How do you find the gradient of a parametric equation at a certain point?
The easiest way to see this is with an example. So let us take the curve:
x = cos(2t) , y = sin(t) with a point P where t = pi/6
The gradient at any point is equal to the derivative of y with respect to x at the given point. But in this situation we cannot simply take dy/dx. However under closer inspection and treating dy/dx as a fraction we see that dy/dx = (dy/dt)(dt/dx). However, this dt/dx may also present problems, but it also can be seen as 1/(dx/dt) in treating it as a fraction as well.
So all we need to calculate is dy/dt and dx/dt:
dy/dt = cos(t) , dx/dt = -2sin(2t)
Now dt/dx = 1/(dx/dt) = -1/(2sin(2t))
Then dy/dx = (dy/dt)(dt/dx) = -cos(t)/2sin(2t)
So we now have the derivative at all points in the curve. To find the gradient at the point P we simply substitute in the value of t and calculate.
This gives:
-cos(pi/6)/2sin(pi/3) = -(31/2/2)/2(31/2/2) = -1/2
Thus the gradient of this parametric equation at P is -1/2. This method can easily be extended and adapted to all equations of parametric form.
