Find the length of the arc of the curve
$$x= \frac{y^5}{5} + \frac{1}{12y^3}$$ over the interval $[2,4]$.
I know I need to find $$\int_{2}^{4}\sqrt{ 1+\Big(\frac{dx}{dy}\Big)^2} dy$$ because that's the 'formula' to find arc length. I want to make sure I am starting the problem off correctly by checking if I did the derivative correctly: $$\frac{dx}{dy}=y^4 – \frac{1}{4y^4}.$$ If that is correct then I have to square $dy/dx$ and when I did this I got
$$\Big(\frac{dx}{dy}\Big)^2 = y^8 – \frac{1}{2} + \frac{1}{16y^8}.$$ Then I plugged this into the formula above and got
$$\int_{2}^{4}\sqrt{y^8+\frac{1}{2}+\frac{1}{16y^2}} \, dy$$ which is a perfect square. So now I have $$\int_{2}^{4}\sqrt{\Big(y^4+\frac{1}{4y^4}\Big)^2} dy$$ and this leaves me with $$\int_{2}^{4} \Big(y^4+\frac{1}{4y^4}\Big) \,dy.$$ When I integrate this i got y^5/5-1/12y^3 invaluated at 4 and 2. once plugging in the 4 and then the 2 i got the answer 992/5 – 1/1500 + 1/96. is this correct?
[Math] Find the length of the arc of the curve $x= y^5/5 + 1/(12y^3)$ over $[2,4]$
calculusintegration
Best Answer
Just a word of the wise: I don't know what went through your head, but remember that your final equation should be equal to the absolute value of the previous. This means you should probably be looking for points of intersection with the y-axis, but you might have just seen the equation and thought "Oh, this is a fourth degree equation, so the result will always be positive," but I can't really tell.
Other than that, if you just evaluated the integral at 4 and 2 and subtracted the two numbers, you should be golden :)