## 17Calculus Integrals - Arc Length

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This page covers the topic of arc length of an explicitly defined smooth curve in the xy-plane in cartesian (rectangular) coordinates. [Arc length can also be calculated in polar coordinates and for parametric curves.]

Setting Up The Integral

Setting up these integrals is very straight-forward and do not require any new techniques. However, the difficulty comes in evaluating them. We will show you a few tricks to evaluating them but first let's talk about how to set them up.

There are two sets of equations, both of which accomplish the same purpose. The reason you need two of them is because sometimes the equations describing the curve are in a form that require one or the other. Additionally, you may be able to set up both integrals but often it is possible to evaluate only one of them.

Here is a video clip deriving the integral in terms of x. This will help you understand where these equations come from.

### MIP4U - Arc Length - Part 1 of 2 [3min-49secs]

video by MIP4U

 $$\displaystyle{s = \int_{a}^{b}{\sqrt{1 + [f'(x)]^2}dx}}$$ To determine the arc length of a smooth curve defined by $$y = f(x)$$ between the points $$x=a$$ and $$x=b$$, we can use the integral where $$\displaystyle{ f'(x) = \frac{dy}{dx} }$$
 $$\displaystyle{s = \int_{c}^{d}{\sqrt{1 + [g'(y)]^2}dy}}$$ When the curve is given by an equation in the form $$x=g(y)$$, we need to determine the endpoints, $$y=c$$ and $$y=d$$ on the y-axis and use the integral where $$\displaystyle{ g'(y) = \frac{dx}{dy} }$$

Okay, let's watch a quick video clip going through these equations again.

### PatrickJMT - Arc Length [1min-35secs]

video by PatrickJMT

Notes:
1. Notice that we use a small s to represent the arc length. Later, we will use a capital S to represent surface area. Many, if not most, mathematicians follow this standard.
2. Sometimes these equations are represented a little differently. We may write the arc length integral as $$s=\int{ds}$$, where $$ds$$ is defined below. When you study surface area, you will see how these are used.

 $$ds = \sqrt{1 + [f'(x)]^2} ~dx$$ $$ds = \sqrt{1 + [g'(y)]^2} ~dy$$

Tricks To Evaluate Integrals

Here are some unusual situation you may encounter when evaluating integrals that have not been covered elsewhere. We will show some specific examples but we will not evaluate the integrals completely. We will just get them into a form that can be evaluated with other techniques that you are already know, like substitution or parts.

Perfect Square

Okay, so I know that you already know about perfect squares but they show up a lot in these problems, so I thought I would remind you. Notice in the integral, we have $$\sqrt{1+[f'(x)]^2}$$.   After expanding the squared derivative and adding one, the easiest problems end up with a perfect square under the square root. So, even with complicated expressions, it is usually a good use of your time to check for a perfect square first, even if it isn't obvious. Here is an example. After expanding the squared term and adding one, we have $$\displaystyle{\sqrt{ x^2 + \frac{1}{2} + \frac{1}{16x^2}}}$$

We can try to set up a perfect square using the square root of the first and last term and see if we get the middle term under the square root when we expand. So we investigate $$\displaystyle{ x + \frac{1}{4x} }$$. When we square this, sure enough we get $$1/2$$ as the middle term. So we can simplify the equation as follows
$$\displaystyle{\sqrt{ x^2 + \frac{1}{2} + \frac{1}{16x^2} } = }$$ $$\displaystyle{ \sqrt{ \left[ x + \frac{1}{4x} \right]^2 } = }$$ $$\displaystyle{ x + \frac{1}{4x} }$$
This last form is easily integrable with the techniques you know.

Getting A Common Denominator

Okay, so you have determined that you do not have a perfect square under the square root. Your next step might be to combine all terms into one fraction and then take the square root of the numerator and denominator separately. This will often allow you to move some terms outside of the square root and make the square root easier and may lead to integration by substitution or trig substitution. Here is an example.

Let's say we have an integral with the integrand $$\displaystyle{ \sqrt{ 1+\frac{y^2-2y+1}{4y} } }$$. In order to simplify this so that we can integrate it, we need to combine the one with the fraction. Doing this we get $$\displaystyle{ \sqrt{ \frac{y^2+2y+1}{4y} } }$$. Now, the numerator is a perfect square, so we factor it and simplify to get $$\displaystyle{ \frac{ \sqrt{ (y+1)^2 } }{\sqrt{4y} } =\frac{y+1}{2\sqrt{y}}}$$. So, now we have an integrand that can be integrated with basic techniques.

Practice

Unless otherwise instructed, calculate the arc length. Give all answers in exact form.
Note: Although some of these problems are very similar, they are worked by different instructors. So going through each solution will give you a broader point of view and help you better understand how to work these types problems.

$$y = 3x+2$$; $$-1 \leq x \leq 2$$

Problem Statement

Calculate the length of the arc $$y = 3x+2$$ for $$-1 \leq x \leq 2$$. Give your answer in exact form.

$$s = 3\sqrt{10}$$

Problem Statement

Calculate the length of the arc $$y = 3x+2$$ for $$-1 \leq x \leq 2$$. Give your answer in exact form.

Solution built with GeoGebra

This is a pretty easy one where the 'arc' is a straight line, shown in the plot. We can therefore calculate the arc length directly using geometry to check our answer to the integral solution. Let's work the integral first using $$\displaystyle{s = \int_{a}^{b}{ \sqrt{1 + [f'(x)]^2} ~dx } }$$

 $$f(x)=3x+2 \to f'(x)=3$$ $$\displaystyle{s = \int_{-1}^{2}{ \sqrt{1 + ^2} ~dx } }$$ $$s = [ x\sqrt{10} ]_{-1}^{2} = 3\sqrt{10}$$

Double Check
Now, let's check our answer using the Pythagorean Theorem. The x-length is $$2-(-1) = 3$$. The y-length is $$8-(-1) = 9$$. So the length of the hypotenuse is $$s = \sqrt{ 3^2 + 9^2} = \sqrt{ 9 + 81 } =$$ $$\sqrt{90} = \sqrt{9(10)} = 3\sqrt{10}$$, which confirms our answer from the integral.

Triple Check
Although we were not asked to in the problem, let's evaluate the integral in the other direction, i.e. with respect to y using the integral $$\displaystyle{s = \int_{c}^{d}{ \sqrt{1 + [g'(y)]^2} ~dy } }$$
We should get the same answer.

 $$f(x)=3x+2 \to$$ $$g(y)=(y-2)/3 \to g'(y)=1/3$$ $$\displaystyle{s = \int_{-1}^{8}{ \sqrt{1 + [1/3]^2} ~dx } }$$ $$s = [ (y/3)\sqrt{10} ]_{-1}^{8}$$ $$(8/3)\sqrt{10} - (-1/3)\sqrt{10} = 3\sqrt{10}$$

Note - This is one of the very few times that we can evaluate the integral in both the x and y-directions. Most of the time, we will get an integral in one of the directions that cannot be evaluated.

$$s = 3\sqrt{10}$$

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$$\displaystyle{ y = \frac{1}{3}(x^2+2)^{3/2} }$$; $$0 \leq x \leq 1$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ y = \frac{1}{3}(x^2+2)^{3/2} }$$ for $$0 \leq x \leq 1$$. Give your answer in exact form.

Solution

### 1188 video

video by PatrickJMT

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$$\displaystyle{ y = \frac{x^2}{2} - \frac{\ln(x)}{4} }$$; $$2 \leq x \leq 4$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ y = \frac{x^2}{2} - \frac{\ln(x)}{4} }$$ for $$2 \leq x \leq 4$$. Give your answer in exact form.

Solution

### 1189 video

video by PatrickJMT

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$$\displaystyle{y = \frac{e^x + e^{-x}}{2} }$$; $$0 \leq x \leq \ln 2$$

Problem Statement

Calculate the length of the arc $$\displaystyle{y = \frac{e^x + e^{-x}}{2} }$$ for $$0 \leq x \leq \ln 2$$. Give your answer in exact form.

3/4

Problem Statement

Calculate the length of the arc $$\displaystyle{y = \frac{e^x + e^{-x}}{2} }$$ for $$0 \leq x \leq \ln 2$$. Give your answer in exact form.

Solution

Since the curve is given in the form $$y=f(x)$$, we will use the equation $$\displaystyle{s = \int_{a}^{b}{\sqrt{1 + [f'(x)]^2} ~dx}}$$ to calculate the arc length.

 $$\displaystyle{y = \frac{e^x + e^{-x}}{2} \to y' = \frac{e^x - e^{-x}}{2} }$$ $$\displaystyle{s = \int_{0}^{\ln2}{\sqrt{1 + [y']^2} ~dx} }$$ $$\displaystyle{s = \int_{0}^{\ln2}{\sqrt{1 + \left[ \frac{e^x - e^{-x}}{2} \right]^2 } ~dx} }$$ $$\displaystyle{s = \int_{0}^{\ln2}{\sqrt{\frac{4}{4} + \left[ \frac{(e^x - e^{-x})^2}{4} \right] } ~dx} }$$ $$\displaystyle{s = \frac{1}{2} \int_{0}^{\ln2}{\sqrt{ 4 + e^{2x} - 2 + e^{-2x} } ~dx } }$$ $$\displaystyle{s = \frac{1}{2} \int_{0}^{\ln2}{\sqrt{ e^{2x} + 2 + e^{-2x} } ~dx } }$$ $$\displaystyle{s = \frac{1}{2} \int_{0}^{\ln2}{\sqrt{ [ e^x + e^{-x} ]^2 } ~dx } }$$ $$\displaystyle{s = \frac{1}{2} \int_{0}^{\ln2}{e^x + e^{-x} ~dx } }$$ $$\displaystyle{s = \frac{1}{2} [ e^x - e^{-x} ]_0^{\ln 2} }$$ $$\displaystyle{s = \frac{1}{2} [ e^{\ln 2} - e^{-\ln 2} - 1 + 1 ] }$$ $$\displaystyle{s = \frac{1}{2} [ 2 - 1/2 ] }$$ $$\displaystyle{s = \frac{1}{2} [ 3/2 ] = 3/4 }$$

3/4

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$$y = x^{3/2}$$; $$0 \leq x \leq 4$$

Problem Statement

Calculate the length of the arc $$y = x^{3/2}$$ for $$0 \leq x \leq 4$$. Give your answer in exact form.

$$(8/27)(10^{3/2}-1)$$

Problem Statement

Calculate the length of the arc $$y = x^{3/2}$$ for $$0 \leq x \leq 4$$. Give your answer in exact form.

Solution

### 1191 video

video by MIT OCW

$$(8/27)(10^{3/2}-1)$$

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$$y = \ln(\cos x)$$; $$0 \leq x \leq \pi/3$$

Problem Statement

Calculate the length of the arc $$y = \ln(\cos x)$$ for $$0 \leq x \leq \pi/3$$. Give your answer in exact form.

$$\ln(2 + \sqrt{3})$$

Problem Statement

Calculate the length of the arc $$y = \ln(\cos x)$$ for $$0 \leq x \leq \pi/3$$. Give your answer in exact form.

Solution

Since the curve is given in the form $$y=f(x)$$, we will use the equation $$\displaystyle{s = \int_{a}^{b}{\sqrt{1 + [f'(x)]^2} ~dx}}$$ to calculate the arc length.

 $$\displaystyle{y = \ln(\cos x) \to }$$ $$\displaystyle{ y' = \frac{1}{\cos x} (-\sin x) = \frac{-\sin x}{\cos x} }$$ $$\displaystyle{s = \int_{0}^{\pi/3}{\sqrt{1 + \left[ \frac{-\sin x}{\cos x} \right]^2 } ~dx } }$$ $$\displaystyle{s = \int_{0}^{\pi/3}{\sqrt{\frac{\cos^2 x}{\cos^2 x} + \frac{\sin^2 x}{\cos^2 x} } ~dx } }$$ $$\displaystyle{s = \int_{0}^{\pi/3}{\sqrt{\frac{\cos^2 x + \sin^2 x}{\cos^2 x} } ~dx } }$$ $$\displaystyle{s = \int_{0}^{\pi/3}{\frac{1}{\cos x} ~dx } }$$ $$\displaystyle{s = \int_{0}^{\pi/3}{\sec x ~dx } }$$ $$\ln |\sec x + \tan x |_{0}^{\pi/3}$$ $$\ln |\sec(\pi/3) + \tan(\pi/3) | - \ln |\sec(0) + \tan(0) |$$ $$\ln |2 + \sqrt{3}|$$

$$\ln(2 + \sqrt{3})$$

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$$\displaystyle{ y = \frac{3}{2} x^{2/3} }$$; $$0 \leq x \leq 1$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ y = \frac{3}{2} x^{2/3} }$$ for $$0 \leq x \leq 1$$. Give your answer in exact form.

$$2^{3/2} - 1$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ y = \frac{3}{2} x^{2/3} }$$ for $$0 \leq x \leq 1$$. Give your answer in exact form.

Solution built with GeoGebra

We will use the integral $$\displaystyle{s = \int_{a}^{b}{\sqrt{1 + [f'(x)]^2} ~dx} }$$ where $$f(x)=(3/2)x^{2/3}$$

 $$f'(x)=(3/2)(2/3)x^{-1/3} = x^{-1/3}$$ $$\displaystyle{s = \int_{0}^{1}{\sqrt{1 + [x^{-1/3}]^2} ~dx} }$$ $$\displaystyle{s = \int_{0}^{1}{\sqrt{1 + \frac{1}{x^{2/3}}} ~dx} }$$ $$\displaystyle{s = \int_{0}^{1}{\sqrt{ \frac{x^{2/3}+1}{x^{2/3}} } ~dx} }$$ $$\displaystyle{s = \int_{0}^{1}{\frac{\sqrt{x^{2/3}+1}}{x^{1/3}} ~dx } }$$ Use substitution with $$u = x^{2/3} + 1 \to du = (2/3)x^{-1/3} ~dx$$ Convert the limits of integration. lower limit: $$x=0 \to u=1$$ upper limit: $$x=1 \to u=2$$ $$\displaystyle{s = \int_{1}^{2}{(3/2) u^{1/2} ~du } }$$ $$\displaystyle{s = \left[ (3/2)\frac{u^{3/2}}{3/2} \right]_1^2 }$$ $$\displaystyle{2^{3/2} - 1^{3/2} }$$

$$2^{3/2} - 1$$

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$$\displaystyle{y = 3 + \frac{4}{5}x^{5/4} }$$; $$0 \leq x \leq 1$$

Problem Statement

Calculate the length of the arc $$\displaystyle{y = 3 + \frac{4}{5}x^{5/4} }$$ for $$0 \leq x \leq 1$$. Give your answer in exact form.

$$\displaystyle{\frac{8}{15}(1+\sqrt{2}) }$$

Problem Statement

Calculate the length of the arc $$\displaystyle{y = 3 + \frac{4}{5}x^{5/4} }$$ for $$0 \leq x \leq 1$$. Give your answer in exact form.

Solution

 $$\displaystyle{y = 3 + \frac{4}{5}x^{5/4} \to y' = x^{1/4} }$$ $$\displaystyle{\int_0^1{\sqrt{1+x^{1/2}} ~dx } }$$ $$u = 1+x^{1/2} \to du = (1/2)x^{-1/2} ~dx$$ $$dx = 2x^{1/2} ~du$$ $$\displaystyle{\int_{x=0}^{x=1}{u^{1/2} (2x^{1/2} ~du) } }$$ Before we can integrate, we need to do two things. First, we need to convert the $$x^{1/2}$$ term in the integrand to a term involving u. Second, we need to convert the limits of integration in terms of u. Since we set $$u = 1+x^{1/2}$$, we can solve this for $$x^{1/2}$$, giving us $$x^{1/2} = u-1$$. Lower limit: $$x=0 \to u=1$$ Upper limit: $$x=1 \to u=2$$ Now we can write the integral completely in terms of u, which allows us to integrate. $$\displaystyle{\int_1^2{2 u^{1/2} (u-1) ~ du } }$$ $$\displaystyle{2 \int_1^2{u^{3/2} - u^{1/2} ~ du } }$$ $$\displaystyle{2 \left[ \frac{u^{5/2}}{5/2} - \frac{u^{3/2}}{3/2} \right]_1^2 }$$ Normally, we would evaluate with the limits of integration at this point. But the fractions look like they could get messy. So let's simplify those first. $$\displaystyle{2 \left[ \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} \right]_1^2 }$$ $$\displaystyle{2 \left[ \frac{6}{15}u^{5/2} - \frac{10}{15}u^{3/2} \right]_1^2 }$$ $$\displaystyle{\frac{4}{15} \left[ 3u^{5/2} - 5u^{3/2} \right]_1^2 }$$ $$\displaystyle{\frac{4}{15} [ ( 3(2)^{5/2} - 5(2)^{3/2} ) - ( 3 - 5 ) ] }$$ $$\displaystyle{\frac{4}{15} [ 12\sqrt{2} - 10\sqrt{2} + 2 ] }$$ $$\displaystyle{\frac{8}{15} [ \sqrt{2} + 1 ] }$$

$$\displaystyle{\frac{8}{15}(1+\sqrt{2}) }$$

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$$\displaystyle{y = \frac{x^4}{4}+\frac{1}{8x^2} }$$; $$2 \leq x \leq 4$$

Problem Statement

Calculate the length of the arc $$\displaystyle{y = \frac{x^4}{4}+\frac{1}{8x^2} }$$ for $$2 \leq x \leq 4$$. Give your answer in exact form.

$$\displaystyle{\frac{7683}{128} }$$

Problem Statement

Calculate the length of the arc $$\displaystyle{y = \frac{x^4}{4}+\frac{1}{8x^2} }$$ for $$2 \leq x \leq 4$$. Give your answer in exact form.

Solution built with GeoGebra

 $$\displaystyle{\int_2^4{\sqrt{1+[y']^2} ~dx } }$$ $$\displaystyle{y = \frac{x^4}{4}+\frac{1}{8x^2} \to y' = x^3-\frac{x^{-3}}{4} }$$ $$\displaystyle{\int_2^4{\sqrt{ 1 + \left[ x^3 - \frac{1}{4x^3} \right]^2 } ~dx } }$$ $$\displaystyle{\int_2^4{\sqrt{ 1 + \left[ \frac{4x^6-1}{4x^3} \right]^2 } ~dx } }$$ $$\displaystyle{\int_2^4{\sqrt{ \frac{16x^6}{16x^6} + \frac{(4x^6-1)^2}{16x^6} } ~dx } }$$ $$\displaystyle{\int_2^4{\sqrt{ \frac{16x^6 +16x^{12}-8x^6+1}{16x^6} } ~dx } }$$ $$\displaystyle{\int_2^4{\sqrt{ \frac{16x^{12}+8x^6+1}{16x^6} } ~dx } }$$ $$\displaystyle{\int_2^4{\sqrt{ \frac{(4x^6+1)^2}{(4x^3)^2} } ~dx } }$$ $$\displaystyle{\int_2^4{\frac{4x^6+1}{4x^3} ~dx } }$$ $$\displaystyle{\int_2^4{\frac{4x^6}{4x^3} + \frac{1}{4x^3} ~dx } }$$ $$\displaystyle{\int_2^4{x^3 + \frac{1}{4}x^{-3} ~dx } }$$ $$\displaystyle{\left[ \frac{x^4}{4} + \frac{1}{4} \frac{x^{-2}}{-2} \right]_2^4 }$$ $$\displaystyle{\frac{1}{4} \left[ 4^4 - 2^4 - \frac{( 4^{-2} - 2^{-2} )}{2} \right] }$$ $$\displaystyle{\frac{1}{4} \left[ 256 - 16 - \frac{1}{32} + \frac{1}{8} \right] }$$ $$\displaystyle{\frac{1}{4} \left[ 240 + \frac{3}{32} \right] }$$ $$\displaystyle{\frac{1}{128} \left[ 7680 + 3 \right] }$$ $$\displaystyle{\frac{7683}{128} }$$

$$\displaystyle{\frac{7683}{128} }$$

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$$\displaystyle{y = \frac{x^3}{6} + \frac{1}{2x} }$$; $$1 \leq x \leq 3$$

Problem Statement

Calculate the length of the arc $$\displaystyle{y = \frac{x^3}{6} + \frac{1}{2x} }$$ for $$1 \leq x \leq 3$$. Give your answer in exact form.

$$\displaystyle{\frac{14}{3} }$$

Problem Statement

Calculate the length of the arc $$\displaystyle{y = \frac{x^3}{6} + \frac{1}{2x} }$$ for $$1 \leq x \leq 3$$. Give your answer in exact form.

Solution

 $$\displaystyle{y = \frac{x^3}{6} + \frac{1}{2x} \to y' = \frac{x^2}{2} - \frac{1}{2x^2} = \frac{x^4-1}{2x^2} }$$ $$\displaystyle{\int_1^3{\sqrt{1 + \left[ \frac{x^4-1}{2x^2} \right]^2 } ~dx } }$$ $$\displaystyle{\int_1^3{\sqrt{\frac{4x^4 + (x^4-1)^2}{4x^4} } ~dx } }$$ $$\displaystyle{\int_1^3{\sqrt{\frac{4x^4 + x^8 - 2x^4+1}{4x^4} } ~dx } }$$ $$\displaystyle{\int_1^3{\sqrt{\frac{x^8 + 2x^4 + 1}{4x^4} } ~dx } }$$ $$\displaystyle{\int_1^3{\sqrt{\frac{(x^4+1)^2}{(2x^2)^2} } ~dx } }$$ $$\displaystyle{\int_1^3{\frac{x^4+1}{2x^2} ~dx } }$$ $$\displaystyle{\frac{1}{2} \int_1^3{x^2 + x^{-2} ~dx } }$$ $$\displaystyle{\frac{1}{2} \left[\frac{x^3}{3} + \frac{x^{-1}}{-1} \right]_1^3 }$$ $$\displaystyle{\frac{1}{2} \left[\frac{x^3}{3} - \frac{1}{x} \right]_1^3 }$$ $$\displaystyle{\frac{1}{2} \left[\frac{3^3}{3} - \frac{1}{3} - \frac{1}{3} + 1 \right] }$$ $$\displaystyle{\frac{1}{2} \left[10 - \frac{2}{3} \right] }$$ $$\displaystyle{\frac{1}{2} \left[\frac{30}{3} - \frac{2}{3} \right] }$$ $$\displaystyle{\frac{1}{2} \left[\frac{28}{3} \right] = \frac{14}{3} }$$

$$\displaystyle{\frac{14}{3} }$$

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$$\displaystyle{ y = \frac{2}{3}(x^2+1)^{3/2} }$$; $$0 \leq x \leq 2$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ y = \frac{2}{3}(x^2+1)^{3/2} }$$ for $$0 \leq x \leq 2$$. Give your answer in exact form.

Solution

For the written out solution to this problem by a different instructor, see this page.

### 1192 video

video by Krista King Math

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$$\displaystyle{ f(x) = \frac{1}{3}x^{3/2} }$$; $$[0,4]$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ f(x) = \frac{1}{3}x^{3/2} }$$ for $$[0,4]$$. Give your answer in exact form.

Solution

### 1193 video

video by MIP4U

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$$\displaystyle{ f(x) = \frac{x^2}{8} - \ln(x) }$$; $$[1,e]$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ f(x) = \frac{x^2}{8} - \ln(x) }$$ for $$[1,e]$$. Give your answer in exact form.

Solution

### 1194 video

video by MIP4U

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$$\displaystyle{ x = \frac{1}{3} \sqrt{y}(y-3) }$$; $$1 \leq y \leq 9$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ x = \frac{1}{3} \sqrt{y}(y-3) }$$ for $$1 \leq y \leq 9$$. Give your answer in exact form.

$$32/3$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ x = \frac{1}{3} \sqrt{y}(y-3) }$$ for $$1 \leq y \leq 9$$. Give your answer in exact form.

Solution

### 1187 video

video by Krista King Math

$$32/3$$

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$$\displaystyle{ x = \frac{y^4}{8} + \frac{1}{4y^2} }$$; $$1 \leq y \leq 2$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ x = \frac{y^4}{8} + \frac{1}{4y^2} }$$ for $$1 \leq y \leq 2$$. Give your answer in exact form.

$$33/16$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ x = \frac{y^4}{8} + \frac{1}{4y^2} }$$ for $$1 \leq y \leq 2$$. Give your answer in exact form.

Solution

Since the curve is given in the form $$x=g(y)$$, we will use the equation $$\displaystyle{ s = \int_{c}^{d}{ \sqrt{ 1 + [g'(y)]^2 } ~dy } }$$ to calculate the arc length.

 $$\displaystyle{ \frac{dx}{dy} = g'(y) = \frac{y^3}{2} + \frac{1}{2}(-y^{-3}) }$$ $$\displaystyle{ s = \int_{1}^{2}{ \sqrt{ \frac{4}{4} + \frac{[ y^3 - y^{-3} ]^2}{4} } ~dy } }$$ $$\displaystyle{ s = \frac{1}{2}\int_{1}^{2}{ \sqrt{ 4 + y^6 - 2 + y^{-6} } ~dy } }$$ $$\displaystyle{ s = \frac{1}{2}\int_{1}^{2}{ \sqrt{ y^6 + 2 + y^{-6} } ~dy } }$$ $$\displaystyle{ s = \frac{1}{2}\int_{1}^{2}{ \sqrt{ [ y^3 + y^{-3} ]^2 } ~dy } }$$ $$\displaystyle{ s = \frac{1}{2}\int_{1}^{2}{ y^3 + y^{-3} ~dy } }$$ $$\displaystyle{ s = \frac{1}{2}\left[ \frac{y^4}{4} + \frac{y^{-2}}{-2} \right]_1^2 }$$ $$\displaystyle{ s = \frac{1}{2}\left[ \frac{2^4}{4} - \frac{2^{-2}}{2} - \frac{1}{4} + \frac{1}{2} \right] }$$ $$\displaystyle{ s = 2 - \frac{1}{16} + \frac{1}{8} }$$ $$\displaystyle{ s = \frac{32}{16} - \frac{1}{16} + \frac{2}{16} = \frac{33}{16} }$$

$$33/16$$

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$$\displaystyle{x = \frac{y^5}{5} + \frac{1}{12y^3} }$$; $$1/2 \leq y \leq 1$$

Problem Statement

Calculate the length of the arc $$\displaystyle{x = \frac{y^5}{5} + \frac{1}{12y^3} }$$ for $$1/2 \leq y \leq 1$$. Give your answer in exact form.

$$373/480$$

Problem Statement

Calculate the length of the arc $$\displaystyle{x = \frac{y^5}{5} + \frac{1}{12y^3} }$$ for $$1/2 \leq y \leq 1$$. Give your answer in exact form.

Solution

Since the curve is given in the form $$x=g(y)$$, we will use the equation $$\displaystyle{s = \int_{c}^{d}{\sqrt{1 + [g'(y)]^2 } ~dy } }$$ to calculate the arc length.
First, let\'s set up the integral.

 $$\displaystyle{\frac{dx}{dy} = g'(y) = y^4 + \frac{1}{12}(-3)y^{-4} = y^4 - \frac{y^{-4}}{4} }$$ $$\displaystyle{s = \int_{1/2}^{1}{\sqrt{1 + \left[ y^4 - \frac{y^{-4}}{4} \right]^2 } ~dy } }$$ Now, let's simplify the term under the square root and integrate. $$\displaystyle{s = \int_{1/2}^{1}{\sqrt{1 + y^8 - 2\frac{1}{4} + \frac{y^{-8}}{16} } ~dy } }$$ $$\displaystyle{s = \int_{1/2}^{1}{\sqrt{y^8 + \frac{1}{2} + \frac{y^{-8}}{16} } ~dy } }$$ $$\displaystyle{s = \int_{1/2}^{1}{\sqrt{\frac{16y^8}{16} + \frac{8}{16} + \frac{y^{-8}}{16} } ~dy } }$$ $$\displaystyle{s = \frac{1}{4} \int_{1/2}^{1}{\sqrt{16y^8 + 8 + y^{-8} } ~dy } }$$ $$\displaystyle{s = \frac{1}{4} \int_{1/2}^{1}{\sqrt{(4y^4 + y^{-4})^2 } ~dy } }$$ $$\displaystyle{s = \frac{1}{4} \int_{1/2}^{1}{4y^4 + y^{-4} ~dy } }$$ $$\displaystyle{s = \frac{1}{4} \left[ \frac{4y^5}{5} + \frac{y^{-3}}{-3} \right]_{1/2}^1 }$$ $$\displaystyle{s = \frac{1}{4} \left[ \frac{4}{5} - \frac{1}{3} - \frac{4(1/2)^5}{5} + \frac{(1/2)^{-3}}{3} \right] }$$ $$\displaystyle{s = \frac{1}{4} \left[ \frac{4}{5} - \frac{1}{3} - \frac{1}{40} + \frac{8}{3} \right] }$$ $$\displaystyle{s = \frac{1}{4} \left[ \frac{4}{5} - \frac{1}{40} + \frac{7}{3} \right] }$$ $$\displaystyle{s = \frac{1}{4} \left[ \frac{96}{120} - \frac{3}{120} + \frac{280}{120} \right] }$$ $$\displaystyle{s = \frac{1}{4} \left[ \frac{373}{120} \right] }$$ $$\displaystyle{s = \frac{373}{480} }$$

$$373/480$$

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$$\displaystyle{ x = \ln y - \frac{y^2}{8} }$$; $$1 \leq y \leq 2$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ x = \ln y - \frac{y^2}{8} }$$ for $$1 \leq y \leq 2$$. Give your answer in exact form.

$$3/8 + \ln 2$$

Problem Statement

Calculate the length of the arc $$\displaystyle{ x = \ln y - \frac{y^2}{8} }$$ for $$1 \leq y \leq 2$$. Give your answer in exact form.

Solution

Since the curve is given in the form $$x=g(y)$$, we will use the equation $$\displaystyle{ s = \int_{c}^{d}{ \sqrt{ 1 + [g'(y)]^2 } ~dy } }$$ to calculate the arc length.

 First, let's set up the integral. $$\displaystyle{ \frac{dx}{dy} = g'(y) = \frac{1}{y} - \frac{1}{8}(2y) = \frac{1}{y} - \frac{y}{4} }$$ $$\displaystyle{ s = \int_{1}^{2}{ \sqrt{ 1 + \left[ \frac{1}{y} - \frac{y}{4} \right]^2 } ~dy } }$$ Now simplify the integrand. $$\displaystyle{s = \int_{1}^{2}{\sqrt{1 + \frac{1}{y^2} + \frac{y^2}{16} - \frac{1}{2} } ~dy } }$$ $$\displaystyle{s = \int_{1}^{2}{\sqrt{\frac{1}{y^2} + \frac{y^2}{16} + \frac{1}{2} } ~dy } }$$ $$\displaystyle{s = \int_{1}^{2}{\sqrt{\frac{16}{16y^2} + \frac{y^4}{16y^2} + \frac{8y^2}{16y^2} } ~dy } }$$ $$\displaystyle{s = \int_{1}^{2}{\sqrt{\frac{16 + y^4 + 8y^2}{16y^2} } ~dy } }$$ $$\displaystyle{s = \int_{1}^{2}{\sqrt{\frac{(y^2+4)^2}{16y^2} } ~dy } }$$ $$\displaystyle{s = \int_{1}^{2}{\frac{(y^2+4)}{4y} ~dy } }$$ $$\displaystyle{s = \frac{1}{4} \int_{1}^{2}{y + \frac{4}{y} ~dy } }$$ Now we can integrate. $$\displaystyle{s = \frac{1}{4} \left[ \frac{y^2}{2} + 4\ln y \right]_1^2 }$$ $$\displaystyle{s = \frac{1}{4} \left[ \frac{2^2}{2} + 4\ln 2 - \frac{1^2}{2} - 4\ln 1 \right] }$$ $$\displaystyle{s = \frac{1}{2} + \ln 2 - \frac{1}{8} - 0 }$$ $$\displaystyle{s = \frac{3}{8} + \ln 2 }$$

$$3/8 + \ln 2$$

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You CAN Ace Calculus

related topics on other pages

For arc length of parametric curves, see the parametrics calculus page.

For the arc length of polar curves, see the polar calculus page.

Wikipedia - Arc Length

Pauls Online Notes - Arc Length

UC Davis Duane Kouba - Arc Length

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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Single Variable Calculus

Multi-Variable Calculus

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Precalculus

Engineering

Circuits

Semiconductors

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