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A steepest tangent vector to the explicit surface $z=f(x,y)$ at the fixed point $(x,y,f(x,y))$ is given by the formula: $$ \mbox{steepTan} = (?)(?)(?) $$ You can replace the $?$s with formulas like $$\frac{\partial}{\partial x}[f(x, y)]$$ One of the hints is that the steepest tan is the $(dx,dy,?)$. The hardest part is figuring out the third variable. Which can be figured out by $$dz = mdx +ndy$$

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What exactly is the question? – Matt Mar 11 '11 at 2:13
I guess the question would be what are the 3 parts of the steepest tangent vector to an explicit surface. I understand that it would be the gradient but what would the 3rd component be? (f/x[x,y],f/y[x,y],______) – user8093 Mar 11 '11 at 2:28
up vote 1 down vote accepted

The surface tangent vector you're looking for is the tangent vector to the curve on the surface that you get by travelling in the direction of the gradient. That curve is given by

$$\left(x + \lambda\frac{\partial f}{\partial x},y + \lambda\frac{\partial f}{\partial y},z\left(x + \lambda\frac{\partial f}{\partial x},y + \lambda\frac{\partial f}{\partial y}\right)\right)\;,$$

so the tangent vector, the derivative with respect to $\lambda$, is

$$\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial x}\cdot\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\cdot \frac{\partial f}{\partial y}\right)=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\lvert\nabla f\rvert^2\right)\;.$$

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