Find the length of normal to the curve `x=a(theta+sintheta),y=a(1-costheta)` at `theta=pi/2dot`

Options

A. 2a
B. `a sqrt(2)`
C. `a//2`
D. `a// sqrt(2)`

Answer

B

Explanation

We have,
`x=a(theta+sintheta),y=a(1-cos theta)`
`rArr (dx)/(d theta)=a(1+cos theta), (dy)/(dx)=a sin theta`
`therefore (dy)/(dx) = (asin theta)/(a(1+cos theta))="tan"(theta)/(2)`
Length of the normal ` = y sqrt(1+((dy)/(dx))^(2))`
`=a(1-cos theta) sqrt(1+"tan"^(2)(theta)/(2))`
`=2a "sin"^(2)(theta)/(2) xx "sec"(theta)/(2)=2a "tan"(theta)/(2) "sin"(theta)/(2)`
`therefore " Length of the normal at" theta = (pi)/(2) " is " 2a "tan"(pi)/(4)"sin"(pi)/(4)=sqrt(2)a`