# AP Calc BC in a week - Part 1

## Introduction

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.

- 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 \(\theta\).

- 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 vĂ˛ila, 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.

### Example

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

## Conclusion

So far so good.