Question

If x = a `(cosθ + log tan  θ/2) and y = sin θ`, then find `(d^2y)/(dx^2)  at  θ = pi/4`.

Answer 1

x = a `(cosθ + log tan  θ/2) and y = sin θ`

Differentiate w.r.t., θ

`(dx)/(dθ) = a(-sin θ + 1/tan(θ/2)  sec^2  θ/2 xx 1/2)`

= `a(-sinθ + (cos  θ/2)/(2sin  θ/2  cos^2  θ/2))`

= `a(-sinθ + (1)/(sinθ))`

= `a((1 - sin^2θ)/sinθ)`

= `a(cos^2θ)/sinθ`   ...(i)

Now, y = sinθ 

`(dy)/(dθ) = cosθ`   ...(ii)

Dividing equation (ii) by equation (i),

`(dy)/(dx) = (dy//dθ)/(dx//dθ) = (cosθ)/((a(cos^2θ)/sinθ))`

= `sinθ/(a cosθ)`

`(dy)/(dx) = tan θ/a`

Again differentiate w.r.t. x,

`(d^2y)/(dx^2) = 1/a sec^2 θ (dθ)/(dx)`

= `1/a sec^2 θ. sin θ/(a cos^2θ)   ...["From equation (i)" (dθ)/(dx) = (sin θ)/(a cos^2 θ)]`

at θ = `pi/4`

`(d^2y)/(dx^2) = 1/a^2 ((sec^2  pi/4)(sin  pi/4))/((cos^2  pi/4))    ...[∵ sin  pi/4 = 1/sqrt2 and cos  pi/4 = 1/sqrt2]`

= `(1/a^2 xx (sqrt2)^2 xx 1/sqrt2)/(1/sqrt2)^2`

= `(1/a^2 xx 2/sqrt2)/(1/2)`

= `4/(a^2sqrt2)`

= `(2sqrt2)/a^2`