# MandysNotes

### 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.