Question

Use implicit differentiation to find an equation of the line tangent to the curve x^3+2xy+y^3=13 at the point (1,2)

Answer 1

its pretty basic, what issue are you coming across?

Answer 2

all i've done is find the implicit derivative. after that, i don't know what i'm supposed to do

Answer 3

show me your implicit and let me see if I agree with it

Answer 4

\[dy divdx=-3x ^{2}\div(2x+3y ^{2})\]

Answer 5

\[x^3+2xy+y^3=13\]
\[3x^2X'+2yX'+2xY'+3y^2Y'=0\]
\[3x^2+2y+2xY'+3y^2Y'=0\]
\[3x^2+2y+Y'(2x+3y^2)=0\]
\[Y'(2x+3y^2)=-3x^2-2y\]
\[Y'=\frac{-3x^2-2y}{2x+3y^2}\]
this is what I get, havent tried to simplfy it tho

Answer 6

this here is the equation for the slope of the line at any given point on the graph

Answer 7

when you knw a point and a slope, you can form a line

Answer 8

i thought that when you applied the product rule to 2xy, the 2 would go off to 0 when differentiating x, causing the first part to be 0 and just be left with 2y(dy/dx)?

Answer 9

okay, so the equation we just got is the slope. all i have to do is insert it into the point slope formula then?

Answer 10

well, the product rule itself doesnt change :)

[fg]' = f'g + fg' regardless of what variables we choose right

Answer 11

well, lets use the point given to find a numerical value for the slope, then it will be easier to use

Answer 12

so y-2=(slope we found)(x-1)?

Answer 13

\[Y'=\frac{-3x^2-2y}{2x+3y^2};(1,2)\]
\[Y'=\frac{-3.1^2-2.2}{2.1+3.2^2}\]
\[Y'=\frac{-3-4}{2+12}\]
\[Y'=\frac{-7}{14}=-\frac{1}{2}\]

and yes, the rest of what you got there is good for the line

Answer 14

okay. is y=-1/2 the equation?

Answer 15

and as a clarification, if we product rule the 2xy we get:

[2xy]' = 2'xy +2x'y +2xy'
= 0xy +2x' y +2x y'
= 2y +2x y'

Answer 16

no, the slope is -1/2

Answer 17

ohhh okay, got it! i was thinking that the 2 went along with the x. I didnt realize theey were 3 seperate components! It all makes sense now. thank you!

Answer 18

youre welcome, and good luck :)

Answer 19

thank you! :)