Math 280 homework notes
- Section 10.1 #26
- You seem to have deduced this result geometrically (which is
fine). You should also be able to deduce or prove this
algebraically.
- 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.
- Section 10.1 #34
- Be careful with the directions of these inequalities. On and
outside the sphere of radius 1 is given by
x2+y2+z2≥1. On and
inside the sphere of radius 2 is given by
x2+y2+z2≤4.
- The two separate inequalities can also be written in one string as
1≤x2+y2+z2≤4
- Section 10.6 #42
- The y=0 cross-section is empty. For values y0<-1
and y0>1, each y=y0
cross-section is a circle.
- In drawing a hyperbola, including asymptote lines will help
you get a more accurate sketch.
- Section 10.2 #50
- To keep a distinction between coordinates of points and
components of vectors, be consistent in using the notation
of (a,b,c) for coordinates of points and
〈a,b,c〉 for components of vectors. The coordinates
of a point are the same numbers as the components of the vector
from the origin to that point. So, if a point has coordinates
(2/3,2/3,4/3), then the vector from the origin to that point
is 〈2/3,2/3,4/3〉 and vice versa.
- 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.
- Note that
\[
\frac{1}{x^2+y^2}=(x^2+y^2)^{-1}
\textrm{ but }
(x^2+y^2)^{-1}\neq x^{-2}+y^{-2}
\]
- Note that
\[
\frac{1}{A+B}\neq \frac{1}{A}+\frac{1}{B}
\]
- Section 12.7 #38
- The second-derivative test is not needed here since we
are after global minimum and maximum. So, we do not
need to classify each interior critical point. We only need to
compute the output for each interior critical point and add
that output to our list of outputs to compare at the end.
- For each piece of the boundary, we need to analyze the
boundary function. This includes finding boundary critical
points. For example, on the hypotenuse of the triangle, we
have the boundary function
b(x)=f(x,1-x)=8x2−6x+3. To find critical
points, differentiate to get b&prime(x)=16x−6. Setting
this equal to zero and solving, we find x=3/8. So, we
compute b(3/8)=f(3/8,5/8)=15/8 and add this to our list
of outputs to compare at the end.
- Applied Optimization #2
- Look for a simple and uniform way to express your
results. In this case, a good form for the minimizing
dimensions is
\[
x=\Bigl(\frac{2c}{a+b}\Bigr)^{1/3}V^{\,1/3}
\qquad\qquad
y=\Bigl(\frac{2c}{a+b}\Bigr)^{1/3}V^{\,1/3}
\qquad\qquad
z=\Bigl(\frac{a+b}{4c}\Bigr)^{2/3}V^{\,1/3}
\]
- You can use units to do a consistency check on your
results. The dimensions of the box should all be in units of
centimeters.
- Section 12.8 #10
- Note that the problem also asks you to compute the
maximum surface area.
- You can approach this problem in at least two ways, one
coordinate-free and the other using coordinates. Here's a
handout with solutions using each approach.
- Curve integration #10
- Be careful here: \(\sqrt{x^2}=|x|\).
- With the incorrect integrand
\(\displaystyle x\sqrt{1+\frac{4b^2}{a^4}x^2}\), this integral evaluates to
0. With the correct integrand
\(\displaystyle |x|\sqrt{1+\frac{4b^2}{a^4}x^2}\), the
integral is not 0. See Note 3.
- Here are two options for handling the absolute value in
\(\displaystyle
\int_{-a}^{a}|x|\sqrt{1+\frac{4b^2}{a^4}x^2}\,dx\):
- Split the interval into two pieces to get
\[
\int_{-a}^{a}|x|\sqrt{1+\frac{4b^2}{a^4}x^2}\,dx
=\int_{-a}^{0}(-x)\sqrt{1+\frac{4b^2}{a^4}x^2}\,dx
+\int_{0}^{a}x\sqrt{1+\frac{4b^2}{a^4}x^2}\,dx.
\]
- Note that the integrand is an even function and use
this symmetry to write
\[
\int_{-a}^{a}|x|\sqrt{1+\frac{4b^2}{a^4}x^2}\,dx
=2\int_{0}^{a}x\sqrt{1+\frac{4b^2}{a^4}x^2}\,dx.
\]
In either case, you can use a substitution to evaluate the
integral(s).