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AP Calc BC in a week - Part 1


Today we'll be doing:

  • 1F Parametrically Defined Functions
  • 1G Polar Functions
  • 3F Derivatives of Parametrically Defined Funcitons
  • 3J Indeterminate Forms and L'Hopital's Rule

Let's start!

1F Parametrically Defined Functions

So these are basically pairs of functions with $$x=f(t), y=f(t)$$. \(t\) is the parameter.

Parametric to Cartesian

To turn a pair of parametric equations into a Cartesian equation, make \(t\) the subject of the \(y\) parametric equation, ie $$t=f(y)$$, then you can substitute \(f(y)\) in for \(t\) in \(x\)'s parametric equation: $$x=f(t)=f(f(y))$$.

For example:

\begin{equation} x = 2t + 1, y = t^2 \\ x = 2t + 1, t = \sqrt{y} \\ x = 2(\sqrt{y}) + 1 \end{equation}

Parametric to Cartesian (relation)

A modified example from the book, which follows basically the same method as above, but \(t\) is made the subjects of both equations along with a trig function. Then a trig identity can be used to form a relation.

For example:

\begin{equation} x = 6\sin t, y = 7\cos t \\ \sin t = \frac{x}{6}, \cos t = \frac{y}{7} \\ \sin^2 t + \cos^2 t = 1 \\ \frac{x^2}{36} + \frac{y^2}{49} = 1 \end{equation}

Inverse Functions

This is some pretty cool magic. When you find \(f^{-1}(x)\) of \(f(x)\), you are simply switching \(x\) and \(y\) in the equation.

So for instance, if you had an equation $$f(x) = x^2 + x + 1$$, remember that \(x=t\) is the independent variable and \(f(x)=y\), and you can let $$x=t, y=t^2+t+1$$. Now, just switch \(x\) and \(y\) ! $$x=t^2+t+1, y=t$$ And you've got your inverse function! Which you could convert back to a Cartesian equation by just plugging in \(y\) for \(t\).

Calculator Magic

On Casio we got Graph (5) -> Type (F3) -> Param (F3, on an empty equation entry) and then we can easily input a parametric equation.

The book also suggests graphing relations with the subject \(x\) equal to polynomials of \(y\) with parametric equations, but us Casio people can disregard that.

Take-home message

  • Get one of the parametric equations to be equal to \(t\), then plug into the other (Parametric to Cartesian)
  • Get both equal to \(t\) something, and equate to each other or use a trig identity (Parametric to Cartesian trig)
    • \(x=t\), \(y=f(x)\), swap \(x\) and \(y\) so that \(x=f(x)\) and \(y=t\), and easy inversions!
  • Easy to graph.

1G Polar Functions

Woah. Coordinates are \((r,\theta)\), where \(r\) is the distance from the origin and \(\theta\) is the angle of rotation. Polar functions are $$r=f(\theta)$$, so I guess that means that the distance depends on the angle.

Oh hey this section looks short.

Calm down, it's just an equation

No really it is.

Consider a polar function \(r=5+6\sin\theta\).

The book only gives this (modified) example, so I'm guessing that's all we have to worry about.

  1. For what values of \(r\) does the curve pass through the origin?

    Remember that it's just an equation. At the origin, the distance from the origin is 0, so \(r=0\), and that means $$0=5+6\sin\theta$$. Now it's just simple algebra magic to find \(\theta\).

  2. For what values of \(r\) does the curve pass through the circle \(r=2\)?

    Still just an equation. $$r=2=5+6\sin\theta$$, and then the usual.

    That was easy. I'm guessing we won't have to worry about intersecting with other functions, so that's convenient.

Polar to Parametric

This isn't that hard either. Just remember: $$x=r\cos\theta, y=r\sin\theta$$. This means that if you need to convert a polar equation to parametric, just plug in the whole polar function in \(r\) and then multiply it by \(\cos\theta\) or \(\sin\theta\) !

That also makes finding \((x,y)\) coordinates a lot easier. Just convert to parametric and vòila, you've already got your \((x,y)\) !

Calculator Magic

Casio's got a mode for this too. Graph -> Type -> r=

Take-home message

  • Calm down, it's just an equation.
  • \(x=r\cos\theta\) and \(y=r\sin\theta\)

3F Derivatives of Parametrically Defined Functions

This looks pretty simple as well. $$\frac{dy}{dx}=\frac{y'}{x'}$$, or more formally: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$.

Also, $$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{\frac{d}{dt}\frac{dy}{dx}}{\frac{dx}{dt}}$$ I suppose informally that would be $$\frac{d^2y}{dx^2} = \frac{\frac{dy}{dx}'}{x'}$$

Assuming that \(x=f(t), y=g(t)\) are diffrentiable.


Find the first and second derivative of \(x=2t^3,y=t^4\). (I know the book uses trig functions of \(t\))

$$\frac{dy}{dx} = \frac{y'}{x'} = \frac{4t^3}{6t^2} = \frac{2}{3}t$$

\begin{equation} \frac{d^2y}{dx^2} = \frac{\frac{dy}{dx}'}{x'} \\ = \frac{\frac{2}{3}}{6t^2} \\ = \frac{2}{18t^2} \\ = \frac{1}{9t^2} \end{equation}

Equation of tangent

I guess they give you the parameter, so just plug that in to your \(\frac{dy}{dx}\) and you got the \(m\) part of the linear equation.

For the \((x,y)\), plug in the parameter to the equations for \(x\) and \(y\), and you've got a point to get the \(c\) from for the linear equation.

Easy peasy.

Collision with Parametric equations

Equate \(x_1\) and \(x_2\) and get the resulting \(t\). Then test that \(t\) while equating \(y_1\) and \(y_2\), and the the \(y\) s are equal, you got a true collision point!

Take-home message

  • For parametric equations, \(\frac{dy}{dx}=\frac{y'}{x'}\)
  • And \(\frac{d^2y}{dx^2}=\frac{\frac{dy}{dx}'}{x'}\)
  • Plug in parameters for tangent lines (easy algebra)
  • Test equality of \(x\) s and \(y\) s for a given \(t\) resulting from the equivalence of \(t\) from the \(x\) s.

3J Indeterminate Forms and L'HĂ´pital's Rule

Indeterminate forms

Uhh, $$\frac{0}{0}, \frac{\infty}{\infty}, 0 \times \infty, \infty-\infty, 0^0, 1^\infty, \infty^0$$

L'HĂ´pital's Rule

This rule is used to find limits of indeterminates of the form \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). It won't work if the limits are not indeterminates of these two forms.

Looks like the book has four rules, but basically if your limit to a finite number or to infinity results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then you can differentiate the numerator and the denominator functions and try again. $$\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$$ $$\lim_{x\to \infty} \frac{f(x)}{g(x)} = \lim_{x\to \infty} \frac{f'(x)}{g'(x)}$$

The rule can be repeated. Of course, if either \(f(x)\) or \(g(x)\) cannot be differentiated, then the rule does not work.

I don't think I need to give examples, this is pretty intuitive.

Other indeterminate forms

If you get indeterminates of the form \(0\times\infty\), then try to convert it to an equivalent indeterminate of the form \(\frac{0}{0}\). It should work.

Example (this is from the book): $$ \lim_{x \to \infty} x\sin \frac{1}{x} = \lim_{x \to \infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}} $$

For the indeterminates that are exponents, like \(1^\infty\) and \(\infty^0\), we can use the natural log to "remove" the exponent.

Example with \(1^\infty\) (from the book):

\begin{equation} \lim_{x\to 0} (1+x)^\frac{1}{x} \\ \text{let} y = (1+x)^\frac{1}{x} \\ \ln y = \frac{1}{x}\ln (1+x) \\ \lim_{x\to 0} \ln y = \lim_{x\to 0}\frac{1}{x}\ln (1+x) \end{equation}

And now it's \(0\times\infty\), so we can just do what we did earlier. Don't forget to remove the \(\ln\) at the end!

Example with \(\infty^0\) (also from the book):

\begin{equation} \lim_{x\to\infty} x^\frac{1}{x} \\ \text{let} y = x^\frac{1}{x} \\ \ln y = \frac{1}{x}\ln x = \frac{ln x}{x} \\ \end{equation}

Basically it's the same as the previous example.

Take-home message

  • If your limit end up as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then you can differentiate the numerator and the denominator functions and try again to get the limit
  • If you get \(0\times\infty\), rearrange the limit to \(\frac{0}{0}\) and try the above method
  • If you get exponents, then remove the exponent with a natural log, and then combine the above methods


So far so good.

2016-03-20 Paul Elder


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