# Maths

a) A curve is defined by the following parametric equations
x= 3t/t^2 +1 , y=1/t^2 +1

(i) Find an expression, in terms of t for the gradient of the curve.

(ii) Determine the value of the gradient at the point where t=0

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1. Assuming the usual carelessness with parentheses, I assume you mean
x = 3t/(t^2+1)
y = 1/(t^2+1)
Now, the gradient is just dy/dx = (dy/dt) / (dx/dt)
dy/dt = -2t/(t^2+1)^2
dx/dt = -3(t^2-1)/(t^2+1)^2
so,
dy/dx = -2t/(t^2+1)^2 * (t^2+1)^2/(-3(t^2-1) = 2t/(3(t^2-1))
at t=0, dy/dx = 0

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oobleck

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