Question

find the radius of curvature at the given point.
y=4sinx-sin2x ,at x=π|2

Answer 1

Answer:

Radius of curvature is

$K=\frac{4}{5 \sqrt{5} }$

Step-by-step explanation:

To find the radius of curvature of the curve

$y = 4 \: sinx - sin \: 2x \:$

at point x = π/2

Step1: find y'

$\frac{dy}{dx} =y^{'}= 4cos \: x - 2cos \: 2x$

Step2:Find y''

$\frac{ {d}^{2}y }{d {x}^{2} } =y^{''}= - 4sin\: x + 4 sin \: 2x$

Step3: Apply the formula

$K = \frac{ |y^{''}| }{ {(1 + {y^{'}}^{2}) }^{ \frac{3}{2} } } \\ \\ K= \frac{ |4sin \: x + 4sin2x| }{ {(1 + {( - 4cos \: x - 2cos \: 2x)}^{2}) }^{ \frac{3}{2} } } \: \\ \\ K_{ x = \frac{\pi}{2} }= \frac{ |4sin \: \frac{\pi}{2}+ 4sin2\frac{\pi}{2}| }{ {(1 + {( - 4cos \: \frac{\pi}{2} - 2cos \: 2\frac{\pi}{2})}^{2}) }^{ \frac{3}{2} } } \: \\ \\ K= \frac{ |4(1)+ 4(0)| }{ {(1 + {( - 4(0) - 2( - 1))}^{2}) }^{ \frac{3}{2} } } \: \\ \\ K= \frac{ |4| }{ {(5) }^{ \frac{3}{2} } } \: \\ \\ K= \frac{4}{5 \sqrt{5} }$

Hope it helps you.