Question

What curve passes through the point ​(1 ​,2​) and has an arc length on the interval​ [2,6] given by Integral from 2 to 6 StartRoot 1 plus 64 x Superscript negative 6 EndRoot dx ​? What is the​ curve?

Answer 1

Answer:

Step-by-step explanation:

Given

Length of curve

$L=\int_{2}^{6}\sqrt{1+64x^{-6}}dx$

Length of curve is given by

$L=\int_{a}^{b}\sqrt{1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x}\right )^2}dx$ over interval a to b

comparing two we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=8x^{-3}$

$dy=8x^{-3}dx$

integrating

$\int dy=\int 8x^{-3}dx$

$y=-4x^{-2}+C$

Curve Passes through (1,2)

$1=-4+C$

$C=5$

curve is

$y+\frac{4}{x^2}=5$