
How do you use implicit differentiation to find the slope of the curve given ${{x}^{2}}+xy+{{y}^{2}}=7$ at $\left( 3,2 \right)$ ?
Answer
525k+ views
Hint: To solve the above equation we have to use implicit differentiation. Implicit differentiation is a way of differentiating when you have a function in terms of both x and y. To find the slope of the given equation we follow some steps which are as:
1. Take the derivative of the given function.
2. Evaluate the derivative at the given point to find the slope of the tangent line.
3. Plug the slope of the tangent line and the given point into the point-slope formula for the equation of the line, $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ and then simplify it.
Complete step-by-step answer:
Now the given equation is:
${{x}^{2}}+xy+{{y}^{2}}=7$ at$\left( 3,2 \right)$
To solve the above equation by using implicit differentiation, we will also use some basic formulas of differentiation. We will use product rule which is as $\dfrac{d\left( u.v \right)}{dx}=u\grave{\ }v+v\grave{\ }u$ and power rule $\dfrac{d\left( {{u}^{n}} \right)}{dx}=n{{u}^{n-1}}$ .
Now we will differentiate the above equation with respect to x
$\begin{align}
& \Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}}+xy=7 \right) \\
& \Rightarrow \dfrac{d}{dx}{{x}^{2}}+\dfrac{d}{dx}{{y}^{2}}+\dfrac{d}{dx}xy=0 \\
& \Rightarrow 2x+2y\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}=0 \\
\end{align}$
Now rearranging them, we get
$\begin{align}
& \Rightarrow x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=-y-2x \\
& \Rightarrow \dfrac{dy}{dx}\left( x+2y \right)=-y-2x \\
& \Rightarrow \dfrac{dy}{dx}=\dfrac{-y-2x}{x+2y} \\
\end{align}$
And now we know that $\dfrac{dy}{dx}$ is nothing but the slope of the given equation. And the points which have been given $\left( 3,2 \right)$ are actually not in graph so we will pick the another points $\left( 3,-1 \right)$
Now we know
$\begin{align}
& \Rightarrow m=\dfrac{-y-2x}{x+2y} \\
& \Rightarrow m=\dfrac{-\left( -1 \right)-2\left( 3 \right)}{3+2\left( -1 \right)} \\
\end{align}$
Now by more simplifying the above equation we get,
$\begin{align}
& \Rightarrow m=\dfrac{1-6}{3-2} \\
& \Rightarrow m=-5 \\
\end{align}$
Hence we get the slope of the given equation which is $m=-5$ .
Note: We can go wrong by directly putting the points which we have been given $\left( 3,-1 \right)$ in the slope. First we have to check whether this point exists on a graph or not. Let me show how we have to check.
The given equation is ${{x}^{2}}+xy+{{y}^{2}}=7$, now put the given point in this equation $\left( 3,-1 \right)$,we get
$\begin{align}
& \Rightarrow {{\left( 3 \right)}^{2}}+{{\left( 2 \right)}^{2}}+3\times 2 \\
& \Rightarrow 9+4+6=7 \\
& \Rightarrow 19\ne 7 \\
\end{align}$
It means the given points are not correct. Then we choose a number which is one less or one greater than 2. Hence we choose -1.
1. Take the derivative of the given function.
2. Evaluate the derivative at the given point to find the slope of the tangent line.
3. Plug the slope of the tangent line and the given point into the point-slope formula for the equation of the line, $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ and then simplify it.
Complete step-by-step answer:
Now the given equation is:
${{x}^{2}}+xy+{{y}^{2}}=7$ at$\left( 3,2 \right)$
To solve the above equation by using implicit differentiation, we will also use some basic formulas of differentiation. We will use product rule which is as $\dfrac{d\left( u.v \right)}{dx}=u\grave{\ }v+v\grave{\ }u$ and power rule $\dfrac{d\left( {{u}^{n}} \right)}{dx}=n{{u}^{n-1}}$ .
Now we will differentiate the above equation with respect to x
$\begin{align}
& \Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}}+xy=7 \right) \\
& \Rightarrow \dfrac{d}{dx}{{x}^{2}}+\dfrac{d}{dx}{{y}^{2}}+\dfrac{d}{dx}xy=0 \\
& \Rightarrow 2x+2y\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}=0 \\
\end{align}$
Now rearranging them, we get
$\begin{align}
& \Rightarrow x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=-y-2x \\
& \Rightarrow \dfrac{dy}{dx}\left( x+2y \right)=-y-2x \\
& \Rightarrow \dfrac{dy}{dx}=\dfrac{-y-2x}{x+2y} \\
\end{align}$
And now we know that $\dfrac{dy}{dx}$ is nothing but the slope of the given equation. And the points which have been given $\left( 3,2 \right)$ are actually not in graph so we will pick the another points $\left( 3,-1 \right)$
Now we know
$\begin{align}
& \Rightarrow m=\dfrac{-y-2x}{x+2y} \\
& \Rightarrow m=\dfrac{-\left( -1 \right)-2\left( 3 \right)}{3+2\left( -1 \right)} \\
\end{align}$
Now by more simplifying the above equation we get,
$\begin{align}
& \Rightarrow m=\dfrac{1-6}{3-2} \\
& \Rightarrow m=-5 \\
\end{align}$
Hence we get the slope of the given equation which is $m=-5$ .
Note: We can go wrong by directly putting the points which we have been given $\left( 3,-1 \right)$ in the slope. First we have to check whether this point exists on a graph or not. Let me show how we have to check.
The given equation is ${{x}^{2}}+xy+{{y}^{2}}=7$, now put the given point in this equation $\left( 3,-1 \right)$,we get
$\begin{align}
& \Rightarrow {{\left( 3 \right)}^{2}}+{{\left( 2 \right)}^{2}}+3\times 2 \\
& \Rightarrow 9+4+6=7 \\
& \Rightarrow 19\ne 7 \\
\end{align}$
It means the given points are not correct. Then we choose a number which is one less or one greater than 2. Hence we choose -1.
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