Math 280 homework notes
- Section 10.1 #26
- You have found a specific point that satisfies the given
condition. There are many others. You should fit the set of
all points that satisfy the condition.
- Give some indication of how you arrive at this result.
- You seem to have deduced this result geometrically (which is
fine). You should also be able to deduce or prove this
algebraically.
- Section 9.4 #28
- Draw the branches of the hyperbola so that they asymptote to
the relevant lines.
- Section 10.6 #42
- The surface is a hyperboloid of two sheets. To give a more
complete verbal description, you should also state that the
central axis is the y-axis and that the cross-sections
perpendicular to the central axis are circles.
- In drawing a hyperbola, including asymptote lines will help
you get a more accurate sketch.
- The y=0 cross-section is empty. For values
y0<-1 and y0>1, each
y=y0 cross-section is a circle.
- Section 12.2 #56
- A table of values gives useful evidence. You can give more
a more complete analysis. For example, we can first analyze the path
\(y=0\) to get
\[
\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}
=\lim_{(x,0)\to(0,0)}\frac{x^2-0^2}{x^2+0^2}
=\lim_{x\to 0}\frac{x^2}{x^2}
=\lim_{x\to 0}1
=1.
\]
Now analyze the path \(x=0\) to get
\[
\qquad
\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}
=\lim_{(0,y)\to(0,0)}\frac{0^2-y^2}{0^2+y^2}
=\lim_{y\to 0}\frac{-y^2}{y^2}
=\lim_{y\to 0}-1
=-1.
\]
Since these two path limits differ, the limit does not exist.
(Note that there are many other paths that can be analyzed here.)
- Section 12.3 #46
- The first partial derivatives simplify considerably:
\[
\frac{\partial s}{\partial x}
=\frac{-y/x^2}{1+y^2/x^2}
=-\frac{y/x^2}{1+y^2/x^2}\frac{x^2}{x^2}
=-\frac{y}{x^2+y^2}
\]
and
\[
\frac{\partial s}{\partial x}
=\frac{1/x}{1+y^2/x^2}
=\frac{1/x}{1+y^2/x^2}\frac{x^2}{x^2}
=\frac{x}{x^2+y^2}
\]
Computing the second derivatives is much easier with these
simplified expressions.