Recent content by Black Orpheus

The factors of a are n and y-x. This means that a, b, and a+b all have a common factor. Since a and b are relatively prime, the common factor must be 1.

1. Suppose that a and b are positive integers. Show that the following are equivalent: 1) a and b are relatively prime 2) a+b and b are relatively prime 3) a and a+b are relatively prime. 2. I know that for a and b to be relatively prime, (a,b) = 1 (that is, their greatest common divisor...

If I assume that q is prime, then that implies that there exist two integers ab such that q divides ab. It also must mean that a and b are both greater than 1 and less than q. If this is the case, then q does not divide a and q does not divide b (contrapositive: if q is not prime, then q divides...

1. Here is the problem I'm stuck on: Let q be a positive integer, q is greater than or equal to 2, let a and b be integers such that if q divides ab, then q divides a or q divides b. Show that q is a prime number. 2. I know that q is prime if and only if 1 divides q and q divides q...

I need to take a surface integral where S is x^2 + y^2 + 2z^2 = 10. I need help with the parametrization of the curve. Letting x=u and y=v makes the problem too complicated. Can you let x=cos(u), y=sin(u) and z=3/sqrt(2)?

I need to find the path, c(t), to represent the set of all (x,y) such that 4x^2 + y^2 = 1. This seems like such a simple question but I don't know where to begin (I know that a path is a function or map over an interval whose image is the function I have). Can anyone offer some help?

If you have time, could you walk me through the dz/dx method? I'm not following this... If I try to take dz/dx with the formula mentioned above, I get -(y + dz/dx + 3z^5)/(dxy/dz + 1 + 15xz^4). Do dz/dx and dxy/dz go to 0? What about the extra zs?

Alternatively, if I use the fact that the partial of z with respect to x equals -(partial of F with respect to x)/(partial of F with respect to z), don't I still end up with a z (and I need it in terms of x and y so I can plug in my point)?

How do I take z as a function of x and y when I can't solve for it?

Do I start by taking d/dx of xy + z + 3xz^4 - 4 = 0, giving y + dz/dx + 3z^5 = 0, or dz/dx = -y - 3z^5..?

I need to compute the partials of z with respect to x and y of: xy + z + 3xz^5 = 4 at (1,0). I already showed that the equation is solvable for z as a function of (x,y) near (1,0,1) with the special implicit function theorem, but that's the easy part. Could someone explain to me how to begin...

I need to find the extrema of f(x,y,z)=x+y+z subject to the restraints of x^2 - y^2 = 1 and 2x+z = 1. So the gradient of f equals (1,1,1) = lambda1(2x,-2y,0) + lambda2(2,0,1). Solving for the lambdas I found that lambda1 = -1/(2x) = -1/(2y), or x=y. But this isn't possible if x^2 - y^2 = 1...

||x-a|| = sqrt[(xsub1 - a)^2 +...+ (xsubn - a)^2] so gradient = (partial derivative of sqrt[(xsub1 - a)^2] , ... , partial derivative of sqrt[(xsubn - a)^2])?

forgot to add that it's for all x not equal to a

One last question for tonight... If you let f: R^n ---> R (Euclidean n-space to real numbers) and f(x) = ||x-a|| for some fixed a, how would you define the gradient in terms of symbols and numbers (not words)?