mathcalculus
Help:implicit differentiation with a fractions :(
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mathcalculus
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myko
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multiplying everything by 36 you get;
2x^2+y^2=36
which maybe more clear for you, :)
mathcalculus
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hey! :) i tried this: x²/16 + y²/36 = 1
(1/16)x² + (1/36)y² = 1
2(1/16)x + 2(1/36)yy' = 0
mathcalculus
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but so far im not sure what to do after because i know i get x/8 + ?? but then i can't really multiply (1/36) 2y*dy/dx ?
mathcalculus
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isn't the point here to find dy/dx?
myko
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let \(F(x,y)=2x^2+y^2=36\)
then:
\(dF(x,y)=4xdx+2ydy=0\)
from here:
\(y'=dy/dx=-4x/2y=-2x/y\)
mathcalculus
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huh? o.o
mathcalculus
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howd you get this??
let F(x,y)=2x2+y2=36
myko
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\(dF(x,y)\) is colled full differential
it is: \(dF(x,y)=F_xdx+F_ydy\)
mathcalculus
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i never seen that before
myko
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\(F_x\) is partial derivative respect to x
mathcalculus
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ok
mathcalculus
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i know those, but where are the numbers coming from?
mathcalculus
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in what order, where did you take them?
mathcalculus
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i know they're given but how did you get 36 on the other side. and 2x^2+y^2let F(x,y)=2x2+y2=36
myko
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ok, other way:
think that y is actualy a function of x, so it is y=y(x). Then:
\(2x^2+y(x)^2=36\)
now we differentiate keeping this in mind
4x+2yy'=0
now just solve for y'
mathcalculus
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2x2+y(x)2=36?
mathcalculus
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yes i know when to differentiate but how did you change the function to that
myko
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Ya you right I made a mistake. I was treating 36 like 32, :). Shold be. Rest of steps same.
dw:1381788857534:dw|
myko
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|dw:1381789088155:dw|
mathcalculus
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i know I was going to tell you.
mathcalculus
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you can't have 36 at the end.
mathcalculus
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i saw a different way on yahoo, would be be able to help me with this one
x²/16 + y²/36 = 1
(1/16)x² + (1/36)y² = 1
2(1/16)x + 2(1/36)yy' = 0
(x/8) + (y/18)y' = 0
(y/18)y' = -(x/8)
y' = -(x/8)/(y/18)
y' = -(x/8)(18/y)
y' = -(9x/4y)
y'(2) = -(9(2))/(4(5.2))
y'(2) = -18/20.8
y'(2) = -9/10.4
y'(2) = -0.865
mathcalculus
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I'm not really sure how they got the 4th step :/.... this part----> + (y/18)*dy/dx ??
myko
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2(1/16)x + 2(1/36)yy' = 0
(2/16)x+(2/36)yy'=0
(x/8) + (y/18)y' = 0
mathcalculus
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yes my question is exactly what you wrote up there .... |dw:1381789520628:dw|(x/8) + (y/18)y' = 0
myko
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(2/36)=1/18
mathcalculus
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how did it get simplified to y/18*dy/dx?
myko
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your answer is same, just simplify 2 in numerator with 36 in denominator
mathcalculus
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thanks! :) i dontkknow why i missed that silly stuff.
myko
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:)
mathcalculus
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then i plug in for x and y and find the common denominators rights?
myko
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yes
mathcalculus
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wait we're not done yet! :(
mathcalculus
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i got -.005379
mathcalculus
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the question is asking y'(1)
myko
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y' = -(9x/4y)
so: substitute x=1, y=5.80948
y'(1)=-9/4(5.80948)
mathcalculus
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okay i got dy/dx=-.387298
mathcalculus
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got it thanks @myko :))!
myko
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yw