Question

f(x)=([(x^4)/4] + [1/(8x^2)] on the interval 1,2
find the arc length. i need help on this please

Answer 1

I am not sure (has been a long time). But the answer may be:

\[ds = \sqrt{dx^{2}+dy^{2}}\]
\[\frac{ds}{dx} = \sqrt{1+\left(\frac{dy^{2}}{dx^{2}}\right)}\]
\[= \sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\]

\[\int f(x)\cdot ds = \int\sqrt{1+\left(\frac{df}{dx}\right)^{2}}dx\]
\[f(x) = \frac{x^{4}}{4}+\frac{1}{8x^{2}}\]
\[\frac{df}{dx} = x^{3}-\frac{1}{4x^{3}}\]
\[1+\left(\frac{df}{dx}\right)^{2} = 1+x^{6}-2\cdot x^{3}\cdot\frac{1}{4x^{3}}+\frac{1}{16x^{6}}\]
\[ = x^{6}+\frac{1}{2}+\frac{1}{16x^{6}}\]
\[\sqrt{1+\left(\frac{df}{dx}\right)^{2}} = \sqrt{x^{6}+\frac{1}{2}+\frac{1}{16x^{6}}}\]
\[ = \sqrt{\frac{16x^{12}+8x^{6}+1}{16x^{6}}}\]
\[ = \frac{4x^{6}+1}{4x^{3}}\]
\[ = 4x^{3}+4x^{-3}\]
\[\int\sqrt{1+\left(\frac{df}{dx}\right)^{2}} = \frac{4x^{4}}{4}+4\cdot\left(-\frac{1}{2}\right)\cdot x^{-2}\]
\[ = x^{4}-2x^{-2}\]
\[\int_{1}^{2}\sqrt{1+\left(\frac{df}{dx}\right)^{2}} = 2^{4}-2\cdot2^{-2}-\left(1^{4}-2\cdot1^{-2}\right)\]
\[ = 16-\frac{1}{4}-1+2\]
\[ = 17.25\]