# Can Anyone solve this Calculus Problem!?!?!

Discussion in 'Archived Threads 2001-2004' started by Jin E, Jun 8, 2002.

1. ### Jin E Second Unit

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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)

2. ### AviTevet Stunt Coordinator

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3. ### PhilipG Cinematographer

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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

4. ### Jonathan Smith Stunt Coordinator

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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

5. ### Jeff Kleist Executive Producer

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Sorry, Calculus is HARD!
(see Jack Briggs' science post)

6. ### Jed M Cinematographer

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Now I remember why I majored in Political Science.

7. ### Chuck C Cinematographer

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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

8. ### Mike St.Louis Supporting Actor

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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