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bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1Find the equation of the tangent line to the curve \[x^{2}y^{2}1=3(x+y)\] at the points (1,1)

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 years ago
Best ResponseYou've already chosen the best response.1I figure I have to do this implicitly.

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1yes  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)

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1that's not the answer.... the answer is: y=1/5x+4/5

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1okay 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
 2 years ago
Best ResponseYou've already chosen the best response.1that's not how i would start the implicit differentiation.

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1what your prof did is correct  he is using the product rule, see here: http://www.1728.org/chainrul.htm

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1how would you have done this?

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1haha no idea... i suck at math : (

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1so i use the product rule for x^2*y^2 ??

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1that is just straight differentiation since it can be written as: 3x + 3y

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1so that will just equal 3y' ??

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1so you can effectively just ignore the 3 outside the braces and just differentiate what is inside the braces

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1okay so now i have the same first line as he does haha.

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1i ignored the 3 outside the brackets because it's a constant right?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1yes. thats good  you are making progress  maybe you are a lot better at maths than you think. :)

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1haha maybe, maybe not...

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1so now i can take the y^2(2x) over to the other side?

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1and the right hand side over to the left?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1yes  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
 2 years ago
Best ResponseYou've already chosen the best response.1first expand the expression on the right

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1do i leave my 3 on the right hand side?

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1okay so now i have: (x^2)(2y*y')(1y')=3(2xy^2)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1you 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:\[x^{2}(2y*y')3y'=3y^{2}(2x)\]

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1then factorise y' on the left

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1\[y'=\frac{ 32xy^{2} }{ 2x^{2}y3 }\]

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.1woooo. so now that part is done, right?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1yup  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
 2 years ago
Best ResponseYou've already chosen the best response.1the 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
 2 years ago
Best ResponseYou've already chosen the best response.1i got the rest of it now ; ) thanks a bunch!

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1thats great! and you are more than welcome! :)
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