{8}-1\right] \\ \\
L & =\frac{1}{243}\left[(85)^{3 / 2}-8\right]
\end{aligned}
$$