SSTjumbo
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Hey Everybody, It's Time To Help Me With Calc!

Fri Nov 30, 2001 12:32 pm

If y=f(x) and m=Dx/Dy, how do I find Dm/Dy? For instance, say y=x^2+3x+2 making Dy/Dx=2x+3 => m=2x+3, how would I find Dm/Dy? Does it involve natural logs maybe?
I don't know, so this is my signature.
 
PanAm747
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RE: Hey Everybody, It's Time To Help Me With Calc!

Fri Nov 30, 2001 1:50 pm

Dy=2x+3

Dm=2

Dm/Dy=(2)/(2x+3)

I believe you have two equations and you just divide them:

1. y=f(x)=x^2+3x+2=
Dy=2x+3

2. m=f(x)=2x+3
Dm=2

Then:

Dm/Dy=(2)/(2x+3)
Pan Am:The World's Most Experienced Airline - P(oor) S(ailor's) A(irline): San Diego's Hometown Airline-Catch Our Smile!
 
Guest

RE: Hey Everybody, It's Time To Help Me With Calc!

Fri Nov 30, 2001 2:00 pm

No, this is what it is...

weopfjW+EPIJHSWIFJP
-/EP'IFHwp[evgjwrpsvg

pdihnifjvw[PG
23-4098T-3
'PGJP

 Big thumbs up

Just holla holla if you need some more help.

 
docpepz
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RE: Hey Everybody, It's Time To Help Me With Calc!

Fri Nov 30, 2001 2:07 pm

You do have to realise that dy/dx is NOT a division, even though in many cases it may seemingly look like one.

anyway you said that m = dx/dy. y=x^2+3x+2 making Dy/Dx=2x+3 => m=2x+3

why is m = 2x + 3? because if dy/dx = 2x +3, m is not 2x + 3, since you defined m as dx/dy. which makes m
1/(2x+3)

unless it's a typo on your part. However, assuming it's not a typo:

dm/dy = (dx/dy) x (dm/dx)

dy/dx = 2x + 3
dx/dy = m = 1/(2x+3)
dm/dx = 2/(2x+3)xY2

multiply dx/dy by dm/dx to get the answer.

By this i am assuming that you correctly defined m as dx/dy and not dy/dx.
 
SSTjumbo
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RE: Hey Everybody, It's Time To Help Me With Calc!

Sat Dec 01, 2001 1:31 pm

OOPS, I meant m=Dy/Dx!!! my bad
I don't know, so this is my signature.
 
docpepz
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RE: Hey Everybody, It's Time To Help Me With Calc!

Sat Dec 01, 2001 1:49 pm

SSTjumbo: even if you meant dx/dy, the question can still be done, and is interesting. You should solve the question the same way.

(dm/dy) = (dx/dy) x (dm/dx)

You should still solve it that way. Except, since m is defined as dy/dx, you should define m, and use the formula (dx/dy) = (inverse of dy/dx) to solve it.

Hope it helps.

and regarding what PanAm747 said, I may be wrong,

but dy= 2x+3 dx, and not 2x+3.

It is wrong to write dy=2X+3 alone.

thus dy/dx= 2X+3.

It's quite important we learn not to forget the 'dx' term, especially in integration. Have you learnt integration by parts or by substitution? If you have, you'll know why.
 
JetService
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RE: Hey Everybody, It's Time To Help Me With Calc!

Sat Dec 01, 2001 10:17 pm

Simplify this way to find Dm/Dy..
(to find Dm/Dy, assuming y=f(x) and m=Dx/Dy...
using variable 'S')
Muliply both sides by 'd'
S=d(y)/fm(x/y)

Now variable use 'S' again to bring 1/m over

SS=Dm(y)/f(x/y)

Simplify 'y' (using variable 'k')

SS=Dm/fk(x)

Now using your 'for instance' result (=> m=2x+3), reduce 'x' by variable 'T' divided by (b/c) and you get the answer.

SST = Dmb/Fck  Big grin

Does it involve natural logs maybe?

Save your natural logs for Sunday Morning.
"Shaddap you!"
 
delta-flyer
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RE: Hey Everybody, It's Time To Help Me With Calc!

Sun Dec 02, 2001 4:50 am

SSTjumbo,

I think you have misstated the problem. I suspect the question is to solve for dm/dx, which is nothing but the second derivative of y with respect to x.

If the problem is correctly stated, then you must first find x as a function of y.

Since m = dy/dx, you can easily get that by differentiating f(x). Next, substitute for x so you get m as a function of y, not x. Then you can differentiate with respect to y to get dm/dy. But the problem is that getting x as a function of y gets tough if y=f(x) is greater than a first order equation, as in your example.

Cheers,
Pete

"In God we trust, everyone else bring data"
 
KROC
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RE: Hey Everybody, It's Time To Help Me With Calc!

Sun Dec 02, 2001 4:55 am

What do you mean "Hey Everybody, It's Time To Help Me With Calc!" You make it sould like we have no chice. We help you or die. Well son, here is my offering...

http://www.casio.com/calculators/product.cfm?section=24&market=0&product=1865
 
Bernard Shakey
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RE: Hey Everybody, It's Time To Help Me With Calc!

Sun Dec 02, 2001 5:14 am

With all due respect to SST, Jetservice that was classic.
Mindless drifter on the road, Carries such an easy load
 
SSTjumbo
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RE: Kroc

Sun Dec 02, 2001 1:28 pm

That's right, you have to help me or else I'll kill you.  Acting devilish  Acting devilish  Acting devilish
I don't know, so this is my signature.
 
SSTjumbo
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RE: Jetservice

Sun Dec 02, 2001 1:32 pm

SST = Dmb/Fck

Please mathematically prove this.  Big thumbs up
I don't know, so this is my signature.
 
JetService
Posts: 4611
Joined: Mon Feb 21, 2000 1:12 pm

RE: Hey Everybody, It's Time To Help Me With Calc!

Sun Dec 02, 2001 1:36 pm

LOL, SST!!!!! I can't! That batteries in my abacus just ran dry.  Big grin

"Shaddap you!"

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