bmelyk
I need someone to explain to me step by step the following problem:
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bmelyk
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Find the equation of the tangent line to the curve
\[x^{2}y^{2}-1=3(x+y)\] at the points (1,-1)
bmelyk
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I have the problem worked out in front of me (prof gave us the answers) but, his work is disorganized and I don't understand what it is he's doing.
bmelyk
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I figure I have to do this implicitly.
asnaseer
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yes - you are correct - use implicit differentiation to get an expression for slope of the tangent line at any point (x,y).
then substitute x=1, y=-1 to find the value of the slope of the tangent line at (1, -1)
then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, -1)
Rezz5
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is the answer y=4-5x?
bmelyk
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that's not the answer.... the answer is: y=-1/5x+4/5
Rezz5
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its a plane....?
bmelyk
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okay and the way my prof did his differentiation was really weird.
his first step looks like: x^{2}2y*y'+y^{2}(2x)=3(1+y')
bmelyk
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a plane? huh?
bmelyk
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that's not how i would start the implicit differentiation.
bmelyk
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\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]
asnaseer
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how would you have done this?
bmelyk
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haha no idea... i suck at math : (
bmelyk
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so i use the product rule for x^2*y^2 ??
asnaseer
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correct
bmelyk
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and for the 3*(x+y) ??
asnaseer
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that is just straight differentiation since it can be written as: 3x + 3y
bmelyk
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so that will just equal 3y' ??
bmelyk
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or 3+3y'
asnaseer
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so you can effectively just ignore the 3 outside the braces and just differentiate what is inside the braces
bmelyk
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so it's 3(1+y')
asnaseer
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yes - correct
bmelyk
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okay so now i have the same first line as he does haha.
bmelyk
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i ignored the 3 outside the brackets because it's a constant right?
asnaseer
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yes.
thats good - you are making progress - maybe you are a lot better at maths than you think. :)
bmelyk
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haha maybe, maybe not...
bmelyk
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so now i can take the y^2(2x) over to the other side?
bmelyk
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and the right hand side over to the left?
asnaseer
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yes - what you need to do is to get an expression for y'.
so move all terms involving y' to the left and the rest to the right
asnaseer
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first expand the expression on the right
bmelyk
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do i leave my 3 on the right hand side?
asnaseer
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snap! yes :)
bmelyk
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okay so now i have: (x^2)(2y*y')-(1-y')=3(-2xy^2)
bmelyk
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that look right to you?
asnaseer
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not quite :(
bmelyk
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oh snappp
asnaseer
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you started with this:\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]so first expand the right to get:\[x^{2}(2y*y')+y^{2}(2x)=3+3y'\]then move terms involving y' to the left and the rest to the right to get:\[x^{2}(2y*y')-3y'=3-y^{2}(2x)\]
asnaseer
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then factorise y' on the left
bmelyk
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\[y'=\frac{ 3-2xy^{2} }{ 2x^{2}y-3 }\]
asnaseer
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perfect!
bmelyk
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woooo. so now that part is done, right?
asnaseer
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yup - the rest should be straight forward. I need to go for dinner now. I'll check back afterwards to see if you had any further problems.
asnaseer
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the steps left are:
then substitute x=1, y=-1 to find the value of the slope of the tangent line at (1, -1)
then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, -1)
bmelyk
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okay thanks. : )
asnaseer
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yw :) and good luck!
bmelyk
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i got the rest of it now ; ) thanks a bunch!
asnaseer
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thats great!
and you are more than welcome! :)