# Find the arclength of the function $x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$.

I'm trying to find the arclength of the function $x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$.

I can find the equation for the length pretty easily but I'm looking at thow to solve for the actual length. It looks like a very complex integral so I'm assuming I made a mistep or theres some easy reduction I can make.

After determining the area of an incredibly small section of the function: $$ds = \sqrt{\left(\frac{dx}{dy}\right)^2 + 1}$$ $$\frac{dx}{dy} = \frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2}$$ $$ds = \sqrt{\left(\frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2}\right)^2 + 1}$$ This leaves me with the integral $$\int{\sqrt{\left(\frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2}\right)^2 + 1}} dy$$ I do still have to calculate from 1 - 25 but I like to plug in after I solve my integral. Anyway, I can't tell how to solve this, but I have a feeling I need to play with the squared term. Perhaps a substitution or maybe the reciprocals simplify into something. If anyone has any tips I appreciate it!

EDIT: Problem solved! (I think)

In the answer to the question I can make the simplification $$\int{\sqrt{\left(\frac{y^{\frac{5}{2}} + y^{\frac{-5}{2}}}{2}\right)^2}} dy$$ $$\int{\left(\frac{y^{\frac{5}{2}} + y^{\frac{-5}{2}}}{2}\right)} dy$$ $$\frac{1}{2}\left(\int{y^{\frac{5}{2}}}dy + \int{y^{\frac{-5}{2}}}dy\right)$$ $$\frac{y^{\frac{7}{2}}}{7} + \frac{1}{-3y^{\frac{3}{2}}}$$

We plug in our bounds here and the answer is $F(25) - F(1)$

• You can use \left( and \right) to make the parentheses automatically tall enough to enclose the whole fraction, integral, radical, etc. – gen-z ready to perish Oct 4 '17 at 17:09

HINT:

$$1+\left(\frac{y^{5/2}-y^{-5/2}}{2}\right)^2=\left(\frac{y^{5/2}+y^{-5/2}}2\right)^2$$

• @MyGlasses Thank you! – Mark Viola Oct 4 '17 at 16:45
• Wow! this deflates the problem like a balloon. Thank you. – Howard P Oct 4 '17 at 16:58
• How did you arrive at this? – gen-z ready to perish Oct 4 '17 at 17:08
• @chaseryantaylor I applied simple arithmetic. – Mark Viola Oct 4 '17 at 17:11
• @MarkViola I put the answer using this step in the question, could you verify it? – Howard P Oct 4 '17 at 17:15

Hint: Let $y=e^{\frac25t}$ and use $\sinh^2t+1=\cosh^2t$.

note that $$\left(\frac{-1}{2}y^{-5/2}+\frac{1}{2}y^{5/2}\right)^2+1=1/4\,{\frac { \left( y+1 \right) ^{2} \left( {y}^{4}-{y}^{3}+{y}^{2}-y +1 \right) ^{2}}{{y}^{5}}}$$

$$\int \sqrt{1+(x'(y))^2}\,dy=\int \sqrt{1+\left(\frac{y^{5/2}}{2}-\frac{1}{2 y^{5/2}}\right)^2}\,dy=\int\sqrt{\frac{1+\left(y^5-1\right)^2}{4 y^5}}\,dy=$$ $$=\int\frac{1}{2} \sqrt{\frac{\left(y^5+1\right)^2}{y^5}}\,dy=\frac12\int\frac{y^5+1}{\sqrt{y^5}}\,dy=\frac12\int\left(y^{-5/2}+y^{5/2}\right)\,dy=\frac{y^{7/2}}{7}-3 y^{-3/2}+C$$

The arc length is $$\left[\frac{y^{7/2}}{7}-3 y^{-3/2}\right]_1^{25}=\frac{29297368}{2625}\approx 11160.9$$