Question

If x=cost(3 - 2 cos^2t) and y= sint(3 - 2 sin^2t) find dy/dx at t = pi / 4

Answer 1

HELLO DEAR,

GIVEN:- x = cost(3 - 2cos²t)
so,
dx/dt = (-3sint + 6cos²t . sint)

dx/dt = [-3sint + 6sint(1 - sin²t)]

dx/dt = [-3sint + 6sint - 6sin³t]

dx/dt = [3sint - 6sin³t]--------( 1 )

also, y = sint(3 - 2sin²t)

similarly, dy/dt = (cost - 6sin²t . cost)

dy/dt = [3cost - 6(1 - cos²t) . cost)]

dy/dt = [3cost - 6cost + 6cos³t]

dy/dt = [6cos³t - 3cost]--------( 2 )

now, dy/dx = {dy/dt}/{dx/dt}

so, $\bold{dy/dx = \{6cos^3t - 3cost\}/\{3sint - 6sin^3t\}}$

$\bold{[dy/dx] = cost\{2cos^2t - 1\}/sint\{1 - 2sin^2t\}}$

$\bold{[dy/dx]_{t = \pi/4} = cot(\pi/4) . \{2(1/\sqrt{2})^2 - 1\}/\{1 - 2(1/\sqrt{2})^2}\}$

$\bold{[dy/dx]_{t = \pi/4} = 1.\frac{1 - 1}{1 - 1}}$

HENCE, dy/dx = 0

I HOPE IT'S HELP YOU DEAR,
THANKS

Answer 2

Answer:

1

Step-by-step explanation:

Hi,

Given x = cost(3 - 2cos²t)

=> x = (3 - 2cos²t)*cost

=> x = 3cost - cos³t

so,

dx/dt = -3sint + 6cos²t . sint

dx/dt = -3sint + 6sint(1 - sin²t)

dx/dt = -3sint + 6sint - 6sin³t

dx/dt = 3sint - 6sin³t

dx/dt = 3sint(1-2sin²t)--------( * )

Also, given

y = sint(3 - 2sin²t)

=> y = (3 - 2sin²t)sint

=> y = 3sint-2sin³t

So,

dy/dt = 3cost - 6sin²t. cost

dy/dt = 3cost(1-2sin²t)---------( ** )

But we Know that

dy/dx = {dy/dt}/{dx/dt}

=>from equations(*) and (**), we get

dy/dx = 3cost(1-2sin²t)/3sint(1-2sin²t)

=> dy/dx = cot(t),

Now at t = π/4

dy/dx = cot(π/4) = 1

Hope, it helped !