P3 Differentiation Help Please

Hi, could someone help, or talk me through the following questions, or the key points of the following questions, I'm really struggling. Thank you!

1

The curve C has parametric equations:
x=4cos2t, y=3sint, t is between -pi/2 and pi/2
A is the point 2,1.5 and lies on C.

a) find the value of t at the point A.
b) Find dy/dx in terms of t (I got dx/dt= -8sin2t and dy/dt=3cost)
c) Show that an equation of the normal to C at A is 6y-16x+23=0

The normal at A cuts C again at the point B

d) Find the y-coordinate of the point B.

2

A curve C is given by the equations:

x=2cost+sin2t, y=cost-2sin2t
where t is a parameter.

a) find the value of dy/dx at the point P on C where t=pi/4
b) find an equation of the normal to the curve at P.

Thanks for any help

Reply 1

swishmasta

quicky skimming through, for 1(a) just sub in x=2 and y=1.5 into x=4cos2t and y=3sint . remember to put your calculator in radian mode

Reply 2

idiopathic

StuartYates

1

The curve C has parametric equations:
x=4cos2t, y=3sint, t is between -pi/2 and pi/2
A is the point 2,1.5 and lies on C.

a) find the value of t at the point A.
b) Find dy/dx in terms of t (I got dx/dt= -8sin2t and dy/dt=3cost)
c) Show that an equation of the normal to C at A is 6y-16x+23=0

a)
A(2, 1.5)
So when x=2, y=1.5
2 = 4cos2t
cos2t=1/2
t=pi/6

b) So you've got dx/dt and dy/dt
remember that dy/dx=dy/dt * dt/dx
or dy/dx = dy/dt * (1/(dx/dt))

Reply 3

StuartYates

so can I leave the answer to b) in the form -(8sin2t)/(3cost) ??

and how would I go about c) and d)

Thanks

Reply 4

dvs

b) It's actually -(3cost)/(8sin2t) = -(3/8) . cost/2sintcost = -3/16sint
c)
Gradient of the normal is: -(dx/dy) = (16/3)sint
Using t=pi/6 you get that it's (8/3). So the eqn of the normal is:
y - 1.5 = (8/3) (x - 2)
3y - 4.5 = 8x - 16
Multiply by 2:
6y - 9 = 16x - 32
6y - 16x + 23 = 0

Reply 5

StuartYates

Thanks everyone, anyone have any ideas for me on question 2? Thanks so much!

Reply 6

username9816

StuartYates

2.) A curve C is given by the equations:

x = 2cost + sin2t, y = cost - 2sin2t
where t is a parameter.

a) find the value of dy/dx at the point P on C where t = pi/4
b) find an equation of the normal to the curve at P.

a.) dx/dt = 2cos2t - 2sint
dy/dt = -sint - 4cos2t
--> dy/dx = dy/dt * dt/dx = (-sint - 4cos2t)(-2sint + 2cos2t) = 2sin^2t + 8sint.cos2t - 2sint.cos2t - 8cos^2(2t) = 2sin^2t - 8cos^2(2t) + 6sint.cos2t.
At P: dy/dx = 2[sin(Pi/4)]^2 - 8[cos(Pi/2)]^2 + 6[sin(Pi/4).cos(Pi/2)] = 1

b.) Grad. (Normal) = -1.
Where t = Pi/4: x = Sqrt(2) + 1, y = [Sqrt(2)]/2 - 2
--> Equation Of Normal: y - (Sqrt2)/2 + 2 = -1(x - Sqrt2 - 1)
--> y - (Sqrt2) + 2 = -x + Sqrt2 + 1
--> y = -x + (2Sqrt2 - 1)