[H] Calc 3

Meta

United States6225 Posts

February 12 2008 04:14 GMT

Sort of conceptual questions:

Does the function
f(x,y,z) = xy - z
have a minimum value on the line:
x = t - 1
y = t - 2
z = t + 7

and if so, what is it?

also there's this limit problem that's sort of conceptual:

Does knowing that |cos(1/y)| <= 1 tell you anything about lim as (x,y) approaches (0,0) of xcos(1/y)?

man i hate this conceptual stuff i'm struggling with the easy problems. thanks to anybody who wants to help me, thought i'd throw it out there. =]

Purind

Canada3562 Posts

February 12 2008 04:36 GMT

I think that if you plug in those values, it gives you a parameterization of a curve that kinda walks on the f(x,y,z) function, and you can find the minimum of that parameterization using calc 1 stuff.

As for the other question, I dunno, my math kinda sucks, but if you have two functions multipled by each other, one of them's magnitude is bounded by 1, the other is going to 0, intuitively, I think the function should go to 0.

(my intuition is more often than not wrong in math, but does my logic make sense?)

Dromar

United States2145 Posts

February 12 2008 04:38 GMT

1. If
x = t - 1
y = t - 2
z = t + 7

then f = t^2 - 4t - 5 = (t - 5)(t + 1) so the minimum must be between 0 < t < 4 (because t = 0,4 are the zeroes of t^2 - 4t). Since this is a quadratic, the min must be halfway between 0 and 4*, aka 2.
So the min is at t = 2. Thus x = 1, y = 0, z = 9, and f(1,0,9) = min(f) = -9.

EDIT: The above answer isn't right. you need to differentiate to find the min by setting f' = 0. I'll probably re-edit in a bit.
RE-EDIT: uh, I guess it is right. f' = 2t - 4 = 0 -> t = 2. f(2) = -9, so -9 is the min.

2. yes. Since |cos(1/y)| <= 1, and 1 is a finite number (that is, for all y in the system of real numbers, |cos(1/y)| < infinity), so xcos(1/y) must approach 0 as x approaches 0.
ANOTHER EDIT: To be more clear,
There exists an M such that, for all y, |cos(1/y)| < M. (Namely, any M > 1) Then |cos(1/y)| is bound (doesn't go to infinity or -infinity).

This also is a relatively loose proof. But this one can be patched up easily. Since |cos(1/y)| <=1, then |cos(1/y)| is at most 1 and at least -1. Then show that as x -> 0, both -x -> 0 and x -> 0 are true. Then quote the pinching theorem if needed.

infinity21

Canada6683 Posts

February 12 2008 06:51 GMT

My answer's the same as Dromar's. What I did:

x = t - 1
y = t - 2
z = t + 7
xy-z = t^2 - 4t - 5
= (t-2)^2 - 9 (completing the square is useful for simple quadratic equationslike this)
So the min value is -9 at t = 2 since (t-2)^2 >= 0

"Does knowing that |cos(1/y)| <= 1 tell you anything about lim as (x,y) approaches (0,0) of xcos(1/y)?"
Yeah. + Show Spoiler [informal babble] +

Since cos(1/y) is bounded by -1 and 1 and x approaches 0, xcos(1/y) --> 0 as (x,y) --> (0,0)

Visualize it like this:
You know that as x -->0, x * 1 --> 0
Then you have |cos(1/y)| which is less than 1 so
as x -->0, x * 1 >= x * |cos(1/y)| --> 0 ==> x * cos (1/y) --> 0 since alternating signs doesn't affect convergence.
If you multiply a number converging to 0 by something less than 1, the product will still converge to 0.

Or a formal proof would be:
x * (-1) <= x * -|cos(1/y)| <= x*cos(1/y) <= x*|cos(1/y)| <= x * 1
Now take the limit of these functions as x --> 0
Then by squeeze theorem, you get
0 <= x * -|cos(1/y)| <= x*cos(1/y) <= x*|cos(1/y)| <= 0
So x*cos(1/y) = 0 as x --> 0

This is Calc 3? :s
edit:I love the conceptual stuff. That's why I'm a math major :D
I'm still in 1st year though
(edited for clarity)

infinity21

Canada6683 Posts

February 12 2008 07:04 GMT

On February 12 2008 13:36 Purind wrote:
I think that if you plug in those values, it gives you a parameterization of a curve that kinda walks on the f(x,y,z) function, and you can find the minimum of that parameterization using calc 1 stuff.

As for the other question, I dunno, my math kinda sucks, but if you have two functions multipled by each other, one of them's magnitude is bounded by 1, the other is going to 0, intuitively, I think the function should go to 0.

(my intuition is more often than not wrong in math, but does my logic make sense?)

Heh, if you plug it in, it becomes a really simple quadratic

Your intuition's correct. Also, note that if you have a product of two functions, one which appraoches zero and the other whose magnitude is bounded by some M (it can be 10^100000000000000000000000000), it will approach 0.

thoraxe

United States1449 Posts

February 12 2008 23:59 GMT

Teamliquid must be like the smartest community, EVA!!!!
aside from Mensa of couse.

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