I am given the following:

"Consider the curve

$$ g: t \mapsto \begin{bmatrix} t\cos{t}\\ t\sin{t}\\ t^{2} \end{bmatrix} $$

and the paraboloid

$$ F(x,y,z) : z = x^{2} + y^{2} $$

containing $g$. Verify that

$$ g'(t) \cdot \nabla F = 0 $$

where the dot repredents the dot product."

I don't know what $\nabla F$ means, nor can I come up with a possible meaning for it. I know what a gradient is and how to compute it, I can calculate the partial derivatives of the arguments(?) $x,y,z$ of $F$ with respect to one another but I wouldn't know how to put them together into a vector. I've had a few ideas, but none of them seemed quite right.

EDIT: I am getting the impression that this question was badly phrased. Anyone has any idea of what could have been initially intended?

EDIT2: For the record, the matter was settled. This answer explains it also. Let $F(x,y,z) = x^{2} + y^{2} - z$. Then, for any $(x_0,y_0,z_0)\in\mathbf{R}^{3}$ with $F(x_0,y_0,z_0) = C $ for some $C\in\mathbf{R}$ we have that the gradient of $F$ at $(x_0,y_0,z_0)$ is orthogonal to the plane tangent to the surface $S=\{(x,y,z)\in\mathbf{R}^3:F(x,y,z) = C\}$ at the point $(x_0,y_0,z_0)$ (in a manner analogous to the gradient being perpendicular to contour lines of a two variable function). Under this definition one can indeed verify that $g'(t) \cdot \nabla F(t\cos{t},t\sin{t},t^{2})=0$ for any $t$.

  • $\begingroup$ What is written makes no sense to me because $g' \in \mathbb{R}^3$ whereas the gradient of $F$ (written $\nabla F$) is in $\mathbb{R}^2$. I have not heard of a dot product between elements of two different spaces. $\endgroup$ – Ahmed S. Attaalla Sep 27 '17 at 0:54
  • $\begingroup$ By $f_{x}$ do you mean $\frac{\partial}{\partial x} F$? I am not sure how to compute this, $F$ is no function. Would $\frac{\partial}{\partial x} F$ be equivalent to $\frac{\partial}{\partial x} z$? (This was in response to your answer before the edit. Re your edit: yes, my thoughts exactly -- hence my confusion. The question however is given as I wrotr here, almost verbatim) $\endgroup$ – wet Sep 27 '17 at 0:55

\begin{align} g(t) &=t\,(\cos t, \sin t, t)\\ g'(t) &=(\cos t, \sin t, t) +t\,(-\sin t, \cos t, 1)\\ &=(\cos t - t\sin t, \sin t + t\cos t, 2t) \end{align}

For $F=x^2+y^2-z$, we have

$$ \nabla F = (2x, 2y, -1) =(2t\cos t, 2t\sin t, -1) $$


\begin{align} g'(t)\cdot \nabla F &= 2t\cos t\times(\cos t-t\sin t)+2t\sin t\times(\sin t +t\cos t) -2t\\ &= 0 \end{align}

  • $\begingroup$ Why is the third component of $\nabla F$ equal to $0$? $\endgroup$ – wet Sep 27 '17 at 4:26
  • $\begingroup$ This is wrong. The idea behind this problem is to first express your surface as a zero level surface i.e $F(x,y,z) = x^2+y^2-z$. And now, since $g(t)$ parametrizes the paraboloid, we have $F(g(t)) = 0$. Now just apply the chain rule. $\endgroup$ – Faraad Armwood Sep 27 '17 at 13:44
  • $\begingroup$ Given any surface, you can express it locally as the graph of smooth function. Since we considering our surface to be embedded in $\mathbb{R}^3$, one of $x,y$ or $z$ can be written as a smooth function on two variables of the others. Let us suppose $z = f(x,y)$. Then we can $G(x,y,z):=f(x,y) - z$ i.e we've expressed a surface patch $S$ as a zero level surface i.e if $(x,y,z) \in S$, then $G(x,y,z) = 0$. What's nice about this is that, when you parametrize the surface patch coordinates by a differentiable function $g(t)$ (I.F.T), then $G(g(t)) = 0$ and so by chain rule $\nabla G \cdot g' =0$. $\endgroup$ – Faraad Armwood Sep 27 '17 at 14:21
  • $\begingroup$ I.F.T means, Implicit function theorem. This shows us that such a $g$ exists. $\endgroup$ – Faraad Armwood Sep 27 '17 at 14:23
  • $\begingroup$ @Faraad, I don't quite get everything that you are saying, but I would like to understand it. Is what you're saying the same I mentioned in my second edit? A few questions: "given any surface, we can express it locally as the graph of a smooth function". In our context, would our surface be the graph of the given $F(x,y,z)$, and "expressing it locally" be fixing one $F(x,y,z)=C$ for some $C$ and then looking at the graph of $x,y,z\mapsto x^{2}+y^{2}=z+C$? $\endgroup$ – wet Sep 27 '17 at 16:34

You just need to use implicit differentiation. You have $F(x,y,x^2+y^2=z)=0$ and $g(t)$ is a differentiable function whose coordinate function have this property i.e we can use them to parametrize a curve on the paraboloid i.e $F(g(t)) = 0$. Now apply the chain rule.

  • $\begingroup$ Yeah, you do! If you differentiate $F(g(t)) = 0$, it's still zero. And by chain rule, you get the result. In any case, I didn't need to do the whole problem. I has to leave something to the OP. $\endgroup$ – Faraad Armwood Sep 27 '17 at 13:47

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