# 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:

$$x = 2t + 1, y = t^2 \\ x = 2t + 1, t = \sqrt{y} \\ x = 2(\sqrt{y}) + 1$$

### 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:

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

### 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.

### Example

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

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

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

### 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'Hôpital's Rule

### Indeterminate forms

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

### L'Hôpital'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):

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

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):

$$\lim_{x\to\infty} x^\frac{1}{x} \\ \text{let} y = x^\frac{1}{x} \\ \ln y = \frac{1}{x}\ln x = \frac{ln x}{x} \\$$

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.

2016-03-20 Paul Elder