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anonymous
 5 years ago
Find the equation of the tangent line at the point (1,1) for the curve x^3+y^32xy=0. But I can't for the life of me figure out how to solve the equation for y.
anonymous
 5 years ago
Find the equation of the tangent line at the point (1,1) for the curve x^3+y^32xy=0. But I can't for the life of me figure out how to solve the equation for y.

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1a tangent line is vector + point

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the tangent vector to the curve matches the slope of the tangent line which is the derivative of the equation at the point right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1D(x^3+y^32xy=0) 3x^2 x' +3y^3 y' 2xy' 2y x' = 0 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1point (x=1,y=1) 3+2 1  =  = 1 32 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1teh tangent vector has the same slope as the tangent line here

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1Tv <1,1> would be suitable

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1so lets define the line by define all magnitudes of this vector from that point x = 1 t y = 1 +t z = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1but 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
 5 years ago
Best ResponseYou've already chosen the best response.1y = mx +b 1 = (1)1 + b 1 = 1 + b 2 = b Eq tan line: y = x +2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1you implicitly differentiate the equation frist; then solve for dy/dx = y'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1then plug in your x and y amounts to determine the slope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whoa  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
 5 years ago
Best ResponseYou've already chosen the best response.1yeah, 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
 5 years ago
Best ResponseYou've already chosen the best response.0hey amistre what happened to that derivative of integral question??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1lol... i left that to smarter people than me ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bt i answered..nd noone replied..lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1kant; if they are equal, then we can derive them both and get the same results right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the thing is; deriving is deriving is deriving regardless of whether you can solve for y forst or not

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.13x +y = 4 y = 3x +4 3 + y' = 0 y' = 3 y' = 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.13xy^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
 5 years ago
Best ResponseYou've already chosen the best response.1Dx(3xy^2) = 3x2y y' + xy^2 ; do you see why?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i get the 3x+y=4 == y=3x+4 i can solve that one but was having trouble with the equation x^3y^32xy 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
 5 years ago
Best ResponseYou've already chosen the best response.1just derive it as tho you could solve for y; becasue it aint the y that you want to find; its y'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1whats the derivative of 3x?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1well, with respect to x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1...... let me re ask that; i type it in wrong lol your right; but..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1what is the derivative with respect to x of: x^3 is what i meant :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1good; what is the derivative with respect to x of: y^2 ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1.....of y^3; i gotta learn to proof my work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since there is no x...y^3 (dx)?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dont 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
 5 years ago
Best ResponseYou've already chosen the best response.0so I do take the derivative of y: 3y^2?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1exactly :).... 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
 5 years ago
Best ResponseYou've already chosen the best response.1lets 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
 5 years ago
Best ResponseYou've already chosen the best response.0those b*st*rds.... then i deal with the mixed item the same way 2x and 2y ( the one im deriving drops off)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1d(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
 5 years ago
Best ResponseYou've already chosen the best response.0right at first i thought the y' notation would be easier but then we started canceling the dy/dx and then that seemed easier

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1now lets derive the y^3 and see what happens: d (y^3) dy  =  3y^2 ; you recognize that dy/dx? dx dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1that 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
 5 years ago
Best ResponseYou've already chosen the best response.0looking at it on paper that is easier for me

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1when you do implicits.... KEEP ALL YOUR DERIVED bits in place till the end

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1derive (2x)(y) for me then; and remember to use the product rule for it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f'g+g'f= so 2y+(1)2x = 2y+12x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1you 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
 5 years ago
Best ResponseYou've already chosen the best response.0akkk not sure where that +1 came from typing math = harder math but the extra y' is just 1 ya?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1lets 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
 5 years ago
Best ResponseYou've already chosen the best response.1teh 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok...i wont jump the gun

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.03x^2+3y^2(dy)2y2x(dy)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1exactly :) 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
 5 years ago
Best ResponseYou've already chosen the best response.1I 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
 5 years ago
Best ResponseYou've already chosen the best response.0once again let me transfer to paper to see it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright 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
 5 years ago
Best ResponseYou've already chosen the best response.1if theres a y involved; it will end up with a dy part with it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there i like hard fast rule like that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
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