EXERCISE 6.4

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  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}

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