# Thread: Cartesian Equation and Parametric Equation

1. ## Cartesian Equation and Parametric Equation

Hey,
I need some help with the following questions:
a) Write down the Cartesian equation of the curve C represented by the parametric equations x=acos(theta) and y=bsin(theta), a>b>0
b) Find an expression for the gradient of the tangent to C
c) Hence show that the equation of the tangent can be written in the form:
bxcos(theta) + aysin(theta) = ab
I would really appreciate it

2. Remember that $\cos^2 \theta + \sin^2 \theta = 1$.
Now using this we can square things and add them
$x^2 + y^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta$
However, this doesn't get rid of the $\sin \theta$ or the $\cos \theta$. We need both of these terms to have the same number in front of them, but they don't right now. So we could have squared the x term then multiplied the y term by $\frac{a}{b}$ squared it and then added them. The following is the result
$x^2 + \frac{a^2}{b^2}y^2 = a^2 \cos^2 \theta + \frac{a^2b^2}{b^2} \sin^2 \theta = a^2$. Finally, the solution to a) is
$x^2 + \frac{a^2}{b^2}y^2 = a^2$.

For the last two, I am unsure how you find the gradient of a function at your level. What has your instructor defined as the gradient and is there a formula?

3. lvleph, thanks for part a.
For part b, I assume they want us to find the derivative of the equation, which I assume would be
2x + 2y (dy/dx) = 0 ... Would this be correct, or would I need to include the a and b values as well?
Thanks again

4. The gradient of a function f(x,y,z) should be $<\frac{df}{dx}, \frac{df}{dy}, \frac{df}{dz} >$.

5. For some reason I thought this was in the pre-calc or trig section. Sorry for that.
The gradient is always a vector and as Math Major pointed out it should be
$\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right)$
In your case $f = x^2 + \frac{a^2}{b^2}y^2 - a^2$