Question

If x and y are connected parametrically by the equation, without eliminating the parameter, find. dy/dx x=a(cost+logtan1/2),y=asint

Answer 1

Given, $x=a(cost+logtan\frac{t}{2}),y=asint$

$x=a\left(\begin{array}{c}cost+logtan\frac{t}{2}\end{array}\right)$
now differentiate x with respect to t
$\frac{dx}{dt}=a\left[\begin{array}{c}(-sint)+\frac{sec^2\frac{t}{2}}{tan\frac{t}{2}}.\frac{1}{2}\end{array}\right]\\\\=a\left[\begin{array}{c}-sint+\frac{sec^2\frac{t}{2}}{2tan\frac{t}{2}}\end{array}\right]$------(1)

y = asint
now differentiate y with respect to t
dy/dt = acost---------(2)

now dividing equations (2) by (1),
dy/dx =$\frac{acost}{a\left[\begin{array}{c}-sint+\frac{sec^2\frac{t}{2}}{2tan\frac{t}{2}}\end{array}\right]}$

now, [-sint + sec²(t/2)/2tan(t/2)]
= [-sint + 1/{2sin(t/2).cos(t/2)]
= [-sint + 1/sint ]
[ we know, 2sin(t/2).cos(t/2) = sint as sin2x = 2sinx.cosx ]
= [(-sin²t + 1)/sint]
= [(cos²t)/sint]
= [cott.cost ] now put it above

dy/dx = (acost)/(acott.cost)
= 1/cott = tant

hence, dy/dx = tant