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consider the parametrized surface X(s,t)=(s,s^2+t,t^2) find and equation for the tangent plan at the point (1,0,1)
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Well to find the normal vector the a surface at any point you simply need to take the gradient. From there you know how to find a plane given its normal vector.
so, Xs= (1,2s,0) and Xt=(0,1,2t) and the normal vector is (4st, -2t, 1)..what is the next step? using the N.(P-P0) does not give me the right answer
Well: \[\vec{\nabla} = \frac{\partial}{\partial x} \hat{x} + \frac{\partial}{\partial y} \hat{y} + \frac{\partial}{\partial z} \hat{z}\] But your x,y,z are functions of s, t, or s & t so the gradient isn't as simple as taking an s and t derivative I don't believe.
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