spiegel Posted June 14, 2004 Share Posted June 14, 2004 i was wondering if anyone can help me with this question thanks a lot guyz :D xSiny + ySinx = 1 i have to derive it using implicit derivation so this is what i have so far...... -Cosy - Cosx dy/dx = 0 not sure if this is right my head hurts :sleep: Link to comment Share on other sites More sharing options...
Brandon Posted June 14, 2004 Share Posted June 14, 2004 (edited) HUH!!! and i thought algebra 1 was confuzing!!!!!! Japaneese to me Edited June 14, 2004 by musicmaster Link to comment Share on other sites More sharing options...
spiegel Posted June 14, 2004 Author Share Posted June 14, 2004 HUH!!! and i thought algebra 1 cas confuzing!!!!!! lol :D Link to comment Share on other sites More sharing options...
hifive Posted June 14, 2004 Share Posted June 14, 2004 I'm not sure what implicit derivation is... Are you just trying to prove it? Link to comment Share on other sites More sharing options...
spiegel Posted June 14, 2004 Author Share Posted June 14, 2004 ^newton's derivation basically you derive both sides of the equation using d/dx but when it comes to a variable besides x (i.e y) you derive it according to that variable and then put a d/dx next to it example x^2 + y^2 = 1 2x + 2y dy/dx = 0 dx/dy = -2x / 2y get it? its kinda wierd ;) Link to comment Share on other sites More sharing options...
hifive Posted June 14, 2004 Share Posted June 14, 2004 WTF. What math is this? Link to comment Share on other sites More sharing options...
kjordan2001 Posted June 14, 2004 Share Posted June 14, 2004 I think the derivative of sin is cos, and derivative of cos is -sin. Link to comment Share on other sites More sharing options...
spiegel Posted June 14, 2004 Author Share Posted June 14, 2004 I think the derivative of sin is cos, and derivative of cos is -sin. it is i know that but the set up is frikkin wierd :blink: Link to comment Share on other sites More sharing options...
hifive Posted June 14, 2004 Share Posted June 14, 2004 Sin x = Cos (90 - x) Cos x = Sin (90 - x) But I don't think that's what he wants. ----- EDIT. NVMD. Forget I even typed that. Hahaha. So what math is this again? Link to comment Share on other sites More sharing options...
I8PP Posted June 14, 2004 Share Posted June 14, 2004 WTF. What math is this? This be calculus. Beautiful as my teacher would say... Link to comment Share on other sites More sharing options...
spiegel Posted June 14, 2004 Author Share Posted June 14, 2004 Sin x = Cos (90 - x)Cos x = Sin (90 - x) But I don't think that's what he wants. ----- EDIT. NVMD. Forget I even typed that. Hahaha. thanks but thats not wat i need This be calculus. Beautiful as my teacher would say... more like BS calc Link to comment Share on other sites More sharing options...
hifive Posted June 14, 2004 Share Posted June 14, 2004 Ahhh. Calculus. I take AP Cal next year. Exciting, eh? Link to comment Share on other sites More sharing options...
kjordan2001 Posted June 14, 2004 Share Posted June 14, 2004 (edited) I think it's actually: 1*sin(y) + x*cos(y)*dy/dx + sin(x)*dy/dx + y*cos(x) = 0 Edited June 14, 2004 by kjordan2001 Link to comment Share on other sites More sharing options...
spiegel Posted June 14, 2004 Author Share Posted June 14, 2004 i took AP calc in high school this is engineering calc AND I STILL DONT FRIKKIN GET THIS ONE i did 25 of these and all of them were easy but this one is driving me nuts :blink: :blink: :blink: :blink: :blink: :blink: Link to comment Share on other sites More sharing options...
hifive Posted June 14, 2004 Share Posted June 14, 2004 Oh wow. Way out of my league.. Link to comment Share on other sites More sharing options...
dreamz Veteran Posted June 14, 2004 Veteran Share Posted June 14, 2004 you need to use the product rule. xsin(y) + ysin(x) = 1 differentiate by term using the product rule: x*cos(y)*dy/dx + sin(y) + y*cos(x) + sin(x)*dy/dx = 0 solve for dy/dx. Link to comment Share on other sites More sharing options...
spiegel Posted June 14, 2004 Author Share Posted June 14, 2004 I think it's actually:1*sin(y) + x*cos(y) + dy/dx *(sin(x) + y*cos(x)) = 0 But I'm not sure, summer mode has set in on my brain. hey thanks a lot i think thats right and i simplified more thanks a lot guys :D :D :D :D Link to comment Share on other sites More sharing options...
session Posted June 14, 2004 Share Posted June 14, 2004 (edited) xSiny + ySinx = 1 [siny + x(cosy)y'] + [y'(sinx) + y(cosx)] = 0 siny + y'(x(cosy)) + y'(sinx) + y(cosx) = 0 y'[x(cosy) + sinx] + [siny + y(cosx)] = 0 y'[x(cosy) + sinx] = -siny - y(cosx) y' = {-siny - y(cosx)}/{[x(cosy) + sinx]} * y' = dy/dx Edited June 14, 2004 by session Link to comment Share on other sites More sharing options...
kjordan2001 Posted June 14, 2004 Share Posted June 14, 2004 you need to use the product rule.xsin(y) + ysin(x) = 1 differentiate by term using the product rule: x*cos(y)*dy/dx + sin(y) + y*cos(x) + sin(x)*dy/dx = 0 solve for dy/dx. Ah yeah, knew I missed a dy/dx somewhere, and for some reason I changed it to dy/dx*(sin(x)+y*cos(x)) :wacko: *quickly goes to change his answer* Link to comment Share on other sites More sharing options...
session Posted June 14, 2004 Share Posted June 14, 2004 use product rule for x*siny and y*sinx Link to comment Share on other sites More sharing options...
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