Problem 1
1. Let \(C\) be the curve with parameterization \(\mathbf{r}(t) = (\cos(t),\sin(t),t)\), \(0\le t\le 2\pi\), so that \(C\) is that portion of the helix of radius one from \((1,0,0)\) to \((1,0,1)\). Find a second parameterization of \(C\) and use this to create a reparameterization of \(C\). Then, check that \(\displaystyle\int_C x+y+z\, ds\) is independent of the two parameterizations. (6 points)
Solution. We see that another parameterization of the curve \(C\) is given by \(\mathbf{s}(t) = (\cos(2t),\sin(2t),2t)\) for \(0\le t\le \pi\). We then calculate the two integrals. Note that \(\mathbf{r}'(t) = (-\sin(t),\cos(t),1)\), so \(||\mathbf{r}'(t)|| = \sqrt{\sin^2(t)+\cos^2(t)+1} = \sqrt{2}\). Then,
\[ \int_C x+y+z\, ds = \sqrt{2}\int_0^{2\pi} \cos(t)+\sin(t)+t\, dt = \sqrt{2}\left[\sin(t)-\cos(t)+\frac{t^2}{2}\right]_0^{2\pi} = 2\sqrt{2}\,\pi^2. \]
On the other hand, one has \(\mathbf{s}'(t) = (-2\sin(2t), 2\cos(2t), 2)\), giving \(||\mathbf{s}'(t)|| = \sqrt{8} = 2\sqrt{2}\). Using this, one has \[ \int_C x+y+z\, ds = 2\sqrt{2}\int_0^\pi \cos(2t)+\sin(2t)+2t\, dt = 2\sqrt{2}\left[\frac{1}{2}\sin(2t)-\frac{1}{2}\cos(2t)+t^2\right]_0^\pi = 2\sqrt{2}\,\pi^2. \]
Problem 2
2. For the sphere \(S: x^2+y^2+z^2=4\), find the plane tangent to \(S\) at \(P=(1,1,\sqrt{2})\). (4 points)
Solution. To find the tangent plane, we first parameterize our surface. Since this is a sphere, we use spherical coordinates with \(\rho=2\). This gives us the parameterization \(S(\theta,\phi) = (2\cos(\theta)\sin(\phi), 2\sin(\theta)\sin(\phi), 2\cos(\phi))\). As \(z=\sqrt{2}=2\cos(\phi)\), we have the corresponding point \((\pi/4,\pi/4)\) in the \((\theta,\phi)\) plane. Then, the tangent vectors are
\[ T_\theta = (-\sqrt{2}\sin(\theta), \sqrt{2}\cos(\theta), 0) \to T_\theta(\pi/4) = (-1,1,0) \]
and \[ T_\phi = (\sqrt{2}\cos(\phi), \sqrt{2}\cos(\phi), -2\sin(\phi)) \to T_\phi(\pi/4) = (1,1,-\sqrt{2}). \]
This allows us to find a normal vector to the plane with a cross product. \[ \begin{vmatrix} \vi & \vj & \vk \\ -1 & 1 & 0 \\ 1 & 1 & -\sqrt{2} \end{vmatrix} = (-\sqrt{2}, -\sqrt{2}, -2). \]
Note that we expect this to point up since we were at \((1,1,\sqrt{2})\), so we flip the sign. Finally, we take the dot product of the normal vector with \((x-1, y-1, z-\sqrt{2})\) to get the plane \[ \sqrt{2}(x-1) + \sqrt{2}(y-1) + 2(z-\sqrt{2}) = 0 \]
or equivalently \[ x+y+\sqrt{2}z = 4. \]