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