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I have a surface parametrised by $$ x = u\sin^3v,\quad y=u\cos^3v,\quad z=u$$ and I'm trying to find the tangent plane at $(x_0, y_0, z_0) = (\sqrt2, -\sqrt2, 4)$ using this formula: $$(x-x_0, y-y_0, z-z_0)\cdot \mathbf n(u_0,v_0)=0$$

I found the normal $ \mathbf n $ but am having difficulty finding $v_0$.

My attempt:

$ x_0=u_0\sin^3v_0 $

$ \sin^3v_0 = \frac{\sqrt2}{4} $

Is this the correct method?

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  • $\begingroup$ the normal vector depends on $v_0$. When you managed to find $\mathbf{n}$ without using $v_0$ then it seems like you don't need to know what $v_0$ is. $\endgroup$ – Nathanael Skrepek Oct 10 '17 at 11:28
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You know that: \begin{align*} \frac{\sqrt{2}}{4} = \sin^3(v)\\ \frac{-\sqrt{2}}{4} = \cos^3(v) \end{align*} Hence we know that $\tan(v)^3 = -1 \Leftrightarrow \tan(v)=-1$. The two solutions to this equation is $-\frac{\pi}{4}$ and $\pi - \frac{\pi}{4}$, where the latter is the solution that you are looking for. The rest you already explained how to do.

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  • $\begingroup$ Thank you very much Martin :) I'm slowly learning to pick out things like this still! $\endgroup$ – Minish Oct 10 '17 at 11:40

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