Answer:
Step-by-step explanation:
Given
Length of curve
[tex]L=\int_{2}^{6}\sqrt{1+64x^{-6}}dx [/tex]
Length of curve is given by
[tex]L=\int_{a}^{b}\sqrt{1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x}\right )^2}dx[/tex] over interval a to b
comparing two we get
[tex]\frac{\mathrm{d} y}{\mathrm{d} x}=8x^{-3}[/tex]
[tex]dy=8x^{-3}dx[/tex]
integrating
[tex]\int dy=\int 8x^{-3}dx[/tex]
[tex]y=-4x^{-2}+C[/tex]
Curve Passes through (1,2)
[tex]1=-4+C[/tex]
[tex]C=5[/tex]
curve is
[tex]y+\frac{4}{x^2}=5[/tex]
