Find the equation of the tangent line at the point (1,1) for the curve x^3+y^3-2xy=0. But I can't for the life of me figure out how to solve the equation for y.

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- amistre64

a tangent line is vector + point

- amistre64

the tangent vector to the curve matches the slope of the tangent line which is the derivative of the equation at the point right?

- amistre64

D(x^3+y^3-2xy=0)
3x^2 x' +3y^3 y' -2xy' -2y x' = 0 right?

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- amistre64

now solve for y'

- amistre64

x' = 1

- amistre64

(y' 3y^2 - y' 2x)+(3x^2 -2y) = 0
y' (3y^2 - 2x) = -(3x^2 -2y)
y' = -3x^2 +2y
---------- now fill in your x and y value from the point
3y^2 - 2x

- amistre64

point (x=1,y=1)
-3+2 -1
----- = --- = -1
3-2 1

- amistre64

teh tangent vector has the same slope as the tangent line here

- amistre64

Tv <1,-1> would be suitable

- amistre64

so lets define the line by define all magnitudes of this vector from that point
x = 1 -t
y = 1 +t
z = 0

- amistre64

but thats only if we want R^3; R^2 can be just the tangent lint thru the point now lol... i read to much into that

- amistre64

y = mx +b
1 = (-1)1 + b
1 = -1 + b
2 = b
Eq tan line: y = -x +2

- amistre64

you implicitly differentiate the equation frist; then solve for dy/dx = y'

- amistre64

then plug in your x and y amounts to determine the slope

- anonymous

whoa - so I don't have to solve for y first.
instead I find the derivative for each part of the product in terms of x;
and for each part in terms of y ----no that is not what is going on sorry the vector thing is way over my head...trying also to get at the notation if everyone could agree to use newton or liebnitz my life would be easier
am working on it thanks so far

- amistre64

yeah, i had the wrong thought in me head at first :)
No, you dont need to solve for y first; tell me:
Does 3x +y = 4 and y = -3x +4 mean the same thing?

- anonymous

hey amistre what happened to that derivative of integral question??

- amistre64

lol... i left that to smarter people than me ;)

- anonymous

bt i answered..nd no-one replied..lol

- amistre64

kant; if they are equal, then we can derive them both and get the same results right?

- amistre64

the thing is; deriving is deriving is deriving regardless of whether you can solve for y forst or not

- amistre64

3x +y = 4 y = -3x +4
3 + y' = 0 y' = -3
y' = -3

- amistre64

3xy^2 +y = 9 cant be solved for y; but it doesnt matter does it? just derive it like normal and solve for y' afterwards

- amistre64

Dx(3xy^2) = 3x2y y' + xy^2 ; do you see why?

- anonymous

well i get the 3x+y=4 == y=-3x+4 i can solve that one
but was having trouble with the equation x^3-y^3-2xy could not figure out how to get the y's to one side. nothing factored out of each element ...
let me fuss with the new item on paper

- amistre64

just derive it as tho you could solve for y; becasue it aint the y that you want to find; its y'

- amistre64

whats the derivative of 3x?

- amistre64

well, with respect to x

- anonymous

3

- amistre64

...... let me re ask that; i type it in wrong lol
your right; but..

- amistre64

what is the derivative with respect to x of: x^3 is what i meant :)

- anonymous

3x^2

- amistre64

good;
what is the derivative with respect to x of: -y^2 ?

- amistre64

.....of -y^3; i gotta learn to proof my work

- anonymous

since there is no x...-y^3 (dx)?

- amistre64

dont worry about the "no x" part; go ahead and 'imply' that there is an x.
It is the same derivative as tho you were doing:
x^3 with respect to x but we got a y in there..

- anonymous

so I do take the derivative of y: 3y^2?

- amistre64

exactly :).... but the thing is, when they first teach you derivatives they have you throw out an important part.... and we actually need to keep that part in to see what happens

- amistre64

lets do x^3 again but keep all the parts and see if you can figure out the secret that they are keeping from you ok?

- anonymous

those b*st*rds....
then i deal with the mixed item the same way 2x and 2y ( the one im deriving drops off)

- amistre64

d(x^3) dx
------ = --- 3x^2 ; you see the dx/dx part? that equals 1
dx dx so they simply tell you to ignore it

- anonymous

right at first i thought the y' notation would be easier but then we started canceling the dy/dx and then that seemed easier

- amistre64

now lets derive the y^3 and see what happens:
d (y^3) dy
------ = --- 3y^2 ; you recognize that dy/dx?
dx dx

- amistre64

that dy/dx doesnt equal 1 so we cant just ignore it; in fact, its what youve always done. watch
d (y) d (x^3)
---- = -------
dx dx
dy dx
-- 1 = -- 3x^2
dx dx
dy/dx = 1 * 3x^2
dy/dx = 3x^2

- anonymous

looking at it on paper that is easier for me

- amistre64

when you do implicits.... KEEP ALL YOUR DERIVED bits in place till the end

- amistre64

derive (-2x)(y) for me then; and remember to use the product rule for it

- anonymous

f'g+g'f=
so
-2y+(1)-2x = -2y+1-2x

- amistre64

you threw out your derived bits.... but you got the idea down, now watch this:
f = -2x ; f' = -2 x'
g = y ; g' = 1 y'
f'g + fg':
-2 x' y + -2x y' ; since x' = dx/dx = 1 we get
-2y - 2x y' rigth?

- anonymous

akkk not sure where that +1 came from typing math = harder math
but the extra y' is just 1 ya?

- amistre64

lets put this into your equation and see if we can peice together the mystery :)
x^3 +y^3 -2xy = 0
3x^2 x' + 3y^2 y' -2y x' -2x y' = 0
now we can take out the derived bits that equal 1:
x' = dx/dx = 1

- amistre64

teh extra y' is just ... y' cant determine that yet becasue we have to solve for it :) it is actually what we are looking for

- anonymous

ok...i wont jump the gun

- anonymous

3x^2+3y^2(dy)-2y-2x(dy)

- amistre64

exactly :) now
3x^2 + 3y^2 dy -2y -2x dy = 0
get all the y's to one side and the rest to the other:
3y^2 dy -2x dy = -3x^2 +2y
dy (3y^2 -2x) = -3x^2 +2y
dy = -3x^2 +2y
--------- now plug in your point (x=1, y=1) to find
3y^2 -2x the slope

- amistre64

I get dy = slope = -1
now we put that into the equation for the line:
y = mx + b and use the point to calibrate it:
1 = -1(1) + b
1 = -1 + b
1+1 = b
the equation of the tangent line at (1,1) is:
y = -x + 2

- anonymous

once again let me transfer to paper to see it

- anonymous

alright i can see that ....one second while i recompute the derivative part im not clear on who got the (dy) part and who did not

- amistre64

if theres a y involved; it will end up with a dy part with it

- anonymous

there i like hard fast rule like that

- anonymous

I think I have it --- find the derivatives for x just like normal; derivative of y just like normal but whenever there is a derivative of y multiply by (dy)....whoa like when using u substitution...thanks for the help

- amistre64

you got it :)

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