Power and periodic functions [1.9-10]

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Proportionality

Proportional: In what follows, $k$ is a non-zero constant. These statements all mean the same thing:

  • $y(x)=kx$
  • "$y$ is proportional to $x$"
  • $y \propto x$
  • "$y$ is a linear function of $x$, passing through the origin."

Inversely proportional: In what follows, $k$ is a non-zero constant. These statements all mean the same thing:

  • $y(x)=\frac{k}{x}$
  • "$y$ is inversely proportional to $x$"
  • $y \propto \frac{1}{x}$
  • "Whenever you double $x$, then $y\to\frac12y$."

Power function

$Q(x)$ is a power function of $x$ if: $$Q(x)=kx^p,$$ where

  • $k$ is the constant of proportionality and
  • $p$ (also a constant) is the power.

Example

  • The graph of $\color{green}{x^{1.5}}$ is between the graphs of $\color{red}{x^1}$ and $\color{blue}x^2$

  • If the power, $p$, is not an integer, $x^p$ is usually undefined for $x\lt 0$
  • As $x\to 0$, $x^p\to 0$. Compare this to any exponential, $\lim_{x\to 0} b^x\to b^0= 1$.

    Problems

    Skin area vs height

    The DuBois formula states that the fourth power of a person's surface area, $s$, is proportional to the cube of her or his height $h$. Find a formula for surface area as a function of height.

    How can this relationship be described in more colloquial terms?

    If a 70 kg person is 180 cm tall and has a surface area of 1.42 m^2, what is the proportionality constant?

    The planets


    Johannes Kepler analyzed the periods of the planets as they related to their distances, enunciating his "third law" in 1619, which involved a power low. This was before Isaac Newton's theory of gravity.

    This is another..."Given two points, find the equation" kind of problem.

    Here's an outline:

    1. Write out the general formula for a power function, using $T(D)$

      $$T=kD^p.$$

    2. Using the table, fill the values into the formula for the two planets that are farthest from each other. This will give you 2 equations.

      Jupiter's $T$ and $D$ tell us that $$11.87=k483.4^p$$ Now, you should go ahead and fill in the values for Mercury to get the 2nd equation...

    3. Now you can eliminate $k$: For example, you can solve one equation for $k$, and then plug in your solution to the other equation. Or use the "divide one equation by the other" approach.
    4. Now you'll have one equation involving one unknown, $p$. At this point you'll probably want to take the logarithm of both sides of your equation, and simplify a bit, and hopefully you'll be able to solve for $p$.

      It turns out that Johannes Kepler's analysis (1609) led him to believe that $p$ should be 1.5. Is this close to what you found??

      He wrote

      I first believed I was dreaming… But it is absolutely certain and exact that the ratio which exists between the period times of any two planets is precisely the ratio of the 3/2th power of the mean distance.

      translated from Harmonies of the World by Kepler (1619)
    5. For your final step: Now that you have a value for $p$ you can substitute that value back into one of your first two equations, and then you'll have one equation with one unknown, $k$, and you can solve for $k$.

    Assuming that period, $T$ is a power function of distance, $D$, then it can be written as: $$T(D)=k*D^p$$ To find the equation from two points, first, find $p$ like this: Write out the power function formula with the values filled into the formula for two of the points (we'll use Jupiter and Mercury), like this: $$11.87=k(483.4)^p$$ $$0.24=k(36.0)^p$$ We don't know $k$. But if we divide the two equations by each other it won't matter. Dividing the top one by the bottom one... $$\begineq \frac{11.87}{0.24}=&\frac{k(483.4)^p}{k(36.0)^p}=\frac{(483.4)^p}{(36.0)^p}\\ 49.46=&\left(\frac{483.4}{36.0}\right)^p=13.43^p\\ \ln(49.46)=&\ln(13.43^p)=p\ln(13.43)\\ \endeq$$

    The solution is: $$p=\frac{\ln(49.46)}{\ln(13.43)}=1.502$$

    Now, find the proportionality constant, $k$ like this: Plug the values from one of the points into the power function formula, together with our known value of $p$. Using Jupiter $$11.87=k(483)^{1.5}.$$ So, $k=11.87/(483^{1.5})=0.00112$. The full power function is $$T(D)=0.00112\,D^{1.5}.$$

    Periodic function

    Given a number $t$, travel that distance counterclockwise around the unit circle starting from the point (1,0). Call the resulting point (x,y). Then $\cos(t)=x$ and $\sin(t)=y$.


    In the diagram: a unit circle of radius $r=1$ has a circumference of $2\pi r=2\pi*1=2\pi$. The red arc runs 1/12 of the way around the circle, so the distance from (1,0) is $2\pi/12=\pi/6$. From the coordinates of the ending point, it appears that $\cos(\pi/6)=0.866$ and $\sin(\pi/6)=0.500$.


    A more familiar view of the unit circle, with a right triangle. The angle of the hypotenuse with the $x$ axis is $\pi/6$ Radians$\equiv 30{}^o$. The opposite side (sin) is $1/2$ and the adjacent side (cos) is $\sqrt{3/2}$.

    A graph of $\sin(t)$: The amplitude of this function is 1 and the period of this function is $2\pi$.

    Use transformations of the $\sin(t)$ to model periodic (repeating) data. $$f(t)=A\sin(B(t-h))+k.$$ [The cosine function works the same way. It can be viewed as a horizontally shifted sine function.]

    See my precalculus Notes on fitting data with sinusoids for pictures and examples. Summary: $$f(t)=A\sin\left[\frac{2\pi}{T}(t-h)\right]+k$$

    • $k$= vertical position of the midline =
      = $\frac{f_\text{max}+f_\text{min}}/2$= average "height" of the function.
    • $A$=amplitude = $f_\text{max}-k$ = maximum distance of the function from its midline =
      =$\frac{f_\text{max}-f_\text{min}}{2}$
    • $T$=period = time for one full cycle of the function = distance between two peaks (or two valleys).
    • $h$="phase" shift = "How far right (in $t$) is the time when the function rises across its midline from the origin?".

      Problem

      Temperature in Goshen

      Consider for Goshen, Indiana, the mean daily temperature (in °C) as a function of days after January 1, 2012.

      Find a function that models the trend of this data. Do this by estimating the amplitude, midline, and the horizontal shift, and figuring out what the period ought to be.

      ACTUALLY This form of the transformation equation, with $f(b(x-c))$, is even more convenient for this problem:

      Fitting a sin function to data (5 minutes)

      Here is a Desmos graph of the function I came up with by roughly estimating the amplitude, midline shift and phase shift, and realizing that the function should have a period of 1 year = 365 days: $$f(t)\approx 13.5*\sin\left[\frac{2\pi}{365}(t-120)\right]+10.5.$$