Recent content by JThompson

P\implies Q was supposed to relate to the "if |x-c|<delta, then |f(x)-L|<epsilon" part of the epsilon-delta definition. Essentially, this statement is equivalent to |x-c|\geq\delta\mbox{ or } |f(x)-L|<\epsilon. So negating this portion of the definition returns |x-c|<\delta\mbox{ and }...

Here are a few hints: P\implies Q is equivalent to \neg P\vee Q. So the negation of the statement is P\wedge\neg Q. Negated universal quantifiers become existential, and negated existential quantifiers become universal.

The closest thing I could find is a Wikipedia table of valid argument forms in propositional logic. http://en.wikipedia.org/wiki/Propositional_logic#Basic_and_derived_argument_forms"

The point is that this means nothing without context- we have absolutely no idea what you are talking about. However- wild guess- are you trying to find a delta such that 0<x<\delta implies \cot (x) > m ?

Firstly, a subset does not necessarily have all of the same properties as any of its supersets, so it's perfectly reasonable that a subset of R would not satisfy the Completeness Axiom. Consider \{x\in \mathbb{Q}: x\leq \sqrt{2}\}. Is there a least upperbound for this set in Q? No, because Q...

This problem is much easier if you use the formula cos^2(x) = 1-sin^2(x). In general, try simple formulas before more complicated ones. (This is of course relative, but I classify a simple formula as one that comes up often and is extremely easy to remember.) Also ∫(cosx)/2dx is 1/2sinx, not...

If you set z=2 you have the parametrization of a circle, not a cylinder. You do know that the cylinder is bounded, but you have to let z vary to obtain a cylinder. Here's the thing. When you do these sorts of problems, the position vector is only part of the parametrization. The position vector...

Based upon the intervals you wrote down, it looks like you are trying to find a single parametrization for the whole surface- don't try that. This will not work because the pieces of the surface, i.e. the cylinder and the two planes, have different normal vectors. 1) Do you see why they...

Your parametrization does not need to include the cone because the problem concerns the surface S, which is just part of the sphere. As Dick was saying, the cone is only relevant to your limits of integration- you do not need to parametrize the cone because it is not part of S.

Just to be clear, the antiderivative of \frac{x}{\sqrt{1-x^2}} is not - \frac{1}{2\sqrt{1-x^2}}. It is -\sqrt{1-x^{2}}.

I'm not sure what you mean by this. If you let h be the height of the house, then the whole thing should be h+4.5 . This should be a side of the triangle with a 59.5 degree interior angle opposite this side. Then you'll have another triangle with an interior angle of 56.3 degrees with an...

I was taught to remove negative exponents when determining the domain of a function, but it is entirely possible my teacher was incorrect. If this were a composite function, say f(x)=x^{-1} and g(x)=\frac{1}{x^2}, then the domain of g(f(x))=\frac{1}{(x^{-1})^{2}} would definitely be x\neq 0...

This is an indeterminant form, 0/0. Have you learned how to handle these?

Yes, \frac{1}{x^{-2}}=x^{2}. By restrictions, do you mean domain restrictions? No it does not have a hole at x=0, since x^{2} does not. In general it is a good idea to convert the denominator to non-negative exponents before analyzing where the equation is defined, but this is mostly because...

Yes. The derivative will be of the opposite sign. v'(x)*u(x)-u'(x)*v(x)= -(u'(x)*v(x)-v'(x)u(x)) Good luck.