Question

help please..

find the length of the curve

Answer 1

o-o

Answer 2

let\[y=\frac{ x^4 }{ 16 } +\frac{ 1 }{ 2x^2 }\]

Answer 3

1 </ x </ 2

Answer 4

so far i have this
\[\int\limits_{1}^{2}\sqrt{\frac{ 1 }{ 2 }+\frac{ x^6 }{ 16 }+x ^{-6}} dx\]

Answer 5

i simplified it which my
derivative was
\[\frac{ dy }{ dx }=\frac{ 1 }{ 4 }x^3-x ^{-3}\]

Answer 6

When you squared the y' you ended up with,\[\large\rm (y')^2\quad=\quad \frac{x^6}{16}\color{orangered}{-\frac12}+x^{-6}\]Right?
And now you have a +1/2 in the middle instead of -1/2 when you add the 1.
Remember this little trick?

Answer 7

uhhhh LOOl which trick D:

Answer 8

Your derivative squared gave you this,\[\large\rm \frac{x^6}{16}\color{orangered}{-\frac12}+x^{-6}\quad=\quad \left(\frac{x^3}{4}-x^{-3}\right)^2\]So then how bout this?\[\large\rm \frac{x^6}{16}\color{royalblue}{+\frac12}+x^{-6}\quad=\quad \]

Answer 9

like this D: