Using parametric differentiation to evaluate the slope of a curve - attempted

[chuffy]

Homework Statement

x(t) = (t^2 -1) / (t^2 +1)

y(t) = (2t) / (t^2 +1)

at the point t=1

Homework Equations

Line equation = y-y1 = m(x-x1)

chan rule = (dy/dt) / (dx/dt) = dy/dx

The Attempt at a Solution

I find the y1 and x1 values by subing in t=1 to the x(t) and y(t) equations
I get the point (0,1) when t=1

I have used the Quotient rule to find dx/dt & dy/dx (Is this right?)
Doing the above I get:

dy/dt = (-2(t^2 -1)) / (t^2+1)^2

dx/dt = (4t) / (t^2 +1)^2

So dy/dx = (dy/dt) / (dx/dt) however when I sub in t=1 to (dy/dt) I get 0 as the numerator

I know that the m of the line equation is equal to dy/dx

does anyone know what I'm doing wrong? I'll try uploading a pic of my work
cheers

 

[chuffy]

pic of my working

 

[pongo38]

What makes you think something is wrong? have you plotted the x v y graph for t having a range of values? say t=-1, 0, 1, 2, 3

 

[chuffy]

if t=1 does this mean that m= 0?

 

[HallsofIvy]

The graph of this function is an ellipse. t= 1 is the point (0, 1) where the tangent line is, in fact, horizontal so the dy/dx= 0 there.

 

[chuffy]

cheers