Length Of Any Curve

Last updated: September 5, 2021 4:37 pm

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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.

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