anonymous
  • anonymous
I need someone to explain to me step by step the following problem:
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Find the equation of the tangent line to the curve \[x^{2}y^{2}-1=3(x+y)\] at the points (1,-1)
anonymous
  • anonymous
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.
anonymous
  • anonymous
I figure I have to do this implicitly.

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asnaseer
  • asnaseer
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)
anonymous
  • anonymous
is the answer y=4-5x?
anonymous
  • anonymous
that's not the answer.... the answer is: y=-1/5x+4/5
anonymous
  • anonymous
its a plane....?
anonymous
  • anonymous
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')
anonymous
  • anonymous
a plane? huh?
anonymous
  • anonymous
that's not how i would start the implicit differentiation.
anonymous
  • anonymous
\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]
asnaseer
  • asnaseer
what your prof did is correct - he is using the product rule, see here: http://www.1728.org/chainrul.htm
asnaseer
  • asnaseer
how would you have done this?
anonymous
  • anonymous
haha no idea... i suck at math : (
anonymous
  • anonymous
so i use the product rule for x^2*y^2 ??
asnaseer
  • asnaseer
correct
anonymous
  • anonymous
and for the 3*(x+y) ??
asnaseer
  • asnaseer
that is just straight differentiation since it can be written as: 3x + 3y
anonymous
  • anonymous
so that will just equal 3y' ??
anonymous
  • anonymous
or 3+3y'
asnaseer
  • asnaseer
so you can effectively just ignore the 3 outside the braces and just differentiate what is inside the braces
anonymous
  • anonymous
so it's 3(1+y')
asnaseer
  • asnaseer
yes - correct
anonymous
  • anonymous
okay so now i have the same first line as he does haha.
anonymous
  • anonymous
i ignored the 3 outside the brackets because it's a constant right?
asnaseer
  • asnaseer
yes. thats good - you are making progress - maybe you are a lot better at maths than you think. :)
anonymous
  • anonymous
haha maybe, maybe not...
anonymous
  • anonymous
so now i can take the y^2(2x) over to the other side?
anonymous
  • anonymous
and the right hand side over to the left?
asnaseer
  • asnaseer
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
  • asnaseer
first expand the expression on the right
anonymous
  • anonymous
do i leave my 3 on the right hand side?
asnaseer
  • asnaseer
snap! yes :)
anonymous
  • anonymous
okay so now i have: (x^2)(2y*y')-(1-y')=3(-2xy^2)
anonymous
  • anonymous
that look right to you?
asnaseer
  • asnaseer
not quite :(
anonymous
  • anonymous
oh snappp
asnaseer
  • asnaseer
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
  • asnaseer
then factorise y' on the left
anonymous
  • anonymous
\[y'=\frac{ 3-2xy^{2} }{ 2x^{2}y-3 }\]
asnaseer
  • asnaseer
perfect!
anonymous
  • anonymous
woooo. so now that part is done, right?
asnaseer
  • asnaseer
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
  • asnaseer
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)
anonymous
  • anonymous
okay thanks. : )
asnaseer
  • asnaseer
yw :) and good luck!
anonymous
  • anonymous
i got the rest of it now ; ) thanks a bunch!
asnaseer
  • asnaseer
thats great! and you are more than welcome! :)

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