course
emailed it, and submitted it as wellHere's what I'd like to go over:
Graphing surfaces in R^3 and things of that nature
I still honestly don't understand the limits with the delta and epsilon. When I was reading through that section I just looked over the examples and figured that I knew it from calculus I. You also said to ask about the tangent plane and the explanation for that. I keep reading over problems and working them, so hopefully I'll be able somewhat successful with the test.
A function is continuous if its value at a point is equal to the limit of its values as you approach the point.
Explanation of the problem I gave you:
The solution is simple if you understand the definition of continuity. Just choose epsilon to be any number less that f(x0, y0), and the result is obvious. That is, if you really understand the definition. At this point, you're just learning it, so you'll need more explanation that this. When the above statement becomes obvious to you, you'll know you're on your way.
Here's the definition:
f(x, y) is continuous at (x0, y0) if for any epsilon > 0 there exists a delta > 0 such that if | (x, y) - (x0, y0) | < delta, we can guarantee that | f(x, y) - f(x0, y0) | < epsilon.
Important observation: | (x, y) - (x0, y0) | = | ( x- x0, y - y0) | = sqrt( (x - x0)^2 + (y - y0)^2 ). This is equivalent to saying that the point (x, y) lies in a circle of radius delta about (x0, y0). Be sure you write this out, sketch it out and understand it.
So, as I said, let epsilon be any number less than f(x0, y0). Since f(x, y) is continuous, by the definition (which is simply quoted here) there exists a delta > 0 such that if | (x, y) - (x0, y0) | < delta, we can guarantee that | f(x, y) - f(x0, y0) | < epsilon.
That is, there exists a delta such that if (x, y) is in a circle of radius delta about (x0, y0), we know that | f(x, y) - f(x0, y0) | < epsilon.
| f(x,y) - f(x0, y0) | < epsilon means
-epsilon < f(x, y) - f(x0, y0) < epsilon so that
f(x0, y0) - epsilon < f(x, y) < f(x0, y0) + epsilon.
Since epsilon < f(x0, y0), it follows that f(x0, y0) - epsilon > 0 so every term in the final string of inequalities above is > 0.
Thus there exists a delta such that if (x, y) is in a circle of radius delta about (x0, y0), we are assured that f(x, y) > 0.
I had wrote down that I wanted you to explain how to use the differential to find the thickness of something (Example: given the equation of the ellipsoid: x^2/a^2 +y^2/b^2+ z^2/c^2= some number, the formula for volume is 4/3 'pi abc, it is known that the shell of the ellipsoid is 1/8 in thick. My best guess is to use the partial derivative of the volume, in the form of 'delta V= dv/da 'delta a +dv/db 'delta b +dv/dc 'delta c, with the delta a, b, and c being 1/8 inch.
You're close.
For small `da, `db and `dc the differential of the volume gives us
`dV = 4/3 pi * (bc `da + ac `db + ab `dc).
From the thickness we know that sqrt( (`da)^2 + (`db)^2 + (`dc)^2 ) = 1/8 inch.
I know how use Lagrange multipliers pretty well now, and the method for solving the equations. I know how to maximize the function using this method, but if we want to find a minimum value, what approach do we do differently?
Lagrange multiplies work to find extrema. They work for minima as well as maxima.
Think about this. If you take the negative of a function with a maximum, you get a function with a minimum. The solutions to the equations would be the same either way. So the method will find minima as well as maxima.
that problem you sent me with the intersection of the sphere and plane, T(x,y,z) = (xy+yz) / (x^2+ y^2 + z^2) and the equation of the sphere: x^2+y^2 + z^2=1 and the plane x +2y + 3z =1
when I completed the square I got this :
(x-1/2)^2 + (y-1)^2 + (z-1.5) ^2 - 4.25 =0 this is my constraint
When I take the partials of the function for T(x,y,z), I honestly don't know if its possible to solve this problem because it would take me three hours to solve these system of equations. I'll show you what I have wrote down for my set up (it would take too long to type out) , but I don't know if it solve able to find values of x, y, z and that give an extremum.
Tx = ( y ( x^2 + y^2 + z^2) - 2 x ( xy + yz) ) / ( x^2 + y^2 + z^2)^2
Ty = ( x + z) ( x^2 + y^2 + z^2) - 2 y (xy + yz) ) / ( x^2 + y^2 + z^2)^2
Tz = (x ( x^2 + y^2 + z^2) - 2 z (xy + yz) ) / ( x^2 + y^2 + z^2)^2.
The gradient is Tx i + Ty j + Tz k.
As long as the gradient has a nonzero component in the direction of the constraint curve the function is changing along the curve, so you can't have an extremum.
Ok, you're right. This one gets too messy ...