# MandysNotes

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## Limits

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Friday, 05 February 2010 00:00

### Trig Values of Special Angles

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Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Wednesday, 27 January 2010 22:43

### Example Problem: Pythagorean Theorem

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Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

This video solves a distance problem concerning ships.  The solution process is based on The Pythagorean Theorem.

Friday, 04 September 2009 17:56

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Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Friday, 04 September 2009 17:54

### WTF is "Soh-Cah-Toa"?

{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Friday, 04 September 2009 17:41

### Basic Trig Functions

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BE ABLE TO RECITE THESE IN YOUR SLEEP!

Here is a chart of the QUADRANTAL ANGLES (angles which lie exactly on either the x-axis or y-axis.

Friday, 04 September 2009 17:57

### The Pythagorean Theorem

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Friday, 04 September 2009 15:33

### The Unit Circle

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Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Friday, 04 September 2009 15:35

### Trig Identities

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This video tutorial uses the acronym "Soh Cah Toa" which made absolutely NO SENSE to me until I watched the full clip.  Turns out it's actually useful in remembering the trig functions!

Friday, 04 September 2009 15:31

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Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Friday, 04 September 2009 15:28

### Solving A Triangle Using Trig Functions

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Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Wednesday, 22 September 2010 21:55

### Evaluating Limits

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Wednesday, 22 September 2010 21:52

### How To Take The Square Root of a Limit Function

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Thursday, 28 January 2010 00:13

### Derivatives of Exponential Functions

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{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Friday, 04 September 2009 15:40

### The Definition of the Derivative

{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

And here is an alternate explanation for the definition of the derivative:

Friday, 04 September 2009 17:29

### Calculating a Derivative by the Definition

{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Wednesday, 27 January 2010 23:20

### Derivatives: The Product Rule

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{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Friday, 04 September 2009 17:28

### Introduction to The Chain Rule

{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Wednesday, 27 January 2010 23:19

### Derivative Example Videos

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{module [79]}

Disclaimer: I did not create nor do I own these videos.  I have simply embedded them, courtesy of YouTube.

Friday, 04 September 2009 17:20

### Derivatives of Inverse Trig Functions

{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Wednesday, 27 January 2010 21:57

### Volumes by Integration: Cylindrical Shells

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{module [79]}

Disclaimer: I did not create nor do I own these videos.  I have simply embedded them, courtesy of YouTube.

Wednesday, 27 January 2010 21:53

### Volumes by Integration: Disk / Washer Method

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{module [79]}

Disclaimer: I did not create nor do I own these videos.  I have simply embedded them, courtesy of YouTube.

Here is an example problem worked through completely:

Here's Example 2!

And Example 3!

Thursday, 28 January 2010 00:05

### Related Rates Pointers

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Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Thursday, 28 January 2010 00:09

### Related Rates: Baseball Problem

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Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

The problem presented here is:

A baseball diamond is a square with side 90 feet.  If a batter hits the ball and runs towards first base with a speed of 20 ft. per sec, at what speed is his distance from second base decreasing when he is halfway to first base?

Wednesday, 27 January 2010 23:42

### Related Rate Problem: Planes Flying

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{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

The problem presented in this video is:

A plane is flying at a constant altitude of 2 miles, and at a constant rate of 180mi/hr.  A camera on the ground is following the plane as it flies away from the camera.  How fast must the camera rotate to keep the plane in view when the camera is pointed up at an angle of pi/3?

Wednesday, 27 January 2010 22:30

### Related Rates: Using Implicit Differentiation

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{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

The problem presented in this video is an example solving by implicit differentiation:

At noon, Ship A is 100km west of Ship B.  Ship A is sailing south at 35 km/h and Ship B is sailing north at 25km/h.  How fast is the distance between the ships changing at 4:00PM?

Wednesday, 27 January 2010 22:07

### Calculating the Work Required to Drain a Tank

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{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Wednesday, 27 January 2010 22:37

### What is Implicit Differentiation?

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Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Tuesday, 13 November 2012 20:24

### Implicit Differentiation

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Implicit Differentiation

Implicit Differentiation involves finding the derivative of a function $y=f(x),$by first finding the derivative of a function $g\left(x, y(x) \right),$and then solving for $\frac{dy}{dx}.$

As such, it involves both the chain rule for differentiation, and (potentially), the rule for differentiating a product.

Wednesday, 22 September 2010 21:47

### First Order Linear Differential Equations

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Wednesday, 27 January 2010 23:13

### Introduction to Differential Equations

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{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

### A Useful Formula

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Let $$f (x)$$ be a smooth (i.e., $$C^{\infty}$$) function in a convex neighborhood $$V$$ of $$\mathbb{R}^{n},$$ then:

$f(x) = f\left( x_{1}, \dots , x_{n} \right) = f\left( a_{1}, \dots , a_{n} \right) + \sum_{i = 1}^{n} \left( x_{i} - a_{i} \right) g_{i} \left( x_{1}, \dots, x_{n} \right)$

for some point $$a = (a_{1}, \dots , a_{n}) \in V,$$ some smooth functions $$g_{i}$$ defined in $$V,$$ and with:

$g_{i}(a_{i}) = \frac{ \partial f}{\partial x_{i}} (a_{i}).$

### Lagrange Multipliers

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Consider a particle that moves in the plane in a circle of radius $$a$$ described by the equation:

$g(x, y) = x^{2} + y^{2} - a^{2} = 0.$

A possible parametrization of this equation is:

$x(t) = a \cos{(t)},$

$y(t) = a \sin{(t)}$

where $$t$$ is time.

Then the particle revolves counterclockwise about the origin on a circle of radius $$a,$$ with a velocity vector:

$v = ( \frac{dx}{dt} , \frac{dy}{dt} ) = ( - a \sin{(t)}, a \cos{(t)} ) = ( - y, x).$

Note that the particle's speed is constant.

$\text{speed } := |v| := \sqrt{ v_{x}^{2} + v_{y}^{2}} = \sqrt{ a^{2} \sin^{2}{(t)} + a^{2} \cos^{2}{(t)} } = \sqrt{ a^{2} } = a.$

Now suppose that there is a temperature gradient in the plane described by the equation:

$T( x, y) = c_{0} + cy.$

Where $$c_{0}$$ and $$c$$ are constants.

We want to know where on the circle the particle will experience the most extreme (greatest, least) temperatures.

We cannot use $$dT=0,$$ because

$dT = \frac{ \partial T}{\partial x} dx + \frac{\partial T}{\partial y}dy = \frac{\partial T}{\partial y}dy = c dy$

and $$\frac{\partial T}{\partial y} = c$$ is a constant.

The technique that allows us to find the extremal values of $$T$$ subject to the constraint $$g = 0,$$ is the use of Lagrange multipliers.