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}}.


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