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Math Help - Gradient Vectors and Tangent Planes to Level Surfaces

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    Gradient Vectors and Tangent Planes to Level Surfaces

    Is the plane that has the normal vector as the gradient vector of a surface at point P perpendicular to the tangent plane at that point?
    Last edited by mr fantastic; October 15th 2010 at 01:48 PM.
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    Yours is the sort of general question that is hard to answer for fear of not having understood exactly what you meant.

    However, yes: The gradient ‘points’ away from the surface in a maximal way.
    It is the normal to the tangent plane at a point.
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    Posted by the OP as edits to the original post:

    Quote Originally Posted by lilaziz1
    So I just found out that the plane that has its normal vector as the gradient vector is the tangent plane at the point P. Doesn't that mean the gradient vector is coming out of the surface? In other words, it does not lay in the tangent plane. Right?

    If the gradient vector is coming out of the surface, how can it be steepest path? Isn't it coming out of the surface?
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    Okay I think I get it now. In this book, they talk about level SURFACES. I was under the assumption that the level surface was actually a surface of two variables that was solved for some constant k. (for example, z = x^2+y^2-3 would be x^2+y^2+z=3). So it wasn't very intuitive when they showed the steepest path (the gradient vector) is coming out (like a binormal vector) of the level surface (once again, I thought it was a two variable surface solved for some constant k).

    However, one can assume a sphere or a ellipsoid (so a two variable function or an implicit function) is a level surface of some 4d function and then we can find the tangent plane of that sphere or ellipsoid at a specific point. But the magnitude of the gradient vector is vain because in this situation because we don't know the 4d function.

    If someone can vouch that my understanding of the gradient vector is correct, I'd be grateful cuz I've been pondering on this for an hour and my head hurts! If my understanding is incorrect, can you explain what's wrong with it?

    Thanks in advance!

    Also, thank you mr fantastic for fixing my first reply. I was not sure if I was allowed to double post.
    Last edited by lilaziz1; October 16th 2010 at 09:14 AM.
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    Quote Originally Posted by lilaziz1 View Post
    Is the plane that has the normal vector as the gradient vector of a surface at point P perpendicular to the tangent plane at that point?
    No, the "plane that has the normal vector as the gradient vector of a surface at point P" is the tangent plane.

    If the function is given implicitely by f(x, y, z)= constant, then its gradient vector \nabla f points in the direction of fastest increase of f. If \vec{u} is a unit vector in the tangent plane to the surface, then the derivative of \vec{v}, D_{\vec{v}}(f)= \nabla f \cdot\vec{v}= 0 because f does not change in that direction.
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    Right so the statements in post #4 are valid right?
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    Quote Originally Posted by lilaziz1 View Post
    Okay I think I get it now. In this book, they talk about level SURFACES. I was under the assumption that the level surface was actually a surface of two variables that was solved for some constant k. (for example, z = x^2+y^2-3 would be x^2+y^2+z=3). So it wasn't very intuitive when they showed the steepest path (the gradient vector) is coming out (like a binormal vector) of the level surface (once again, I thought it was a two variable surface solved for some constant k).

    However, one can assume a sphere or a ellipsoid (so a two variable function or an implicit function) is a level surface of some 4d function and then we can find the tangent plane of that sphere or ellipsoid at a specific point. But the magnitude of the gradient vector is vain because in this situation because we don't know the 4d function.
    No, because we do know the 4d function- at least up to a constant. If, for example, the ellipsoid is given by bc(x- x_0)^2+ ac(y- y_0)^2+ ab(z- z_0)^2= abc then we can take our 4d function to be F(x, y, z)=  bc(x- x_0)^2+ ac(y- y_0)^2+ ab(z- z_0)^2 so that our ellipsoid is the "abc" level surface or G(x, y, z)= bc(x- x_0)^2+ ac(y- y_0)^2+ ab(z- z_0)^2- abc so that our ellipsoid is the "0" level surface. But since they differ by a constant, they have the same gradient.

    If someone can vouch that my understanding of the gradient vector is correct, I'd be grateful cuz I've been pondering on this for an hour and my head hurts! If my understanding is incorrect, can you explain what's wrong with it?

    Thanks in advance!

    Also, thank you mr fantastic for fixing my first reply. I was not sure if I was allowed to double post.
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