Consider the function $\frac{1}{1+x^2+3y^2}$ and the trajectory x(t) = t, y(t) = t(t-2)
I am asked to show that the directional vector in the direction of motion is given by $-\frac{2t(6t^2 - 18t + 13)}{\sqrt{1 + 4(t-1)^2}(1+t^2+3t^2(t-2)^2)^2)}$
I started by finding ∇f(x,y) = $(\frac{-2x}{(1+x^2+3y^2)^2}, \frac{-6y}{(1+x^2+3y^2)^2})$
Now using the definition of directional vector which is ∇f(x,y).$\hat{u}$ I constructed the $\hat{u}$ = $\frac{1}{\sqrt{t^2 + t^2(t-2)^2)}}(t, t(t-2))$
Finally I calculated the dot product and tried to simplify. However, I couldn't reach the expression I am asked to show. I don't know if I made a calculation error or my approach is incorrect. Any help is appreciated.