Mathematics, 07.03.2020 04:27 juhavyen
Find the length of the curve yequalsthree fourths x Superscript 4 divided by 3 Baseline minus three eighths x Superscript 2 divided by 3 Baseline plus 5 for 1less than or equalsxless than or equals27.
Answers: 3
is continous on [a,b], then the length of the curve 
is![L=\int\limits^b_a {\sqrt{1+[\frac{dy}{dx} ]^2} } \, dx](/tpl/images/0537/6591/f2179.png)
![\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}}](/tpl/images/0537/6591/f6f5c.png)
use ![\frac{d}{dx}[x^n]=nx^{n-1}](/tpl/images/0537/6591/cca2b.png)
![(\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}}](/tpl/images/0537/6591/f5eb6.png)
![[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}](/tpl/images/0537/6591/dab4c.png)
![=[x^{\frac{1}{3}}+\frac{1}{4}x^{-\frac{1}{3}}]^2\\\\=x^{\frac{1}{3}}+\frac{1}{4}x^{-\frac{1}{3}}](/tpl/images/0537/6591/97b3f.png)
![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](/tpl/images/0537/6591/79381.png)
![=\frac{3}{4} [27^{\frac{4}{3} }-1]+\frac{3}{8} [27^{\frac{2}{3}}-1]\\\\=63](/tpl/images/0537/6591/d71c1.png)
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Find the length of the curve yequalsthree fourths x Superscript 4 divided by 3 Baseline minus three...
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