Question

How do you find horizontal and vertical tangent lines after using implicit differentiation ${{x}^{2}}+xy+{{y}^{2}}=27$?

Answer 1

First differentiate the whole function with respect to $x$ and then collect all the $dy/dx$ on the same side and solve for $dy/dx$. On finding $dy/dx$ that is the differentiation of $y$ with respect to $x$ is the slope of the tangent at any point $(x,y)$ on the given function $y$. Now put this $dy/dx$ is equal to zero and not defined separately as the slope of a horizontal line is zero and for vertical is not defined then on simplifying, we would get the coordinates of those required points.

Complete step by step solution:
As it is given a function ${{x}^{2}}+xy+{{y}^{2}}=27$
Since it is implicit as it is a mixed variable function and we will not get $y$ directly on knowing $x$
Now differentiating both sides with respect to $x$
$\dfrac{d}{dx}({{x}^{2}}+xy+{{y}^{2}})=\dfrac{d}{dx}(27)$
Since the differentiation of a constant term is zero
$\Rightarrow \dfrac{d}{dx}({{x}^{2}}+xy+{{y}^{2}})=0$
And we know that $\dfrac{d}{dx}(f+g)=\dfrac{df}{dx}+\dfrac{dg}{dx}$
$\Rightarrow \dfrac{d}{dx}({{x}^{2}})+\dfrac{d}{dx}(xy)+\dfrac{d}{dx}({{y}^{2}})=0$
Since the differentiation of a term of form ${{x}^{n}}{{y}^{m}}$ is first differentiate ${{y}^{m}}$ consider as ${{x}^{m}}$ then multiply it with differentiation of $y$ with respect to $x$ then add the term with keeping function of $y$ as it is and differentiate $x$
i.e. $\dfrac{d({{x}^{n}}{{y}^{m}})}{dx}={{x}^{n}}m{{y}^{m-1}}\dfrac{dy}{dx}+{{y}^{m}}n{{x}^{n-1}}$
Now differentiating with respect to $x$
$\Rightarrow 2x+\left( x\dfrac{dy}{dx}+y \right)+2y\dfrac{dy}{dx}=0$
On simplifying
$\dfrac{dy}{dx}=\dfrac{-(2x+y)}{x+2y}$
We know that $\dfrac{dy}{dx}$ is the slope of the tangent at any point $(x,y)$ on given function.
Ans it is given that the tangents are horizontal and vertical
As for horizontal tangent, $\dfrac{dy}{dx}=0$
$\Rightarrow \dfrac{-(2x+y)}{x+2y}=0$
For this the numerator must be zero
$\Rightarrow 2x+y=0$
$\Rightarrow y=-2x$
Now substituting this value of $y$ in our original function
$\Rightarrow {{x}^{2}}+x(-2x)+{{(-2x)}^{2}}=27$
$\Rightarrow 3{{x}^{2}}=27$
$\Rightarrow x=\pm 3$
On putting these values in the given function ${{x}^{2}}+xy+{{y}^{2}}=27$ we get,

& x=3 \\\ & \Rightarrow y=-6,3 \\\ & x=-3 \\\ & \Rightarrow y=3,-6 \\\ \end{aligned}$$ $$\Rightarrow (3,-6),(3,3),(-3,3),(-3,6)$$ This is not the final coordinates, we have to put this in the $$\dfrac{dy}{dx}$$ to check whether it would be zero or not. On substituting we get, Only $$(3,-6)$$ and $$(-3,6)$$ coordinates are satisfying the condition for horizontal tangent. Now for vertical tangent, $$\dfrac{dy}{dx}$$ is not defined $$\Rightarrow \dfrac{dy}{dx}=\dfrac{-(2x+y)}{x+2y}$$ Means for this, $$x+2y=0$$ $$\Rightarrow x=-2y$$ As similar to the above, we get $$\Rightarrow (-6,3),(3,3),(6,-3),(-3,-3)$$ On putting these coordinates, we get **Only $$(-6,3)$$ and $$(6,-3)$$ coordinates satisfy for vertical tangent.** **Note:** The implicit differentiation means the differentiation of an implicit function that is something variable equal to constant means we can’t get the value of a variable directly by knowing the other while the explicit function is the representation of a variable in terms of others.