a curve ahs parametric equations x=t^2
and y= 1-1/2t for t>0.
i)find the co-ordinates of the point P where the curve cuts the x-axis which i found to be P(1/4, 0)
the next part i cant do
ii) find the gradient of the curve at this point.
So far, I have the gradient to be:
is that gradient right and how do I get the value at the point P.
Maths - Reiny, Tuesday, February 22, 2011 at 9:21am
you could change it to cartesian form ...
t^2 = x
t = √x
y = 1 - (1/2)t
2y = 2 - t
t = 2 - 2y
so √x = 2 - 2y
√x + 2y = 2
to cut the x-axis, y = 0
so the x-intercept is (4,0)
>b>Unless ... you meant to type y = 1 - 1/(2t)
in that case the point is correct and the cartesian equation would be
√x = 1/(2-2y)
(but you didn't type it that way)
In either case, differentiate the correct equation implicitly, and sub in the corresponding x-intercept.