Posted by Hannah on Tuesday, February 22, 2011 at 5:24am.
a curve ahs parametric equations x=t^2
and y= 11/2t for t>0.
i)find the coordinates of the point P where the curve cuts the xaxis 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:
2/4t^3
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 xaxis, y = 0
√x=2
x=4
so the xintercept 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/(22y)
(but you didn't type it that way)
In either case, differentiate the correct equation implicitly, and sub in the corresponding xintercept.