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Can Anyone solve this Calculus Problem!?!?! (1 Viewer)

Jin E

Second Unit
Joined
Nov 19, 2000
Messages
452
I admit, my calculus is a bit fuzzy after not having it in over 8 years. But, since my girlfriend is taking a Calculus class I've found that I know a lot more about math and calculus then I thought I did. So far I've been able to answer every question she's had, cept for this one. When all else fails, turn to the Home Theater Forum collective for help.... so here goes.

The problem ask for the values of a and b so that:

Limit [(ax-b)^(1/2) - 3] / x = 2
(x->0)

If I can get her the answer to this problem... she's promised me conjugal relations. Please help a brother out!
 

AviTevet

Stunt Coordinator
Joined
Apr 11, 2002
Messages
110
she's promised me conjugal relations.
I'll only post the answer if I can get in on this. :)
Man I am rusty at algebra!
Here's the deal. Since you have only one equation but two unknowns, there are infinitely many sets of (a, b) pairs that satisfy the conditions set forth in the problem. Which raises the obvious question "if there are infinitely many, why can't I find any of them?" Well, you just have to pick one of them (blindly) by assigning "b" a convenient value and hoping "a" will be easy to find once b is known. YOu can choose any value for b, like 34.7, pi, or 0.004, but personally I chose -9 :).
And don't forget L'Hopital's rule:
For f(x) and y(x) where f(0)=0 and y(0)=0,
lim f(x)/y(x) ==
(x->0)
lim f'(x)/y'(x)
(x->0)
 

PhilipG

Senior HTF Member
Joined
Jan 13, 2000
Messages
2,002
Real Name
PhilipG
Looks like we have a notation difference between US & UK. So I'm assuming you're taking the differential of the term in square brackets and x is greater than 0? Multiply both sides by x. Integrate the rhs. Square both sides. Reorder. I get
x^4 + 6x^2 -ax + ( 9 + b ) = 0
So there's Avi's b=-9 :)
 

Jonathan Smith

Stunt Coordinator
Joined
May 26, 2002
Messages
122
Avi's method should get you a correct answer. I'm pretty sure it is the only one, not one of an infinite set like he suggests.
I get a=12 and b=-9
If anyone can post different values that work, I would be surprised :)
 

Chuck C

Senior HTF Member
Joined
Jan 6, 2001
Messages
2,224
I also get a=12 and b= -9
The work: (assume lim = lim as x approaches 0)
using l'hospital's rule :
lim ½a(ax-b)^(-½) = 2
lim a(ax-b)^(-½) = 4
a(-b)^(-½) = 4
a/(-b^½) = 4
12/3 = 4
yay!
now have some fun with your amiga
 

Mike St.Louis

Supporting Actor
Joined
Sep 22, 1999
Messages
518
I got a = 12 and b = -9

----

(((ax - b)^1/2) - 3)/x = 2
(ax - b)^1/2 - 3 = 2x
(ax - b)^1/2 = 2x + 3
(ax - b) = (2x + 3)^2
(ax - b) = 4x^2 + 12x + 9

lim x -> 0
b = -9

Substituting b = -9 and resolving:

ax + 9 = 4x^2 + 12x + 9
ax = 4x^2 + 12x
a = 4x + 12

lim x ->0
a = 12
 

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