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The solution i have envisioned is that I need to find the point of tangency and the vector that defined as the crossproduct of $R_x$ $R_y$ to be able to solve for the tangent plane where $u=u_0$ and $v=v_0$.

after obtaining a tangent plane equation in terms of $u_0$ and $v_0$, i should make $x=0$ and $z=0$ to be able to get the points that would be parallel to the y axis.

I tried this and got stuck with too many terms wherein i could only assume i made a wrong strategy regarding the solution of this problem.

What should be my next (first) step

*Note R is a Vector

Ross Henderson
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  • So $S$ is a parametric surface parametrized by $u$ and $v$. A normal to the tangent plane is given by $\mathbf{n}=\mathbf{r}_u \times \mathbf{r}_v$. (This is not a unit vector in general.) That plane is parallel to the $y$ axis if it contains some line parallel to the $y$ axis i.e. if $\mathbf{j} \cdot \mathbf{n}=0$ which seems like it is probably a fairly straightforward algebraic equation. Maybe you should write some more details? Also I assume there is some typo in the second component of your $R$. – Ian Oct 15 '20 at 21:26
  • What is $S$? And is R really a map to $\mathbb R^2$? – Arctic Char Oct 15 '20 at 21:28
  • Indeed there was a typo! but i have now fixed it. but! when I use $r_u$ X $r_v$ there are no terms that cancel out and i am left with this bulky equation that does not fit with the problem. Could i perhaps have made a miscalculation with my arithmetic? – Ross Henderson Oct 15 '20 at 22:53
  • As I said what matters is just $\mathbf{j} \cdot \mathbf{n}$, i.e. the $y$ component of $\mathbf{n}$, which actually does not look to be particularly complicated. There is a $e^{u^2+v^2}$ on both sides of the equation that you can divide out (since it isn't zero) and you are then left with some straightforward relationship between $u$ and $v$. If you just focus on $\mathbf{j} \cdot \mathbf{n}$ and ignore the other components then maybe you won't get overwhelmed by the algebra. – Ian Oct 15 '20 at 23:15

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It is asked for the plane to be parallel to the $y$ axis. This means the plane contains some line with direction vector $\mathbf{j}$ which means the normal is perpendicular to $\mathbf{j}$, i.e. its $y$ component should be zero. From what you said, I think you were trying to choose the normal to be parallel to $\mathbf{j}$ which is not correct.

In this case the equation for the $y$ component of the normal being zero is very simple, at least once you divide it through by $e^{u^2+v^2}$.

Ian
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  • Thank you so much for the help! I apologize for my slow pick up ;(( From what I followed, I would need to equate the J component of the defined n to zero. and after simplifying a straight forward $ 2v_0 + 4u_0 v_0 = 0$ however how would i be able to extract the ordered pairs from such an equation since there are no x and y variables? – Ross Henderson Oct 16 '20 at 00:30
  • @TerrenceMatthews They're asking for you to do it all in the $(u,v)$ plane anyway. So it suffices to just find all solutions to that equation. It seems to me that there are infinitely many, but that's fine, you can describe the entire solution set. – Ian Oct 16 '20 at 00:36
  • Oh!! so the solution is is already defined by the equation $2v_0 + 4u_0 v_0 = 0$? I see now i had trouble because i was forcing my solutions to have x and y variables to find a specific value for the ordered pairs :(( thank you so much!!! you are a god sent ;(( – Ross Henderson Oct 16 '20 at 00:42
  • Hi if it's not too late, I was told to further solve my equation. You said that the solution had infinitely many solutions but when I further simplified i got the values $u_0 = -1/2$ and $v_0 = 0$. Aren't these the only solutions for the problem? – Ross Henderson Oct 18 '20 at 01:31
  • @TerrenceMatthews If $v_0=0$ then any value of $u_0$ gives a solution; if $u_0=-1/2$ then any value of $v_0$ gives a solution. – Ian Oct 18 '20 at 02:02