## Calculus Tutorials

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

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.

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

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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.

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!

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

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Here is an example problem worked through completely:

Here's Example 2!

And Example 3!

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?**

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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?**

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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?**

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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.

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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:

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

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**.