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Find all points on the paraboloid $z=x^2+y^2$ where tangent plane is parallel to the plane $x+y+z=1$ and find equations of the corresponding tangent planes. Sketch the graph of these functions.


I have its answer. I don't really understand such type of questions. And I am really willing to learn. Also I added its answer as a picture. Please teach me how to solve.

the image is its answer

Please help me. Thank you so much:)

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  • $\begingroup$ What and where is the answer you have? $\endgroup$
    – DonAntonio
    Commented Jul 14, 2013 at 19:57
  • $\begingroup$ I have its solution manual. I added the answer as a picture @DonAntonio do I make a mistake? $\endgroup$
    – 1190
    Commented Jul 14, 2013 at 19:59
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    $\begingroup$ Since I don't know what's your mathematical background and all that looks pretty straightforward, you must tell us what is it that you don't understand... $\endgroup$
    – DonAntonio
    Commented Jul 14, 2013 at 20:02
  • $\begingroup$ Perhaps it is easier to learn several variables if you try to formulate the theorems/problems in one variable. $\endgroup$ Commented Jul 14, 2013 at 20:08
  • $\begingroup$ In fact, I solve such tangent plane question first time. And I could not produce any idea to solve it, thus, I looked at its solution. And in general, I dont understand. In the solution, first of all, a function is defined as $f(x,y,z)=z-x^2-y^2$ and $\nabla f(x,y,z)=-2xi-2yj+k$ is found. That's the normal direction is (-2x,-2y,1) After there, I took derivatives of $x+y+z=1$ and I got (1,1,1). And so as to find parallel points, $(1,1,1)=(-2x,-2y,1)$ thus, $x=-1/2=y$ and $z=1/2$ that is paralel points are $(-1/2,-1/2,1/2)$ right? And then, what i need to do? @DonAntonio $\endgroup$
    – 1190
    Commented Jul 14, 2013 at 20:11

2 Answers 2

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To get a normal vector to the paraboloid at a point (x,y,z), we can take the gradient $\nabla f(x,y,z)=-2xi-2yj+k$. Since we want the tangent plane at the point to be parallel to the plane $x+y+z=1$, the normal vector $\nabla f(x,y,z)=-2xi-2yj+k$ has to be parallel to the vector $i+j+k$ (since this is a normal vector to $x+y+z=1$). This means that $-2xi-2yj+k$ must be a constant multiple of $i+j+k$, so $-2xi-2yj+k=c(i+j+k)$ for some constant c. Then $-2x=c$, $-2y=c$, and $1=c$, so $x=-1/2$ and $y=-1/2$. Therefore $z=x^2+y^2=1/4+1/4=1/2$ at the point of tangency, and the tangent plane has equation $x+y+z=-1/2$ at this point.

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  • $\begingroup$ Yes!!! Everything is clear:) the answer I want is this. Thank you so much:)) $\endgroup$
    – 1190
    Commented Jul 14, 2013 at 20:56
  • $\begingroup$ Sorry, but I want to ask one thing. Why $c=1$? $\endgroup$
    – 1190
    Commented Jul 14, 2013 at 22:01
  • $\begingroup$ I guess, accourding to sign of $z$ $\endgroup$
    – 1190
    Commented Jul 14, 2013 at 22:05
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    $\begingroup$ You're correct; setting the coefficients of k equal gives that $c=1$. $\endgroup$
    – user84413
    Commented Jul 14, 2013 at 22:11
  • $\begingroup$ Okay, now got it. thanks :) $\endgroup$
    – 1190
    Commented Jul 14, 2013 at 22:13
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You can obtain a parametrisation of your submanifold by viewing it as a graph of a function, and the parametrisation's partial derivates give a basis of the tangent space at each point.

Let the paraboloid, as a submanifold, be denoted by $S$. We have $f(x,y) = (x, y, x^{2} + y^{2})$ as our parametrisation, hence $T_{f(x,y)}S = <(1,0,2x)^{t}, (0,1,2y)^{t}>$.

You now get exactly two unit-length vectors that span the orthogonal complement of $T_{f(x,y)}S$. Chose such a vector, let's call it $\nu$. Now all you have to do is find the plane you want your $T_{f(x,y)}S$ to be parallel to, and check that the line defined by $c(t) = f(x,y) + t\nu$ meets the plane orthogonally.

Edit:

For the last part, we use that in euclidian space, a hyperplane is parallel to another hyperplane iff there exists a line that meets both orthogonally.

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  • $\begingroup$ Thank you for helping but you explained at too advanced level. I am sure, it is true. But I am beginner of calculus. :) $\endgroup$
    – 1190
    Commented Jul 14, 2013 at 21:01
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    $\begingroup$ I am sorry, I should have taken this into account. I think the other answer is more useful to you. In this example, you can visualise it fairly easily, however. Just think of the tangent plane as a sheat of paper tangential at one point to your paraboloid. One orthogonal vector is then that one vector that goes out of this one tangential point and stands, you guessed it, in a right angle on the plane. Now if you want to find a plane that is orthogonal to that sheat of paper, you extend that orth. vector to a line and look at all the planes this lines is orthogonal to as well. That's all. $\endgroup$ Commented Jul 14, 2013 at 21:18
  • $\begingroup$ Not need to apologize :) I read your answer carefully:) thanks. $\endgroup$
    – 1190
    Commented Jul 14, 2013 at 21:31

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