Implicit Diff and an Ellipse

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Oct 15th 2007, 07:08 PM
ebonyscythe

Implicit Diff and an Ellipse

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?
Oct 15th 2007, 07:32 PM
Jhevon

Quote:

Originally Posted by ebonyscythe

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 xy

when you get the correct derivative, solve for y' and set it equal to -1 (the slope of y = -x)
Oct 15th 2007, 07:46 PM
Soroban

Hello, ebonyscythe!

Your differentiation is off . . .

Quote:

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 $y =-x$

$\text{We have: }\;x^2 + \underbrace{xy}_{\text{product}} + y^2\:=\:1\;\;{\color{blue}[1]}$
. . $\text{then: }\;2x + \overbrace{xy' + y} + 2yy' \:=\:0$

Then we have: . $xy' + 2yy' \:=\:-2x - y$

Factor: . $(x + 2y)y' \:=\:-(2x + y)$

. . Hence: . $y' \:=\:-\frac{2x + y}{x + 2y}$

$\text{The slope of }\, y \,= \,\text{-}x\:\text{ is }-1.$

Set: . $-\frac{2x + y}{x + 2y} \:=\:-1\quad\Rightarrow\quad y \,=\,x$

Substitute into [1]: . $x^2 + x\!\cdot\!x + x^2\:=\:1\quad\Rightarrow\quad3x^2\:=\:1\quad\Righ tarrow\quad x \:=\:\pm\frac{\sqrt{3}}{3}$
. . Then: . $y \:=\:\pm\frac{\sqrt{3}}{3}$

Answers: . $\left(\frac{\sqrt{3}}{3},\:\frac{\sqrt{3}}{3}\righ t),\;\left(\text{-}\,\frac{\sqrt{3}}{3},\:\text{-}\,\frac{\sqrt{3}}{3}\right)$