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Consider the graph of x^2+xy+y^2=1. Use implicit differentiation to find all points on the graph where the tangent line is parallel to the line y=-x.
Where do I go after I implicitly differentiate this thing?
2x+y'+2y*y'=0
y'+2y*y'=-2x
y'(1+2y)=-2x
y'=-(2x)/(1+2y)
Did I do that right?
you did not do this correctly. note that you need the product rule to differentiate the xyQuote:
Originally Posted by ebonyscythe
when you get the correct derivative, solve for y' and set it equal to -1 (the slope of y = -x)
Hello, ebonyscythe!
Your differentiation is off . . .
Quote:
Consider the graph of $\displaystyle x^2+xy+y^2\:=\:1$
Use implicit differentiation to find all points on the graph
where the tangent line is parallel to the line $\displaystyle y =-x$
$\displaystyle \text{We have: }\;x^2 + \underbrace{xy}_{\text{product}} + y^2\:=\:1\;\;{\color{blue}[1]}$
. . $\displaystyle \text{then: }\;2x + \overbrace{xy' + y} + 2yy' \:=\:0$
Then we have: .$\displaystyle xy' + 2yy' \:=\:-2x - y$
Factor: .$\displaystyle (x + 2y)y' \:=\:-(2x + y)$
. . Hence: .$\displaystyle y' \:=\:-\frac{2x + y}{x + 2y}$
$\displaystyle \text{The slope of }\, y \,= \,\text{-}x\:\text{ is }-1.$
Set: .$\displaystyle -\frac{2x + y}{x + 2y} \:=\:-1\quad\Rightarrow\quad y \,=\,x$
Substitute into [1]: .$\displaystyle x^2 + x\!\cdot\!x + x^2\:=\:1\quad\Rightarrow\quad3x^2\:=\:1\quad\Righ tarrow\quad x \:=\:\pm\frac{\sqrt{3}}{3}$
. . Then: .$\displaystyle y \:=\:\pm\frac{\sqrt{3}}{3}$
Answers: .$\displaystyle \left(\frac{\sqrt{3}}{3},\:\frac{\sqrt{3}}{3}\righ t),\;\left(\text{-}\,\frac{\sqrt{3}}{3},\:\text{-}\,\frac{\sqrt{3}}{3}\right)$
