Find an equation of the tangent plane to the given surface at the specified point. z = ex2 - y2, (1, -1, 1)
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Originally Posted by acg716 Find an equation of the tangent plane to the given surface at the specified point. z = ex2 - y2, (1, -1, 1) What does ex2 mean? You have to be a whole lot clearer in stating your equations if you expect to get effective help.
i'm really sorry! when i posted it it looked correct, but e is raised to x^2-y^2
Originally Posted by acg716 i'm really sorry! when i posted it it looked correct, but e is raised to x^2-y^2 First get the partial derivatives and . Now construct the vector . Substitute and into this vector. This gives a normal of the form to the surface at . The equation of the tangent plane is therefore . To get the value of , substitute , and .
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