C4 curve and parametric question

My C4 exam is on Monday and i can never do these types of questions.

1)
A curve has equation

x² + 2xy - 3y² + 16 = 0

Find the co-ordinates of the points on the curve where dy/dx = 0

2)
A curve has parametric equations

x = 2 cot t, y = 2sin² t , 0 < t < pi/2

(a) find an expression for dy/dx in terms of the parameter t
(b) Find an equation of the tangent to the curve at the point where t = pi/4
(c) Find a cartesian equation of the curve in the form y = f(x). State the domain on which the curve is defined.

Thanks

Reply 1

Chewwy

x² + 2xy - 3y² + 16 = 0

differentiating each term wrt x...

d(x^2)/dx = 2x
d(2xy)/dx = y(d(2x)/dx) + 2x(dy/dx) = 2y + 2xdy/dx
d(-3y^2)/dx = d(-3y^2)/dy * dy/dx = -6ydy/dx
d(16)/dx = 0

so derivative: 2x + 2y + 2xdy/dx - 6ydy/dx = 0

dy/dx = (-2x-2y)/(2x-6y) = 0

(-2x-2y) = 0
x = y

into original equation...

x^2 + 2x^2 - 3x^2 + 16 = 0

ah ****. it should be y = -x. fix that.

Reply 2

Dabster

2)a) dy/dx=dy/dt x dt/dx

dy/dx=4costsint

dt/dx=1/(dx/dt)
= 2(sin^2)t
dy/dx=-8cost(sin^3)t

b)x=2cot(pi/4) = 2
y=2sin(pi/4)^2= 1

gradient = -1/2 (put t=pi/4 into the differential equation)

y-1=(-1/2)(x-2)
y=(-x/2)+2

c) use 1+cot^2t = cosec^2t
rearrange x=2cott to x/2=cott and the same with
y=2sin^2t which is y/2 = sin^2t
1+(x^2/4) = (2/y)

y=8/(x^2 + 4)

Reply 3

rodriguez99

Original post by Dabster