Blogs, Calculus, Study Help

Arc-length of a function

As part of my Calculus 2 class, I have created a practice exam to help my students study.  In doing so, I felt that providing the questions and the solutions in a blog series would be helpful not just to my students, but also to anyone that may be taking, or teaching, calculus 2.  I hope you find the following solutions helpful in your studies.

You can also view me presenting the solution on YouTube at

Find the length of the curve of \(f(x)=\frac{2}{3}(x+1)^{\frac{3}{2}}\) on the interval \([0,3]\).

We begin by drawing this function on the given interval below.

In order to find the total length, we will again cut cross sections out and add up the lengths of all the cross sections. The cross sections here, will just be line segments along the function.

We now note that the length of the cross section is given by
l_{cs}&=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}} \\
&=\sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^{2}}\Delta x.

Since \(\frac{dy}{dx}=f'(x)\), we can now find that the total length will be

Now, we note that since \(f(x)=\frac{2}{3}(x+1)^{\frac{3}{2}}\), we can use the chain rule to find \(f'(x)\). Here, we let our inside function \(i(x)=x+1\). Then \(o(i)=\frac{2}{3}i^{\frac{3}{2}}\). We then have that
f'(x)&=o'(i)*i'(x) \\
&=i^{\frac{1}{2}}*1 \\

Furthermore, the smallest and largest \(x\) we are interested in are given by the problem as \(x=0\) and \(x=3\), respectively. Therefore,
l_{T}&=\int_{0}^{3}\sqrt{1+\left((x+1)^{\frac{1}{2}}\right)^{2}}dx \\
&=\int_{0}^{3}\sqrt{1+(x+1)}dx \\
&=\int_{0}^{3}\sqrt{2+x}dx \\
&=\frac{2}{3}(2+x)^{\frac{3}{2}}|_{0}^{3} \\
&=\frac{2}{3}(5^{\frac{3}{2}}-2^{\frac{3}{2}}) \\
&=\frac{2}{3}(\sqrt{125}-\sqrt{8}) \\
&\approx 5.57.

Other Resources

I have it linked above, but you can find the practice exam here.  You can find the other solutions to the practice exam in the other posts available here.

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