# Implicit differentiation problem; find normal line to the given graph

PROBLEM:

Heat flows normal to isotherms, curves along which the temperature is constant. Find the line along which heat flows through the point $(2,5)$ when the isotherm is along the graph of $2x^2+y^2=33$.

QUESTION:

Here it gives me the equation of the graph (original function). I know that by finding the derivative of the expression it gives I can substitute $x$ ($x=2$) in and solve for the slope. Then I can use $y = mx + b$ to get the equation of the line as it asks for.

However, it doesn't show the function in $f(x)$ form. Therefore, I'm told that I need to use Implicit Differentiation before I do anything. This is where I'm completely lost. I have no idea what Implicit Differentiation means and how to get this expression in $f(x)$ form so I can find the derivative. Can someone please guide me. I would appreciate it if you could comment on your steps so I can have a more intuitive understanding.

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By differentiating implicitly with respect to $x$ both sides of the implicit equation $$\begin{equation*} 2x^{2}+y^{2}=33,\tag{1} \end{equation*}$$

since the derivatives of both sides should be equal we get successively: $$\begin{eqnarray*} &&\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =\frac{d}{dx}\left( 33\right) \\ &\Rightarrow &\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =0 \\ &\Leftrightarrow &\frac{d}{dx}\left( 2x^{2}\right) +\frac{d}{dx}\left( y^{2}\right) =0 \\ &\Leftrightarrow &4x+2y\frac{dy}{dx}=0,\qquad \frac{d}{dx}\left( y^{2}\right) =2y\frac{dy}{dx}\text{ by the chain rule} \\ &\Leftrightarrow &\frac{dy}{dx}=-\frac{4x}{2y}=-\frac{2x}{y}\tag{2} \\ &\Rightarrow &\left. \frac{dy}{dx}\right\vert _{x=2,y=5}=-\frac{4}{5}.\tag{3} \end{eqnarray*}$$

The equation of the tangent line at $(2,5)$ is $$\begin{equation*} y-5=-\frac{4}{5}(x-2),\tag{4} \end{equation*}$$ while the equation of the normal line to the curve $2x^{2}+y^{2}=33$ at $(2,5)$ is $$\begin{equation*} y-5=\frac{5}{4}(x-2)\Leftrightarrow y=\frac{5}{4}x+\frac{5}{2},\tag{5} \end{equation*}$$

because the slope $m$ of the tangent line and the slope $m^{\prime }$ of the normal line are related by $mm^{\prime }=-1$.

ADDED. In a more general case, when we have a differentiable implicit function $F(x,y)=0$, let $y=f(x)$ denote the function such that $F(x,f(x))\equiv 0\quad$ ($f(x)$ does not need to be explicitly known). If we differentiate both sides of $F(x,y)=0$ and apply the chain rule, we get the following total derivative with respect to $x$:

$$\frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{ \partial F}{\partial y}\frac{dy}{dx}\equiv 0.\tag{A}$$

Solving $(\mathrm{A})$ for $\frac{dy}{dx}$, gives us the following formula

$$\frac{dy}{dx}=-\frac{\partial F}{\partial x}/\frac{ \partial F}{\partial y}.\tag{B}$$

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To differentiate implicitly is easy, you'll see.

Take the equation and differentiate like you usually do, with respect to $x$:

$$2x^2 + y^2 = 33$$

Differentiate $2x^2$ to get $4x$.

Differentiate $y^2$ to get $2y\dfrac{dy}{dx}$. Notice that since there is no $x$ here, but a y, which is dependent on $x$, you add the $\dfrac{dy}{dx}$.

Differentiate $33$ to give zero.

And the you get the derivative that Ron Gordon obtained, that is:

$$4 x+ 2 y \dfrac{dy}{dx} = 0$$

I hope this helps you understanding implicit differentiation!

Additional: Say you had $y$ instead of $y^2$. You differentiate $y$ to get $$1\dfrac{dy}{dx} = \dfrac{dy}{dx}$$ Similarly: $4y^3$ becomes $$12y^2\dfrac{dy}{dx}$$

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You want the normal to the isotherm, which is normal to a tangent. You find the tangent by differentiating:

$$4 x+ 2 y y' = 0 = y' = -2\frac{x}{y}$$

At the point $(2,5)$, that tangent slope is $y' = -4/5$. This means that the slope of the normal to the isotherm at this point is $5/4$. (The negative reciprocal) Now you have a slope and a point - can you find the equation of the line?

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