Course Notes for Calculus III

I start with \(\theta = 0\text{,}\) and then I see what happens to the radius as \(\theta\) increases or decreases. As the angle grows, the radius grows quadratically. This should give a spiral going out at the angle increases, and a mirrored spiral also going out as the angle decreases (since the negative angle will give a positive radius when squared). The double spiral result is shown in Figure 5.3.1

I could look for simplifications if I wished to, but nothing obviously makes the equations more succinct as far as I can tell.

The locus only depends on \(r\text{,}\) so \(\theta\) can take any possible value. In polar coordinates, this should produce circles: shapes where the radius is fixed by the angle is arbitrary. This is a quadratic in the radius with roots \(r=2\) and \(r=5\text{,}\) so there are two cirles, one with radius 2 and one with radius 5.

There may be simplifications that can be made, but I don't see anything that would obviously make the locus more succinct.

First I notice that the left side is positive, so the right side must also be positive; therefore, \(\theta\) is restricted to positive angles. Starting at \(\theta = 0\) and rotating counterclockwise, the radius grows with the angle as \(r = \sqrt{\theta}\text{.}\) Therefore, I expect a spiral with curves slowly getting closer and closer to each other as it expands, since the square root function is slower than linear growth (which would be an evently spaced spiral).

I could try to simplify, but I don't see any obvious simplifications that would make the equation more succinct.

I can re-arrange this equation into \(r = -/theta\text{,}\) which is the negative of the archimedian spiral. Since the radius must be positive, the angle must be negative, so we get a clockwise archimedian spiral. Since the growth is linear, the arms of the spiral are equally spaced.

I could try to simplify, but I don't see any changes that make the equation substantially more succinct.

This locus only has one point, which is the origin. This point happens when \(\theta = -1\) and \(r=0\text{.}\) For all other values of \(/theta\text{,}\) the right side is negative. The the radius cannot be negative, there are no possible radius values that satisfy the equation. I can still translate this locus, thought the cartesian locus is still a complicated way of simply referring to the point \((0,0)\text{.}\)

This is nicely set up, since the cartesian locus includes the term \(x^2 + y^2\text{,}\) which directly translates to the polar term \(r^2\text{.}\) Making that replacement gives the polar locus.

This cartesian locus is in a convenient form. The artcangent term is precisely the cartesian equivalent of the polar \(\theta\text{.}\) If I write \((-x^2 - y^2\) as \(-(x^2+y^2)\text{,}\) I can also remember the fact that \((x^2 + y^2)\) is equivalent to the polar \(r^2\text{.}\) Making those two replacements produces the polar locus.

Here I simply replace \(x\) amd \(y\) with their polar equivalents.

I could try to simplify, but I don't see any particular simplifications that make the equation any more succinct.

Here I simply replace \(x\) amd \(y\) with their polar equivalent.

I could try to simplify if I wanted. One might be tempted to cancel an \(r\) term from both sides of the equations. That would be mostly fine, but there is one tricky subtlety. With the equation as written, the origin (with \(r=0\) and \(\theta\) undetermined) satisfies the equation; if I cancel \(r\) from both sides, I preserve the rest of the locus but I lose the origin. Therefore, the simplification of cancelling out \(r\) needs to be accompanied by a note for the case \(r=0\) and the inclusion of the origin.

1. A function defined on the set of people which outputs each persons age, height and weight. This is a vector valued function: for each person, three values are given as output. This is not a geometric vector, but it can still be represented as a triple: (age, height, weight). That's the algeraic definition of a vector.

2. A function which takes a location described by lattitude and longitude and returns the altitude of that point. This is not a vector-valued function. It depends on multiple inputs (lattitude and input) but outputs a simple value: the altitude. Altitude is a number, not a vector.

3. A function defined on the positive real numbers which outputs the square root, cube root and fourth root of the number. This is a vector valued function. If \(x\) is the variable, I can write \(f(x) = \left( \sqrt{x}. \sqrt[3]{x}, \sqrt[4]{x} \right)\text{.}\) That triple can be interpreted as a vector.

4. A function of time which gives the local windspeed and direction at the top of a sailbost mast as the sailboat moves. This is a vector valued function. It has a single input (time), but outputs windspeed and direction, which is a vector quantity.

5. A function of time which gives the loudness (in decibels) of a recorded music track played on a particular sound system. This is not a vector-valued function. This is an ordinay single-variable functions: the input is a single value (time) and the output is also a single value (decibels).