Hypercube Posted December 28, 2003 Share Posted December 28, 2003 Question on number 6 [see attached], am I missing something or can someone please explain the separation of variables. Question on number 8 [see attached], I know it is the limit definition, but what is the shortcut to solving it [/me doesn't remember]. Thanks in advance. Link to comment Share on other sites More sharing options...
Hypercube Posted December 28, 2003 Author Share Posted December 28, 2003 Problem 8 Link to comment Share on other sites More sharing options...
artichokee Posted December 28, 2003 Share Posted December 28, 2003 Snip... On Number 6, just find the derivative. Link to comment Share on other sites More sharing options...
MackDiesel2010 Posted December 28, 2003 Share Posted December 28, 2003 On Number 6, just find the derivative. yeah you use derivative, but you can't solve for y by itself. you'll have to use dy/dx in place of derivative of y, without forgetting the "chain rule." crap i can't see image i'll post again... Link to comment Share on other sites More sharing options...
Guest Posted December 28, 2003 Share Posted December 28, 2003 6) Differentiate 2x-2y ahh crap... im confused. Link to comment Share on other sites More sharing options...
dtmunir Posted December 28, 2003 Share Posted December 28, 2003 Question 6: x^2 - 2xy + 4y^2 = 64 solution: 2x - [ (2x . dy/dx) + (y.2) ] + (8y . dy/dx) = 0 2x - 2x.dy/dx - 2y +8y.dy/dx =0 2x -2y + (8y-2x).dy/dx = 0 (8y-2x).dy/dx = 2y - 2x dy/dx = (2y-2x)/(8y-2x) I hope you can follow. By the way '.' stands for multiplicatioon Question 8: lim h app 0 [ 5 (1/2 + h)^4 - 5(1/2)^4 ]/h Solution: as h approaches 0, (1/2 + h)^4 approaches (1/2)^4 therefore the expression becomes: [ 5(1/2)^4 - 5(1/2)^4 ]/h now just look at the numerator, and ull see tht it works out to be 0 therefore the solution is tht as h approaches 0, the expression approaches 0 as well lemme know if u need more help Link to comment Share on other sites More sharing options...
Hypercube Posted December 28, 2003 Author Share Posted December 28, 2003 ty; forgot about dy/dx on the first one; the second, I get a calculator answer of 5/4 [iirc]? Thanks once again. Link to comment Share on other sites More sharing options...
adamp2p Posted December 28, 2003 Share Posted December 28, 2003 I highly advise the calculus forums at www.sosmath.com "the best" Link to comment Share on other sites More sharing options...
neowin_hipster Posted December 28, 2003 Share Posted December 28, 2003 For #8, dont' listen to dtmunir, because 0/0 is undetermined. You would either need to use l'hopitals rule or expand it out. Or you could try using delta epsilon :p Link to comment Share on other sites More sharing options...
caustiK Posted December 28, 2003 Share Posted December 28, 2003 8. use l'hopital's: 20(.5+h)^3, h=0 20(.5)^3 = 20(1/8) = 5/2. i think that's right. Link to comment Share on other sites More sharing options...
Hypercube Posted December 28, 2003 Author Share Posted December 28, 2003 For 8, it would seem that 5/2 is correct. @adamp2p, your recommended site helped greatly. ty once again to all that helped. Link to comment Share on other sites More sharing options...
dtmunir Posted December 30, 2003 Share Posted December 30, 2003 8. use l'hopital's:20(.5+h)^3, h=0 20(.5)^3 = 20(1/8) = 5/2. i think that's right. hmmm maybe my eyes are deceiving me but didnt the original question posted up there have a power of 4 instead of 3 for (.5+h) lemme check... oh yes...my eyes are NOT deceiving me :no: danish Link to comment Share on other sites More sharing options...
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