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:
-2/4t^3
is that gradient right and how do I get the value at the point P.
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
√x=2
x=4
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.
To find the gradient of the curve at a specific point, you need to first differentiate both the x and y equations with respect to t. Let's differentiate the given equations:
Given:
x = t^2
y = 1 - (1/2)t
Differentiating x with respect to t:
dx/dt = 2t
Differentiating y with respect to t:
dy/dt = -1/2
To find the gradient at the point P(1/4, 0), substitute t = 1/4 into the gradients:
Gradient of x: dx/dt = 2(1/4) = 1/2
Gradient of y: dy/dt = -1/2
Therefore, the gradient of the curve at point P is 1/2.