Question

Find the length of the curve y = 3/5x^5/3 - 3/4x^1/3 + 6 for 1 < = x < = 8. The length of the curve is . (Type an exact answer, using radicals as needed.)

Answer 1

Answer:

$\sqrt\frac{387}{20}$

Step-by-step explanation:

$Arc Length =\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^2 } } \, dx$

$y=\dfrac{3}{5}x^{\frac{5}{3}}- \dfrac{3}{4}x^{\frac{1}{3}}+6$

$\frac{dy}{dx} =x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}}$

$1+(\frac{dy}{dx})^2 }=1+(x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}})^2\\=1+(x^{\frac{4}{3}}-\dfrac{1}{2}+ \dfrac{1}{16}x^{-\frac{4}{3}})$

$=\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}}$

For the Interval $1\leq x\leq 8$

Length of the Curve $=\int\limits^8_1 {\sqrt{\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}} } } \, dx$

Using T1-Calculator

$=\sqrt\frac{387}{20}$