Thanks for the reply. One thing to clarify, in the e^(3x) case, you have f(u)*u^0 + 1*f'(u)*u^1 +0.5*f''(u)*u^2...
All the derivatives are evaluated at u=0 in the case we are discussing, but could be non-zero. But if u=3x, and you are taking some derivative of u, why don't you have to use the...
In many of my physics classes we have been using Taylor Expansions, and sometimes I get a bit confused. For example, I feel like different things are going on when one expands (1-x)^-2 vs. e^(-Ax^2), where I just have some constant in front of x^2 to help make my point. To keep things simple...
Do you have a choice? If you have a cylinder whose upper and lower surfaces reside somewhere on the z-axis, say z1 and z2, can you project the surface of the cylinder to any value of z (where obviously z=0 is the simplest)?
This is where I am getting confused: If it were the same problem...
For example, you have a surface defined by a cylinder whose circular faces are in the x-y plane with x^2 +y^2=1 and -1<z<3. Then let's say you have a vector field F=(zx, xy, zy) and you want to calculate the the double integral of the curl of F. You could do this directly, or by using Stokes's...
I am a little confused about how to generally go about applying Stokes's Theorem to cylinders, in order to calculate a line integral. If, for example you have a cylinder whose height is about the z axis, I get perfectly well how to parameterize the x and y components, using polar coordinates...
Here is the complete example:
Example 7: Show that the vector field V on R2 defined by V= yi-xj is not a gradient vector field; that is, there is no C1 function f such that V=\nabla f(x,y)= \partial f /\partial x i + \partial f / \partial y j.
Solution Suppose that such an f exists...
In Marsden and Tromba's Vector Calculus (5e, p.289), there is an example calculating the gradient of V(x,y)= yi -xj. In the next step it shows the gradient being equal to the product of \partial f / \partial x and yi, plus the product of \partial f / \partial y the and -xj. It then says...
Thanks for your hep but I am still confused about a couple of things though. Why is \partial / \partial x of yi - xj = y? To me, it makes sense that when you are dealing with the i component and taking its partial with respect to x, you look at the value in front of the component, which is y...
Sorry, every time I attempted to write the partial derivative of V with respect to x, it just ended up saying "partial" in red font. Just so you know what it was intended to be...
I was reading a section on vector fields and realized I am confused about how to take partials of vector quantities. If V(x,y)= yi -xj, I don't understand why the \partialx= y and the \partialy= -x. The problem is showing why the previous equation is not a gradient vector field (because the...