There is a circular gate rotating at a constant angular speed of ω. The circular gate has a tunnel across its diameter. The mission is to pass through the gate. (That is, come in one side of the gate, travel the whole diameter, and exit at the other side.)
Also, craft is neutrally buoyant...
Okay. I'm now confused. What is M? The only thing that would make sense would be if M is a scalar, but then the locus of M would not be a surface, since it lives in the straight line of the real numbers.
The expression \partial x is precisely equal to dx if all other "variables" are kept constant. Thus, in iterated integrals, where you *do* keep all variables constant, dx = \partial x.
If your set is open, then every point is an interior point and can be enclosed inside a ε-disc. This ensures that it can be approached by all directions, as needed by the definition.
If you are working with a closed set as your domain, your boundary points are still being approached in all...
\underset{(x,y)\rightarrow (0,0)}{\lim}=\frac{x^2\cdot \sin^2 y}{x^2+y^2}=\cos^2 \theta \sin ^2 (r\sin \theta)) where θ is the angle of (x,y). Does that help at all? Edit: I put a mult. sign where the + sign should be! o_O
Suppose we have the graph of a function. Let us approximate its shape with vectors whose horizontal component is dx and vertical component is dy. Then, the derivative of the function gives us the number we need so that dx \cdot \frac{dy}{dx}=dy. (Typically, y is dependent on x, so we can...
Well... I'd say that it "makes sense" only if it were a definite integral.
And I wasn't complaining about the uselessness of the proof, so much as the impossibility of computing the integral on the RHS.
I hope you're not using this to "prove" FTC. If you are, you should consider the change of...
If dy and dx have been properly defined (which for some reason is a rare occurrence), then this equality is immediately apparent for differentiable functions. However, this is useless as a means to finding an antiderivative (which is what you're doing, since you have no bounds on your...
Definitely not enough info here for anyone to understand what's going on. For instance, "solve for FOC for the lagrangian function with respect to r(y,a)" is meaningless to me.