Question

if `x=a(costheta+thetasintheta) , y=a(sintheta-thetacostheta)` then `(d^2y)/(dx^2)`
A. `sec^(2)theta`
B. `-"cosec"^(2)theta`
C. `(sec^(3)theta)/(a theta)`
D. None of these

Answer 1

Correct Answer - C
Given `x = a(cos theta + theta sin theta)`
`y = a(sin theta - theta cos theta)"and we have"(dy)/(dx) = tan theta`
According to question `(d^(2)y)/(dx^(2))=sec^(2)theta(d theta)/(dx)`
` =sec^(2)theta((1)/(a theta cos theta))" "[because (dx)/(d theta)=a theta cos theta]`