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Let $S$ be given by the vector valued function $\vec r(u,v)=\langle u\cos v,u\sin v,v\rangle$. Find the tangent plane to the surface at $\vec r\left(1,\frac {\pi}{4}\right)$.

What I did:

$$\vec r_u=\langle\cos v,\sin v, 0\rangle, \vec r_v=\langle-u\sin v,u\cos v,1\rangle$$

$$\vec r_u=\left\langle\frac {\sqrt2}{2},\frac {\sqrt2}{2}, 0\right\rangle, \vec r_v=\left\langle-\frac {\sqrt 2}{2}, \frac {\sqrt2}{2}, 1\right\rangle\text{ at }\left(1, \frac {\pi}{4}\right)$$

$$\vec r(1,\frac {\pi}{4})=\left\langle\frac {\sqrt 2}{2},\frac {\sqrt 2}{2}, \frac {\pi}{4}\right\rangle$$

But how should I continue? Please help, thanks.

Michael Hardy
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user95087
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1 Answers1

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Make use of the fact that a plane in $\mathbb{R}^{3}$ is determined by a base point $P(x, y, z)$ and a normal vector $\vec{n}= \left\langle a, b, c\right \rangle$ attached at the base point.

In this case, your base point is clear: $\vec{r}\left(1, \frac{\pi}{4}\right)$.

The key connection to make is that both $\vec{r}_{u}\left(1, \frac{\pi}{4}\right)$ and $\vec{r}_{v}\left(1, \frac{\pi}{4}\right)$ are tangent to your surface at the point $\vec{r}\left(1, \frac{\pi}{r}\right)$. Thus, in order to get your normal vector at the point $\vec{r}\left(1, \frac{\pi}{4}\right)$, it suffices to take $$ \vec{n}\left(1, \frac{\pi}{4}\right) = \vec{r}_{u}\left(1, \frac{\pi}{4}\right) \times \vec{r}_{v}\left(1, \frac{\pi}{4}\right), $$ where $\times$ indicates the usual Euclidean cross product.

You should now be able to proceed by writing the equation of the plane with the appropriate base point and normal vector.

I hope that this helps.

THW
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  • Thanks! If I use the form $z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$, and using the cross product $<\frac {\sqrt2}{2}, -\frac {\sqrt2}{2},1>$, I got the final answer $\frac {\sqrt2}{2}(u-1)-\frac {\sqrt2}{2}(v-\frac {\pi}{4})=r(u,v)-1$. Does $r_0(u,v)$ on the RHS look right? I got 1 because from the original function the k term equals to $v$. – user95087 Apr 05 '14 at 23:22
  • @user95087 No, you have to distinguish between the case when your surface is parameterized by the graph of a function (i.e., $z = f(x, y)$) and when your surface is a general parametrized surface. If the surface is given by the graph of a function $z = f(x, y)$, then we do have a generic form for the equation of then tangent plane at a particular point $$z = f_{x}(x_0)(x - x_0) + f_{y}(y_0)(y - y_0) + f(x_0, y_0),$$ but this case is a special case of the parametrized surface, not the generic case. – THW Apr 05 '14 at 23:27
  • You need to go back further in multivariable calculus to when you were writing the equation of a plane: You simply need a point $P(x_0, y_0, z_0)$ and a normal vector $\vec{n} = \left \langle a, b, c\right \rangle$.

    The equation of a plane containing the point $P(x_0, y_0, z_0)$ with normal vector $\vec{n} = \left \langle a, b, c\right \rangle$ is $$\vec{n} \cdot \left \langle x - x_0, y - y_0, z - z_0 \right \rangle =0,$$

    or equivalently $$ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0.$$

     – THW Apr 05 '14 at 23:33
  • Now, in this situation, you have the point $ \vec{r}\left(1, \frac{\pi}{4}\right) = \left \langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{\pi}{4} \right \rangle$ and the normal vector coming from the indicated cross product (which I didn't calculate). You should able to make use of the comment above to complete the problem. – THW Apr 05 '14 at 23:36