Question

Find the derivatives of the following:

If x = a(θ + sin θ), y = a(1 – cos θ) then prove that at θ = `pi/2`, yⁿ = `1/"a"`

Answer 1

x = a(θ + sin θ), y = a(1 – cos θ)

`("'d"x)/("d"theta) = "a"(1 + cos theta), ("d"y)/("d"theta) = "a"(0 - (- sin theta))`

`("d"x)/("d"theta) = "a"(1 + cos theta), ("d"y)/("d"theta) = "a" sin theta`

`(("d"y)/("d"theta))/(("d"x)/("d"theta)) = ("a" sin theta)/("a"(1 + cos theta))`

`("d"y)/("d"x) = sin theta/(1 + cos theta)`

y' = `(2 sin  theta/2  cos  theta/2)/(2 cos^2  theta/2)`

y' = `(sin  theta/2)/(cos  theta/2)`

= `tan  theta/2`

Differentiating with respect to x we get

`"d"/("d"x) (("d"y)/("d"x)) = sec^2 theta/2 xx 1/2 xx ("d"theta)/("d"x)`

`("d"^2y)/("d"x^2) = 1/2 sec^2 theta/2 xx 1/("a"( + cos theta))`

y" = `1/(2"a") sec^2  theta/2 xx 1/(2 cos^2  theta/2)`

= `1/(4"a") sec^2  theta/2 xx sec^2  theta/2`

= `1/(4"a") (1 + tan^2  theta/2)(1 + tan^2  theta/2)`

y" at θ = `pi/2` is

y" = `1/(4"a") (1 + tan^2((pi/2)/2)) (1 + tan^2 ((pi/2)/2))`

y" = `1/(4"a") (1 + tan^2 (pi/4)) (1 + tan^2 (pi/4))`

y" = `1/(4"a") (1 + 1^2) (1 + 1^2)`

y" = `1/(4"a") 2 xx 2`

y" = `1/"a"`