Using calculus to find the length of a curve

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In this section we'll learn how to use derivatives and definite integrals to calculate the length of a curve. We'll do this both for functions of the form $y = f(x)$, and for parametric functions, where each point $(x, y)$ is defined by a parameter (like time, t), such as $(x, y) = (x(t), y(t))$.

Arc length

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We begin by defining a function $f(x)$, like in the graph below. To find the length of the curve between $x = x_o$ and $x = x_n$, we'll break the curve up into $n$ small line segments, for which it's easy to find the length just using the Pythagorean theorem, the basis of how we calculate distance on the plane.

The figure shows the basic geometry. Each point $(x, f(x))$ is labeled $P_1, p_2, \dots p_{i-1}, p_i, \dots , p_n$, where we break our curve into $n$ pieces.

The area around two adjacent points, $P_{n-1}$ and $P_n$ is enlarged below to show how the Pythagorean theorem helps us calculate distance. The horizontal distance is $x_i - x_{i-1}$ and the vertical distance is the difference between the function values at those two points, $\Delta y = f(x_i) - f(x_{i-1})$. It's the same for any two adjacent points.

So the distance between successive points is the square root of the sum of the squares of the rise and run from one point to the other.

Example 1: Testing our formula on a line

Calculate the length of   $f(x) = 2x + 1$   between $x = 1$ and $x = 4$

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Solution: We really don't need an integral to find the length of a line, but let's use it to confirm that the method works.

If $f(x) = 2x + 1$, then $f(1) = 3$ and $f(4) = 9$, so our two endpoints are (1, 3) and (4, 9). We can just use the distance formula (really the Pythagorean theorem) to find the distance:

$$ \begin{align} \text{distance} &= \sqrt{(x_i - x_{i - 1})^2 + (y_i - y_{i - 1})^2} \\[5pt] &= \sqrt{(4 - 1)^2 + (9 - 3)^2} \\[5pt] &= \sqrt{9 + 36} = \sqrt{45} = 3 \sqrt{5} \\[5pt] &= 6.708 \; \text{units} \end{align}$$

Example 2

Calculate the length of the circle   $x^2 + y^2 = r^2.$

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Solution: Here again, we already know the answer: The circumference of a circle is $c = 2 \pi r$. It's nice to solve a problem like this, though, to make sure we're doing it correctly — and this one is a little more complicated. We begin with the formula for a circle of radius r, centered at the origin. (If it wasn't centered at the origin, we could always move it there without changing the circumference, right?).

$$x^2 + y^2 = r^2$$

We need to solve for $y$ to get the functional form, and we can just consider the top of the circle (the + solution) for now:

$$y = \sqrt{r^2 - x^2}$$

Now we'll need the derivative,

$$\frac{dy}{dx} = \frac{1}{2}(r^2 - x^2)^{\frac{1}{2}}(2x)$$

which reduces to:

$$\frac{dy}{dx} = \frac{x}{\sqrt{r^2 - x^2}}$$

Now squaring the derivative gets rid of the radical in the denominator:

$$\left(\frac{dy}{dx}\right)^2 = \frac{x^2}{r^2 - x^2}$$

We'll need to add 1, and we'll do that by using the common denominator, $r^2 - x^2$:

$$\begin{align} \left(\frac{dy}{dx}\right)^2 + 1 &= \frac{x^2 + r^2 - x^2}{r^2 - x^2} \\ &= \frac{r^2}{r^2 - x^2} \end{align}$$

So the integral is

$$L = 4 \int_0^r \sqrt{\frac{r^2}{r^2 - x^2}} dx$$

Here I've taken the integral of the top half of the circle (the + square root) only between x = 0 and x = 4, and just multiplied it by 4 to take advantage of the ample symmetry of a circle.

Curves in parametric form

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For a curve in parametric form, $(x, y) = (x(t), y(t)),$ where $t$ is a parameter, like time, it's not too much of a stretch to derive the arc length. In a similar manner, we can divide the curve into chunks of the parameter, $t_o, \; t_1 \; 2_2,$ and so on,

Practice problems

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1. Calculate the length of the parabola $y = x^2$ between x = -1 and x = 1.

   Solution

   
   

   $y = x^2$ between x = -1 and x = 1. First set up the integral, taking advantage of the symmetry of the function (integrate from 0 to 1 and multiply by 2):

   $$ \begin{align} L &= 2 \int_0^1 \sqrt{1 + \left( \frac{d}{dx}x^2 \right)^2} dx \\[5pt] &= 2 \int_0^1 \sqrt{1 + 4x^2} dx \end{align}$$

   This integral can be done by trig. substitution:

   Let   $x = \frac{1}{2} tan(t),$   then   $x^2 = \frac{1}{4} tan^2(t)$   and   $dx = \frac{1}{2} sec^2(t) dt.$

   Additionally, we can find new integration limits:   $t = tan^{-1}(2x),$   so   $t(0) = tan^{-1}(0) = 0,$   and   $t(1) = tan^{-1}(2) = 1.107.$

   Now we have

   $$ \begin{align} L &= 2 \int_0^{1.107} \sqrt{1 + tan^2(t)} \left( \frac{1}{2} sec^2(t)\right) dt \\[5pt] &= \int_0^{1.107} sec^3(t) dt \end{align}$$

   Now solve that integral by parts,

   $$ \begin{align} Let \: u &= sec(t) \: \: \text{and} \: dv = sec^2(t) dt \\[5pt] \text{then} \: \: du &= sect(t)tan(t) dt, \: \: \text{and} \: \: v = tan(t) \end{align}$$

   
   

   $$ \begin{align} \int sec^3(t) dt &= sec(t)tan(t) - \int sec(t) tan^2(t) dt \\[5pt] \int sec^3(t) dt &= sec(t)tan(t) - \int sec^3(t) dt - \int sec(t) dt \\[5pt] 2 \int sec^3(t) dt &= sec(t)tan(t) - ln|sec(t) + tan(t)| \\[5pt] \int sec^3(t) dt &= \frac{1}{2} \left[ sec(t) tan(t) - ln|sec(t) + tan(t) \right]_0^1.107 \\[5pt] &= \frac{1}{2} \left[ \frac{sin(t)}{cos^2(t)} - ln \big| \frac{1}{cos(t)} - \frac{sin(t)}{cos(t)} \big| \right]_0^1.107 \\[5pt] &= \frac{1}{2} \left[ 4.469 - ln |2.235 - 2| - \left( 0 - ln|1 - 0|) \right) \right] \\[5pt] &= \bf{1.51 \: \text{units}} \end{align}$$




2. Calculate the exact length of the curve defined by   $f(x) = \frac{x^2}{2} - \frac{ln(x)}{4}$   for   $2 \le x \le 4$

   Solution

   
   

   $$f(x) = \frac{x^2}{2} - \frac{ln(x)}{4} \phantom{00} x \in [2, 4]$$

   $$f'(x) = x - \frac{1}{4x}$$

   $$[f'(x)]^2 = x^2 - \frac{1}{2} + \frac{1}{16x2} = \frac{16x^4 - 8 x^2 + 1}{16x^2}$$

   $$ \require{cancel} \begin{align} 1 + [f'(x)]^2 &= \frac{16x^2 + 16x^4 - 8x^2 + 1}{16x^2} \\[5pt] &= \frac{16x^4 + 8x^2 + 1}{16x^2} \\[5pt] &= \frac{\cancel{16}(x^2 + \frac{1}{4})^2}{\cancel{16}x^2} \end{align}$$

   
   

   So:

   $$ \begin{align} L &= \int_2^4 \sqrt{\frac{\left( x^2 + \frac{1}{4} \right)}{x^2}} \, dx \\[5pt] &= \int_2^4 \frac{x + \frac{1}{4}}{x} \, dx \\[5pt] &= \int_2^4 1 + \frac{1}{4x} \, dx = x + \frac{1}{4} ln|x| \bigg|_2^4 \\[5pt] &= 4 + \frac{1}{4} ln(4) - 2 - \frac{1}{4} ln(2) \\[5pt] &= 2 + \frac{1}{4} ln \left( \frac{4}{2} \right) \\[5pt] &= 2 + \frac{1}{4} ln(2) = 2.173 \end{align}$$

   (That's a lot of tricky algebra!)




3. Calculate the length of the curve $y = x^{\frac{3}{2}}$ between (0, 0) and (1, 1)

   Solution

   
   

   $$y = x^{\frac{3}{2}} \phantom{00} (0, 0) \text{ to } (1, 1)$$

   $$ \begin{align} y' = \frac{3}{2} x^{\frac{1}{2}} &\rightarrow [y']^2 = \frac{9}{4} x \\[5pt] &\rightarrow L = \int_0^1 \sqrt{1 + \frac{9}{4} x} \, dx \end{align}$$

   Integrate by parts:

   $$\text{let } u = 1 + \frac{9}{4} x, \; du = \frac{9}{4} \, dx$$

   
   

   $$ \begin{align} L &= \frac{4}{9} \int_1^{13/4} u^{\frac{1}{2}} du = \frac{4}{9} \cdot \frac{2}{3} u^{\frac{3}{2}} \bigg|_1^{13/4} \\[5pt] &= \frac{8}{27} \left[ \left( \frac{13}{4} \right)^{\frac{3}{2}} - 1^{\frac{3}{2}} \right] \\[5pt] &= \frac{8}{27} \left[ \frac{13^{\frac{3}{2}}}{8} - 1 \right] \\[5pt] &= \frac{\cancel{8}}{27} \left[ \frac{13^{3/2}-8}{\cancel{8}} \right] \\[5pt] &= \frac{13^{3/2} - 8}{27} = 1.44 \, \text{units} \end{align}$$




4. Calculate the length of the parametric curve $x = t^2 \: , \: y = t^3$ between (1, 1) and (4, 8).

   Solution

   
   

   This one is parametric:

   $$x = t^2, \; \; y = t^3 \phantom{00} (1, 1) \text{ to } (4, 8)$$

   $$ \begin{align} L &= \int_1^2 \left[ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 \right]^{\frac{1}{2}} \, dt \\[5pt] &= \int_1^2 \left[ (2t)^2 + (3t^2)^2 \right]^{\frac{1}{2}} \, dt \\[5pt] &= \int_1^2 (4t^2 + 9t^4)^{\frac{1}{2}} \, dt \\[5pt] &= \int_1^2 t(4 + 9t^2)^{\frac{1}{2}} \, dt \end{align}$$

   
   

   Integrate by parts:

   $$\text{let } u = 4 + 9t^2, \phantom{00} du = 18t \, dt$$

   $$ \begin{align} \rightarrow &\; \frac{1}{18} \int_{13}^{40} \, u^{\frac{1}{2}} du \\[5pt] &= \frac{1}{18} \cdot \frac{3}{2} u^{\frac{3}{2}} \bigg|_{13}^{40} \\[5pt] &= \frac{1}{12} \left( 40^{\frac{2}{2}} - 13^{\frac{3}{2}} \right) = 17.2 \text{ units} \end{align}$$




5. The polar curve $x = r \cdot cos(\theta), \: y = r \cdot sin(\theta)$ is the parameterization (in terms of the polar angle θ) of a circle of radius r. Use calculus to show that the circumference of such a circle is $C = 2\pi r.$

   Solution

   
   

   $$x = r\cdot cos(\theta), \; y = r \cdot sin(\theta) \phantom{00} \theta \in [0,2\pi]$$

   $$x' = -r \cdot sin(\theta), \phantom{00} y' = r \cdot cos(\theta)$$

   $$(x')^2 = r^2 sin^2(\theta) \phantom{00} (y')^2 = r^2 cos^2(\theta)$$

   $$(x')^2 + (y')^2 = r^2(sin^2(\theta) + cos^2(\theta)) = r^2$$

   $$L = \int_0^{2 \pi} \sqrt{r^2} \, d\theta = r\theta \, \bigg|_0^{2 \pi} = 2 \pi r - 0 = 2 \pi r$$

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Video examples

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I'm working on some examples ... stand by (but do check out the problems above)

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