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John's question at Yahoo! Answers regarding implicit differentiation & horizontal/vertical tangents

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MarkFL

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Feb 24, 2012
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John's question at Yahoo! Answers regarding implicit differentiation & horizontal/vertical tangents

Here is the question:

Implicit Differentiation Problem?


For the equation x^2+xy+y^2=y;

(a) Find the points where the tangent line is parallel to the x-axis. (b) Find the points where the tangent line is parallel to the y-axis.
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Re: John's question at Yahoo! Answers regarding implicit differentiation & horizontal/vertical tange

Hello John,

We are given the equation:

\(\displaystyle x^2+xy+y^2=y\)

Implicitly differentiating with respect to $x$, we find:

\(\displaystyle 2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=\frac{dy}{dx}\)

Solving for \(\displaystyle \frac{dy}{dx}\), we obtain:

\(\displaystyle x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}=-2x-y\)

\(\displaystyle \frac{dy}{dx}(x+2y-1)=-2x-y\)

\(\displaystyle \frac{dy}{dx}=\frac{2x+y}{1-x-2y}\)

Now, the tangent line(s) will be horizontal where the numerator is zero, which implies:

\(\displaystyle y=-2x\)

Substituting this into the original equation, we obtain:

\(\displaystyle x^2+x(-2x)+(-2x)^2=(-2x)\)

\(\displaystyle x^2-2x^2+4x^2=-2x\)

\(\displaystyle 3x^2+2x=0\)

\(\displaystyle x(3x+2)=0\)

\(\displaystyle x=0,\,-\frac{2}{3}\)

Hence we have two points from $(x,-2x)$:

\(\displaystyle (0,0),\,\left(-\frac{2}{3},\frac{4}{3} \right)\)

Here is a plot of the curve with its two horizontal tangent lines:

john1.jpg

The tangent line(s) will be vertical where the denominator is zero, which implies:

\(\displaystyle y=\frac{1-x}{2}\)

Substituting this into the original equation, we obtain:

\(\displaystyle x^2+x\left(\frac{1-x}{2} \right)+\left(\frac{1-x}{2} \right)^2=\left(\frac{1-x}{2} \right)\)

\(\displaystyle 4x^2+2x(1-x)+(1-x)^2=2(1-x)\)

\(\displaystyle 4x^2+2x-2x^2+1-2x+x^2=2-2x\)

\(\displaystyle 3x^2+2x-1=0\)

\(\displaystyle (3x-1)(x+1)=0\)

\(\displaystyle x=-1,\,\frac{1}{3}\)

Hence we have two points from \(\displaystyle \left(x,\frac{1-x}{2} \right)\):

\(\displaystyle \left(-1,1 \right),\,\left(\frac{1}{3},\frac{1}{3} \right)\)

Here is a plot of the curve with its two vertical tangent lines:

john2.jpg