Answer:
Length of the curve = 1.207 units
Step-by-step explanation:
y = [tex]\frac{3}{4} x^{4/3} - \frac{3}{8} x^{2/3} + 9[/tex] for 1 ≤x≤8
The formula for finding the length of a curve is:
Length of a curve = [tex]\int\limits^a_b {\sqrt{1 + (\frac{dy}{dx})^{2} } } \, dx[/tex]
To compute the length we need (dy/dx)². So,
dy/dx = [tex]\frac{4}{3} * \frac{3}{4} x^{(4/3) -1} - \frac{2}{3}*\frac{3}{8}x^{(2/3) - 1}[/tex]
dy/dx = [tex]x^{1/3} - \frac{1}{4} x^{-1/3}[/tex]
(dy/dx)² = ([tex]x^{1/3} - \frac{1}{4} x^{-1/3}[/tex])²
(dy/dx)² = [tex]x^{2/3} - \frac{1}{16} x^{-2/3}[/tex]
Length of a curve = [tex]\int\limits^a_b {\sqrt{1 + (\frac{dy}{dx})^{2} } } \, dx[/tex]
= [tex]\int\limits^8_1 {\sqrt{1 + x^{2/3} - \frac{1}{16} x^{-2/3} } } \, dx[/tex]
= [tex]\int\limits^8_1[/tex][(1 + x^(2/3) - (1/16)(x^-2/3)]^1/2 dx
= 2/3 [(1 + x^(2/3) - (1/16)(x^-2/3)]^3/2 * [(2/3)*x^-1/3 + (1/24)*x^(-5/3)]
Applying limits 1 to 8:
(2/3)*[(1 + (8)^(2/3) - (1/16)(8)^(-2/3)]^3/2 * [(2/3)*(8)^(-1/3) + (1/24)*(8)^(-5/3)] - (2/3)*[(1 + (1)^(2/3) - (1/16)(1)^(-2/3)]^3/2 * [(2/3)*(1)^(-1/3) + (1/24)*(1)^(-5/3)]
= (2/3)*(11.12)*(0.3346) - (2/3)*(2.6968)*(0.7083)
= 2.4805 - 1.2734
Length of the curve = 1.207 units
