I have this function:$$f(x,y)=2xy^2$$ and I have to find a vector perpendicular to the the graph of the function in the point $(2,1)$. The gradient is $$\nabla{f(x,y)}=(2y^2,4xy)$$ and $$\nabla{f(2,1)}=(2,8)$$ but the answer is not correct.
Vector perpendicular to a graph in a given point?
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multivariable-calculus
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0It looks like you’re meant to be working in three dimensions, not two. – 2017-02-19
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0So it will be something like $(\nabla{ f(x,y)},z)$ – 2017-02-19
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0Not quite. Write the equation of the surface in the form $F(x,y,z)=0$ and use the fact that the gradient of a function is always normal to its level surfaces. – 2017-02-21