## Advanced Calculus

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**.