# Differentiation along surface

I have this question I got when trying to solve a physics problem and I don't know which topic it belongs to. Please redirect me if anyone asked the same question before.

I have a function $f(x_1,x_2,\ldots,x_{32})$ and want to differentiate the function along a surface defined by the constraint $x^2_1+ x^2_2+\cdots+x^2_{32}=1$. How would one do that?

Thanks alot

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Let's say $\vec{x}(t) \in S^n$ is a curve. Then $|\vec{x}(t)|^2 = \vec{x}(t) \cdot \vec{x}(t) = 1$ and differentiate both sides with respect to time:

$\vec{x}'(t)\cdot \vec{x}(t) = 0$

This equation says the tangent to the sphere is perpendicular to the radius.

How does $F(\vec{x})$ changes as you move along the sphere in the path $\vec{x}(t)$ ?

$\frac{dF}{dt} = \nabla F \cdot \vec{x}'(t)= \sum_{i=1}^{32} \frac{dF}{dx_i} \frac{dx_i}{dt}$

Somehow, you have to write down a curve constrained to the sphere, or a vector tangent to the sphere.

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If you want the differential, then the constraint doesn't change the calculation at all: it simply gives you a way to simplify the result.

For example, the differential of $z = x^3 + y^3$ restricted to $x^2 + y^2 = 1$ is

$$dz = 3x^2 dx + 3y^2 dy = 3x^2 dx + 3(1 - x^2) dy = 3x^2 (dx - dy) + 3 dy$$

(which version is the "simplest" would depend on application)

Depending on the meaning of the constraint, the differential of the constraint can also be a constraint. In that case, we also have the equation

$$2x dx + 2y dy = 0 \qquad \Longrightarrow \qquad dy = -\frac{x}{y} dx$$

and then we could simplify $dz$ by

$$dz = 3x^2 dx + 3y^2 dy = 3x^2 dx - 3xy dx = 3x(x-y) dx$$

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