Posted by **Hannah** on Tuesday, February 22, 2011 at 5:24am.

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.

- 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

√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.

## Answer this Question

## Related Questions

- Calculus - A curve is defined by the parametric equations: x = t2 – t and y = t3...
- Calc. - sketch the curve using the parametric equation to plot the points. use ...
- calc - for the parametric curve defined by x=3-2t^2 and y=5-2t ...sketch the ...
- trig - The parametric equations for a curve in the x-y plane are x=2+t^2 and y=4...
- math - find the eqt. of tangent to the curve y=-x^2+2x-10 @ the point where the ...
- Calc 3 - Find the parametric equations for the tangent line to the curve with ...
- Calc 3 - Find the parametric equations for the tangent line to the curve with ...
- 12th Grade Calculus - 1. a.) Find an equation for the line perpendicular to the ...
- Maths - 1. Sketch the curves y=e^x and y=e^-2x, using the same axes. The line y=...
- AP AB Calculus - Linear approximation: Consider the curve defined by -8x^2 + 5xy...