basic / September 5, 2021

Length Of Any Curve

Let’s suppose is the equation of any curve. Then the length of the curve between and isContentsEXAMPLE 1SOLUTIONSTEP 1STEP 2STEP 3EXAMPLE 2SOLUTIONSTEP 1STEP 2STEP 3 EXAMPLE 1 According to Stroud and Booth (2013)*, “Find the length of the arc of the curve , between and .” SOLUTION Now here the equation of the curve is (1) First of all, I’ll get […]

By senseadmin · September 5, 2021 · 2 min read

Let’s suppose y = f(x) is the equation of any curve. Then the length of the curve between x = a and x = b is

Contents

EXAMPLE 1
SOLUTION

STEP 1
STEP 2
STEP 3

EXAMPLE 2
SOLUTION

STEP 1
STEP 2
STEP 3

\[S = \int_a^b \sqrt{1+\left(\cfrac{dy}{dx}\right)^2}~dx.\]

EXAMPLE 1

According to Stroud and Booth (2013)*, “Find the length of the arc of the curve 6xy = x^4 + 3 , between x = 1 and x = 2 .”

SOLUTION

Now here the equation of the curve is

(1) $\begin{equation*}6xy = x^4 + 3.\end{equation*}$

First of all, I’ll get the value of $\left(\cfrac{dy}{dx}\right)^2$ .

STEP 1

So, from equation (1), I can say that the value of is

\[y = \frac{x^4+3}{6x} \Rightarrow y = \frac{x^3}{6} + \frac{1}{2x}.\]

Next, I’ll differentiate with respect to to get

\[\frac{dy}{dx} = \frac{3x^2}{6} - \frac{1}{2x^2}.\]

Then I’ll simplify it to get

\[\frac{dy}{dx} = \frac{x^2}{2} - \frac{1}{2x^2}.\]

So this means

\[\frac{dy}{dx} = \frac{1}{2} \left(x^2 - \frac{1}{x^2}\right).\]

Therefore the value of $\left(\cfrac{dy}{dx}\right)^2$ is

\[\left(\cfrac{dy}{dx}\right)^2 = \frac{1}{4} \left(x^2 - \frac{1}{x^2}\right)^2.\]

If I simplify it, I’ll get

\begin{eqnarray*} \left(\cfrac{dy}{dx}\right)^2 &=& \frac{1}{4} \left(x^4 + \frac{1}{x^4} - 2.x^2.\frac{1}{x^2}\right)\\ &=& \frac{1}{4} \left(x^4 + \frac{1}{x^4} - 2}\right)\\&=& \frac{1}{4} \left(\frac{1+x^8-2x^4}{x^4}}\right) .\end{eqnarray*}

Next, I’ll get the value of $\sqrt{1+\left(\cfrac{dy}{dx}\right)^2}$ .

STEP 2

At first, I’ll get the value of $1+\left(\cfrac{dy}{dx}\right)^2$ .

Thus it will be

\[1+\left(\cfrac{dy}{dx}\right)^2 = 1 +\frac{1+x^8-2x^4}{4x^4}.\]

Now I’ll simplify it to get

\[1+\left(\cfrac{dy}{dx}\right)^2 = \frac{4x^4+1+x^8-2x^4}{4x^4}.\]

So this gives

\begin{eqnarray*}1+\left(\cfrac{dy}{dx}\right)^2 &=& \frac{2x^4+1+x^8}{4x^4}\\ &=& \frac{(x^4+1)^2}{(2x^2)^2}\\&=& \left(\frac{x^4+1}{2x^2}\right)^2.\end{eqnarray*}

Therefore the value of $\sqrt{1+\left(\cfrac{dy}{dx}\right)^2}$ is

\[\sqrt{1+\left(\cfrac{dy}{dx}\right)^2} = \frac{x^4+1}{2x^2}.\]

And that means

\[\sqrt{1+\left(\cfrac{dy}{dx}\right)^2} = \frac{x^2}{2}+\frac{1}{2x^2}.\]

Now I’ll get the length of the arc of the curve 6xy = x^4 + 3 , between x = 1 and x = 2 .

STEP 3

So it will be

\[S = \int_1^2 \sqrt{1+\left(\cfrac{dy}{dx}\right)^2}~dx.\]

And that means

\[S = \int_1^2 \left(\frac{x^2}{2}+\frac{1}{2x^2}\right)~dx.\]

Next, I’ll integrate it to get

\[S = \left[\frac{x^3}{2.3} - \frac{1}{2x} \right]_1^2 = \left[\frac{x^3}{6} - \frac{1}{2x} \right]_1^2 .\]

Then I’ll substitute the limits to get

\[S = \left[\frac{2^3}{6} - \frac{1}{2.2} \right] - \left[\frac{1^3}{6} - \frac{1}{2.1} \right].\]

Now I’ll simplify it. And that gives

\begin{eqnarray*} S &=& \left[\frac{8}{6} - \frac{1}{4} \right] - \left[\frac{1}{6} - \frac{1}{2} \right]\\ &=& \frac{4}{3} - \frac{1}{4} - \frac{1}{6} + \frac{1}{2}\\ &=& \frac{16-3-2+6}{12}\\ &=& \frac{17}{12}.\end{eqnarray*}

Thus the length of the arc is $\cfrac{17}{12}$ . Hence I can conclude that this is the answer to the given example.

Now I’ll give another example.

EXAMPLE 2

According to Stroud and Booth (2013)*, “Find the length of the curve $8(y + \ln x) = x^2$ between x = 1 and x = e .”

SOLUTION

Now here the equation of the curve is

(2) $\begin{equation*}8(y + \ln x) = x^2.\end{equation*}$

First of all, I’ll get the value of $\left(\cfrac{dy}{dx}\right)^2$ .

STEP 1

So I’ll differentiate equation (2) throughout with respect to . And that gives

\[8\left(\frac{dy}{dx} + \frac{1}{x}\right) = 2x.\]

Next, I’ll simplify it to get

\[4\left(\frac{dy}{dx} + \frac{1}{x}\right) = x.\]

So this means

\[\frac{dy}{dx} + \frac{1}{x} = \frac{x}{4}.\]

And that gives the value of $\cfrac{dy}{dx}$ as

\[\frac{dy}{dx} = \frac{x}{4} - \frac{1}{x} = \frac{x^2 - 4}{4x}.\]

Now I’ll get the value of $\left(\cfrac{dy}{dx}\right)^2$ . So that will be

\[\left(\cfrac{dy}{dx}\right)^2 =\frac{(x^2 - 4)^2}{(4x)^2}.\]

Next, I’ll get the value of $\sqrt{1+\left(\cfrac{dy}{dx}\right)^2}$ .

STEP 2

At first, I’ll get the value of $1+\left(\cfrac{dy}{dx}\right)^2$ .

Thus it will be

\[1+\left(\cfrac{dy}{dx}\right)^2 = 1 + \frac{(x^2 - 4)^2}{(4x)^2}.\]

Now I’ll simplify it to get

\[1+\left(\cfrac{dy}{dx}\right)^2 = \frac{(4x)^2 + (x^2 - 4)^2}{(4x)^2}.\]

So this gives

\begin{eqnarray*} 1+\left(\cfrac{dy}{dx}\right)^2 &=& \frac{(4x)^2 + (x^2 - 4)^2}{(4x)^2}\\ &=& \frac{16x^2 + x^4 +16 -8x^2}{(4x)^2}\\ &=& \frac{x^4 +16 + 8x^2}{(4x)^2}\\ &=& \frac{(x^2 + 4)^2}{(4x)^2} .\end{eqnarray*}

Therefore the value of $\sqrt{1+\left(\cfrac{dy}{dx}\right)^2}$ is

\[\sqrt{1+\left(\cfrac{dy}{dx}\right)^2} = \frac{x^2+4}{4x}.\]

And that means

\[\sqrt{1+\left(\cfrac{dy}{dx}\right)^2} = \frac{x}{4}+ \frac{1}{x}.\]

Now I’ll get the the length of the curve $8(y + \ln x) = x^2$ between x = 1 and x = e .

STEP 3

So it will be

\[S = \int_1^e \sqrt{1+\left(\cfrac{dy}{dx}\right)^2}~dx.\]

And that means

\[S = \int_1^e \left(\frac{x}{4}+ \frac{1}{x}\right)~dx.\]

Next, I’ll integrate it to get

\[S = \left[\frac{x^2}{2.4} + \ln x\right]_1^{e} = \left[\frac{x^2}{8} + \ln x\right]_1^{e}.\]

Then I’ll substitute the limits to get

\[S = \left[\frac{e^2}{8} + \ln e\right] - \left[\frac{1^2}{8} + \ln 1\right].\]

Now I’ll simplify it and that gives

\begin{eqnarray*} S &=& \left[\frac{e^2}{8} + 1\right] - \left[\frac{1}{8} + 0\right]\\ &=& \frac{e^2}{8} + 1 - \frac{1}{8}\\ &=& \frac{e^2}{8} + \frac{7}{8}\\ &=& \frac{e^2 + 7}{8}. \end{eqnarray*}

Thus the length of the curve is $\cfrac{e^2 + 7}{8}$ .

Hence I can conclude that this is the answer to the given example.