Answer:
Step-by-step explanation:
Consider the equation of the curve,
[tex]y(x)=\frac{3}{4}x^{\frac{4}{3}}-\frac{3}{8}x^{\frac{2}{3}}+5:1\leq x \leq 27[/tex]
Find the arc length directly as follows:
Arc length of the curve:
if [tex]f'[/tex] is continous on [a,b], then the length of the curve [tex]y=f(x),a\leq x \leq b[/tex] is
[tex]L=\int\limits^b_a {\sqrt{1+[\frac{dy}{dx} ]^2} } \, dx[/tex]
Differentiate the function with respect to x,
[tex]\frac{dy}{dx}= \frac{d}{dx}[\frac{3}{4}^{\frac{4}{3} }-\frac{3}{8}^{\frac{2}{3} }+5]\\\\=\frac{d}{dx}[\frac{3}{4}^{\frac{4}{3} }]-\frac{d}{dx}[\frac{3}{8}^{\frac{2}{3} }]+\frac{d}{dx}(5)\\\\=\frac{3}{4} \frac{d}{dx}[x^{\frac{4}{3}}] -\frac{3}{8} \frac{d}{dx}[x^{\frac{2}{3}}]+\frac{d}{dx}(5)\\ \\=\frac{3}{4}[\frac{4}{3}x^{\frac{1}{3} } ] +\frac{3}{8}[\frac{2}{3}x^{\frac{1}{3} } ]+0\\ \\=x^{\frac{1}{3}}-\frac{1}{4}x^{\frac{1}{3}}[/tex]use [tex]\frac{d}{dx}[x^n]=nx^{n-1}[/tex]
That is,
[tex](\frac{dy}{dx} )^2=(x^{\frac{1}{3} }-\frac{1}{4}x^\frac{1}{3} )^2\\\\=x^{\frac{1}{3} }+\frac{1}{16}x^{\frac{1}{3} }-\frac{1}{2}\\\\\sqrt{1+[\frac{dy}{dx} ]^2}=\sqrt{1+x^{\frac{1}{3} }+\frac{1}{16}x^{\frac{1}{3} }-\frac{1}{2}} \\\\ =\sqrt{x^{\frac{1}{3} }+\frac{1}{16}x^{\frac{1}{3} }+\frac{1}{2}}[/tex]
since[tex][x^{\frac{1}{3}}+\frac{1}{4}x^{-\frac{1}{3}}]^2=x^{\frac{1}{3}}+\frac{1}{16}x^{-\frac{2}{3}}+\frac{1}{2}[/tex]
[tex]=[x^{\frac{1}{3}}+\frac{1}{4}x^{-\frac{1}{3}}]^2\\\\=x^{\frac{1}{3}}+\frac{1}{4}x^{-\frac{1}{3}}[/tex]
substitute this value in arc length formula to get,
[tex]L=\int\limits^{27}_1 {[x^{\frac{1}{3}}]+\frac{1}{4}x^{-\frac{1}{3}}} \, dx \\\\=\int\limits^{27}_1 {[x^{\frac{1}{3}}]\, dx + \frac{1}{4}\int\limits^{27}_1 [x^{-\frac{1}{3}}}] \, dx\\\\=\frac{3}{4} [x^{\frac{1}{3}}]\limits^{27}_1 + \frac{1}{4} [\frac{3}{2}x^{\frac{1}{3}}]\limits^{27}_1[/tex]
use [tex]\int {x^n}=\frac{x^{n+1}}{n+1}[/tex]
[tex]=\frac{3}{4} [27^{\frac{4}{3} }-1]+\frac{3}{8} [27^{\frac{2}{3}}-1]\\\\=63[/tex]
therefore the length of the curves is L=63
