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    see the attached picture , I was having issues formatting it on here so I just took a screenshot of what I wrote any help would be extremely appreciated.Name:  image.png
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    To use LaTeX on here, enclose your LaTeX code in [tex] {stuff} [/tex].

    e.g. [tex]\dfrac{-b \pm \sqrt{b^2-4ac}}{2}[/tex] will give \dfrac{-b \pm \sqrt{b^2-4ac}}{2}


    Even if you can't use LaTeX, have you not heard of paragraphs? Full stops? Capitalisation?

    Sorry to snipe on presentation, but what you've written is incredibly difficult to read.

    As far as the actual question, it's kind of unclear what you've done so far. What are your thoughts on finding the tangent plane?
    For the last bit, I have no idea what vector you mean by $1/3$ so it's hard to comment, but unless the rate of change was 0 I think it's unlikely to help you in terms of finding a vector where the directional derivative is zero.
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    (Original post by DFranklin)
    To use LaTeX on here, enclose your LaTeX code in [tex] {stuff} [/tex].

    e.g. [tex]\dfrac{-b \pm \sqrt{b^2-4ac}}{2}[/tex] will give \dfrac{-b \pm \sqrt{b^2-4ac}}{2}


    Even if you can't use LaTeX, have you not heard of paragraphs? Full stops? Capitalisation?

    Sorry to snipe on presentation, but what you've written is incredibly difficult to read.

    As far as the actual question, it's kind of unclear what you've done so far. What are your thoughts on finding the tangent plane?
    For the last bit, I have no idea what vector you mean by $1/3$ so it's hard to comment, but unless the rate of change was 0 I think it's unlikely to help you in terms of finding a vector where the directional derivative is zero.
    Sorry about this I didn't really think to check on the latex used on this website I just assumed it would be like others I had also completely forgot to make paragraphs hopefully it is now easier to read
    Anyway to find the tangent plane to the surface I calculated nabla phi at the point p then worked out a normal to the surface at the same point p but I'm unsure what to do next, do I need to use the dot product with the normal and another vector say

    And for the last bit I mean I calculated a directional derivative which was 1/3 at the point p
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    (Original post by Scary)
    Sorry about this I didn't really think to check on the latex used on this website I just assumed it would be like others I had also completely forgot to make paragraphs hopefully it is now easier to read
    Anyway to find the tangent plane to the surface I calculated nabla phi at the point p then worked out a normal to the surface at the same point p but I'm unsure what to do next, do I need to use the dot product with the normal and another vector say
    Unless I am completely forgetting how to do this, \nabla \phi is already normal to the surface (it's the vector along which phi changes most quickly). So you have a vector n normal to the plane, you now just need to find the appropriate constant d so that the plane r.n = d passes through the given point on the surface.

    And for the last bit I mean I calculated a directional derivative which was 1/3 at the point p
    You're going to need the directional derivative to be 0. Which just means you need to find a vector in the xy plane orthogonal to \nabla \phi
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    (Original post by DFranklin)
    Unless I am completely forgetting how to do this, \nabla \phi is already normal to the surface (it's the vector along which phi changes most quickly).
    This is indeed how to approach this.

    So you have a vector n normal to the plane, you now just need to find the appropriate constant d so that the plane r.n = d passes through the given point on the surface.
    and to do this bit, note that:

    a) for any vector \vec{v} lying in the plane, \vec{v} \cdot \nabla \phi = 0
    b) the general position vector of a point in the plane is (x,y,z)
    c) the point with position vector (2,1,\pi/2) is also in the plane
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    (Original post by atsruser)
    This is indeed how to approach this.



    and to do this bit, note that:

    a) for any vector \vec{v} lying in the plane, \vec{v} \cdot \nabla \phi = 0
    b) the general position vector of a point in the plane is (x,y,z)
    c) the point with position vector (2,1,\pi/2) is also in the plane
    Hi thanks for your response in the question it is asking to find the tangent plane to the surface phi=4 so would it be nabla Phi so would it be each term of the nabla phi evaluated at the point so the equation of the tangent would be dphi/dx(X-xo)+dphi/dy(y-yo) +dphi/dz(z-z0)=4 or 0? Then for where X0 and y0 z0 would be coordinates
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    For the equation of the tangent surface at phi=4 at the point (2,1,pi/2) I got the equation X+2y+2z=4+pi is this right?
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    Also how would one go about finding a such vector that is orthogonal to nabla phi, is there an infinite amount? And if so would 2i-j be a sufficient vector?
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    (Original post by Scary)
    Hi thanks for your response in the question it is asking to find the tangent plane to the surface phi=4 so would it be nabla Phi so would it be each term of the nabla phi evaluated at the point so the equation of the tangent would be dphi/dx(X-xo)+dphi/dy(y-yo) +dphi/dz(z-z0)=4 or 0? Then for where X0 and y0 z0 would be coordinates
    Sorry, I can't really think about this now, but I've just noticed something that looks a bit odd: the point (2,1,\pi/2) doesn't lie on the surface \phi=4 - have you copied the question correctly?

    [Aside: please use latex - I rarely respond to questions that don't use it when it's needed]
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    (Original post by atsruser)
    Sorry, I can't really think about this now, but I've just noticed something that looks a bit odd: the point (2,1,\pi/2) doesn't lie on the surface \phi=4 - have you copied the question correctly?

    [Aside: please use latex - I rarely respond to questions that don't use it when it's needed]
    Hi thanks for your response and yeah I'm 100% sure that I copied the question correctly does the point not lie on the Phi(X,y,z)=4?
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    (Original post by Scary)
    Hi thanks for your response and yeah I'm 100% sure that I copied the question correctly does the point not lie on the Phi(X,y,z)=4?
    Well, when x,y,z have those values, what does phi equal? Is it 4?

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    (Original post by DFranklin)
    Well, when x,y,z have those values, what does phi equal? Is it 4?

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    Ahh I just checked , il email my tutor informing him
 
 
 
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