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