# HW 4 Solutions

Problem 1.  It turns out that these two lines do not intersect.  This can be confirmed by setting their parametric functions equal, specifically by setting up the equations

$x(t) = x(s)$

$y(t) = y(s)$

$z(t) = z(s)$

When one attempts to solve all equations at once, impossible equations result!  This tells us that the two lines never intersect.

Because these lines are in $\mathbb{R}^3$ it could be that the lines are “skew” or it could be that the lines are parallel.  If its the second case, then there is a plane that the two lines create.  However, if the lines are not parallel (and they don’t intersect), then they do not lie on a shared plane.

The direction vector for the first line is $\vec{v}_1 = \langle 2, -3, 4 \rangle$ and the direction vector for the second line is $\vec{v}_2 = \langle 2, 2, 1/2 \rangle$.  Since these vectors are not parallel, the given lines are not parallel.

This was, as it turns out, a trick question: the two lines do not lie on a plane!

Problem 2.  (a) The domain is all possible pairs of real numbers, i.e. the entire set $\mathbb{R}^2$.  This function does take a square root, but the expression in the square root is never negative (no matter what values for $(x,y)$ we plug in), and so does not have a restricted domain.

(b) As discussed in class, this level set consists of infinitely many circles — each one centered at $(0,0)$ and with radii of the form $\pi/2 + n\,2\pi$ where $n$ is a positive whole number.  Here is a picture of two of these circles:

(c ) For the $x = 0$ slice, we see a curve in the $yz$-plane given be $z = \sin(\sqrt{y^2}) = \sin(|y|$.  Here is a picture:

When $x = 1$, we see a curve in the $x=1$ plane which is given by $z = \sin(\sqrt{1+y^2})$.  Here is a picture:

(d)  Here is a (computer-generated) plot of the surface graph $z = \sin(\sqrt{x^2+y^2})$.  See if you can see any of the circles in the $z = 1$ slice from part (b) or if you can see the curves from part (c ) in the $x = 0$ and $x = 1$ planes:

(e) The equation $z = f(x,y)$ converts to $z = g(r)$ where $g(r) = \sin r$ when using cylindrical coordinates.  The fact that $\theta$ is absent makes visual sense because, as depicted in the graph above, the surface is rotationally symmetric about the $z$-axis.  In other words, we could have viewed this surface as what results when we take the graph of $z = \sin r$ and revolve it about the $z$-axis.

Problem 3.  We have

$x = r\cos\theta = \rho\sin\varphi\cos\theta$

$y = r\sin\theta = \rho\sin\varphi\sin\theta$

$z = z = \rho\cos\varphi$

and we also have

$r = \sqrt{x^2+y^2}$

$\tan\theta = y/x$

This wasn’t asked, but its also worth mentioning that

$\rho = \sqrt{x^2+y^2+z^2}$.

Problem 4.  The equation

$\displaystyle \rho = \sec\theta\csc\varphi$

is strange-looking and is expressed in spherical coordinates.  If we rearrange the terms in this equation, though, we can get it to look like the following equation:

$\displaystyle \rho \cos\theta\sin\varphi = 1$.

In spherical coordinates this equation is not very clear, but note that the expression on the left in the equation above is precisely the expression for $x$.  That means that, in rectangular coordinate,s this equation becomes

$x = 1.$

The picture for this level set is a (vertical) plane with normal $\vec{n} = \langle 1, 0, 0 \rangle$ and that passes through the point $\vec{x}_0 = (1, 2, 4)$ (for example).

Problem 5.  The area is given by

$\displaystyle \left| \langle 1, 2, 1 \rangle \, \times\,\langle 0, 1, 1\rangle \right| = \left|\langle 1, -1, 1 \rangle\right| = \sqrt{3}$

Problem 6.  The volume is given by

$\left| \, \left(\vec{v}\times\vec{w}\right)\,\cdot \vec{z}\right| = 2$.

Problem 7.

The partial derivative with respect to $x$ is given by

$\displaystyle \frac{\partial f}{\partial x} = \frac{\sqrt{x^2+y^2} - x^2\left(x^2+y^2\right)^{-1/2}}{x^2+y^2}.$

The partial derivative with respect to $y$ is given by

$\displaystyle \frac{\partial f}{\partial y} = \frac{-xy\left(x^2+y^2\right)^{-1/2}}{x^2+y^2} = -\frac{xy}{(x^2+y^2)^{3/2}}.$