1. Parametric Differentiation help....

Need help having trouble with this question....
$x=2cos\theta, y=\frac{1}{\sqrt2}sin\theta$
a) Find gradient at the point where theta=pi
b) Hence find the cartesian co-ordinates
c) Find equations of the tangent and normal at this point.
My workings
a)
$\frac{dx}{d\theta}=-2sin\theta , \frac{dy}{d\theta}=\frac{1}{\sqrt2}cos\theta$
$\frac{d\theta}{dx}=-\frac{1}{2sin\theta}$
$\frac{dy}{dx}=\frac{dy}{d\theta} x \frac{d\theta}{dx}$
$\frac{dy}{dx}=\frac{1}{\sqrt2}cos\theta x -\frac{1}{2sin\theta}$
$\frac{dy}{dx}=-\frac{\frac{1}{\sqrt2}cos\theta}{2sin\theta}$
$\frac{dy}{dx}=-\frac{\frac{1}{\sqrt2}}{2tan\theta}$
$\frac{dy}{dx}=-\frac{\frac{1}{\sqrt2}}{2tan\pi}$
as tan pi = 0, i will get an undefined answer is so how can i do the following parts? :s can someone please help me out

And yes Theta=Pi heres the original question, Question 3 here [/URL]

2. Your working looks fine and I get the same answer. It's probably just undefined

3. at (-2,0)

you have a vertical tangent and a horizontal normal

4. how did you get that co-ordinate of -2,0 just need help on b and c if my a is correct. thanks.

5. x= 2cos(pi) y = 1/(2^1/2)sin(pi)

6. How do i get equations of tangent+normal ? :s im doing this

(y-0)=m(x-(-2)) but what do i put in for m infinity? =/ never had a question like this before t.t

7. remember your old algebra days --the eqn of a vertical line is x = c

in this case x = -2

8. x=-2, and y=-2 but is there any working out needed ? i know its really basic but how would i know that the tangent + normal were going to be horizontal and vertical?

9. Originally Posted by Kevlar
x=-2, and y=-2 but is there any working out needed ? i know its really basic but how would i know that the tangent + normal were going to be horizontal and vertical?
the slope was undefined, this means the tangent is vertical. the normal is perpendicular to the tangent, so it had to be horizontal

10. y = 0