Question

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.

Answer 1

a tangent line is vector + point

Answer 2

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

Answer 3

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

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

Answer 4

now solve for y'

Answer 5

x' = 1

Answer 6

(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

Answer 7

point (x=1,y=1)

-3+2 -1
----- = --- = -1
3-2 1

Answer 8

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

Answer 9

Tv <1,-1> would be suitable

Answer 10

so lets define the line by define all magnitudes of this vector from that point

x = 1 -t
y = 1 +t
z = 0

Answer 11

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

Answer 12

y = mx +b

1 = (-1)1 + b

1 = -1 + b

2 = b
Eq tan line: y = -x +2

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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?

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

3x +y = 4 y = -3x +4

3 + y' = 0 y' = -3

y' = -3

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

whats the derivative of 3x?

Answer 28

well, with respect to x

Answer 29

3

Answer 30

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

your right; but..

Answer 31

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

Answer 32

3x^2

Answer 33

good;

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

Answer 34

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

Answer 35

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

Answer 36

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..

Answer 37

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

Answer 38

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

Answer 39

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?

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

now lets derive the y^3 and see what happens:

d (y^3) dy
------ = --- 3y^2 ; you recognize that dy/dx?
dx dx

Answer 44

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

Answer 45

looking at it on paper that is easier for me

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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?

Answer 50

akkk not sure where that +1 came from typing math = harder math

but the extra y' is just 1 ya?

Answer 51

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

Answer 52

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

Answer 53

ok...i wont jump the gun

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

once again let me transfer to paper to see it

Answer 58

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

Answer 59

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

Answer 60

there i like hard fast rule like that

Answer 61

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

Answer 62

you got it :)