Question

Determine the length of the curve $x=\frac{1}{2}(e^y-e^{-y})$

• A. $\frac{2}{3}(e^3-e^{-3})$
• B. $\frac{1}{2}(e^2-e^{-2})$
• C. $\frac{1}{3}(e^3-e^{-3})$
• D. $\frac{1}{2}(e^3-e^{-3})$

Answer 1

The correct option is D $\frac{1}{2}(e^3-e^{-3})$

Given, $x=\frac{1}{2}(e^y-e^{-y})$

Thus $x^{\prime }=(2\sin hy{)}^{\prime }=2\cos h$ y

Length of the curve $={\int }_0^3\sqrt{2\cos h\left(2y\right)+3}dy$

$=\sin h\left(b\right)-\sin h\left(0\right)$

$=\sin h\left(3\right)$

$=\frac{1}{2}(e^3-e^{-3})$