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a tangent line is vector + point
the tangent vector to the curve matches the slope of the tangent line which is the derivative of the equation at the point right?
D(x^3+y^3-2xy=0) 3x^2 x' +3y^3 y' -2xy' -2y x' = 0 right?
now solve for y'
x' = 1
(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
point (x=1,y=1) -3+2 -1 ----- = --- = -1 3-2 1
teh tangent vector has the same slope as the tangent line here
Tv <1,-1> would be suitable
so lets define the line by define all magnitudes of this vector from that point x = 1 -t y = 1 +t z = 0
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
y = mx +b 1 = (-1)1 + b 1 = -1 + b 2 = b Eq tan line: y = -x +2
you implicitly differentiate the equation frist; then solve for dy/dx = y'
then plug in your x and y amounts to determine the slope
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
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?
hey amistre what happened to that derivative of integral question??
lol... i left that to smarter people than me ;)
bt i answered..nd no-one replied..lol
kant; if they are equal, then we can derive them both and get the same results right?
the thing is; deriving is deriving is deriving regardless of whether you can solve for y forst or not
3x +y = 4 y = -3x +4 3 + y' = 0 y' = -3 y' = -3
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
Dx(3xy^2) = 3x2y y' + xy^2 ; do you see why?
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
just derive it as tho you could solve for y; becasue it aint the y that you want to find; its y'
whats the derivative of 3x?
well, with respect to x
...... let me re ask that; i type it in wrong lol your right; but..
what is the derivative with respect to x of: x^3 is what i meant :)
good; what is the derivative with respect to x of: -y^2 ?
.....of -y^3; i gotta learn to proof my work
since there is no x...-y^3 (dx)?
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..
so I do take the derivative of y: 3y^2?
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
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?
those b*st*rds.... then i deal with the mixed item the same way 2x and 2y ( the one im deriving drops off)
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
right at first i thought the y' notation would be easier but then we started canceling the dy/dx and then that seemed easier
now lets derive the y^3 and see what happens: d (y^3) dy ------ = --- 3y^2 ; you recognize that dy/dx? dx dx
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
looking at it on paper that is easier for me
when you do implicits.... KEEP ALL YOUR DERIVED bits in place till the end
derive (-2x)(y) for me then; and remember to use the product rule for it
f'g+g'f= so -2y+(1)-2x = -2y+1-2x
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?
akkk not sure where that +1 came from typing math = harder math but the extra y' is just 1 ya?
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
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
ok...i wont jump the gun
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
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
once again let me transfer to paper to see it
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
if theres a y involved; it will end up with a dy part with it
there i like hard fast rule like that
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
you got it :)