Consider the point \(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\) on the unit circle. Is it possible to find a line tangent to the unit circle at this point? First thing that is needed, find the derivative of the unit circle equation: \(x^2+y^2=1\). The unit circle is not even a function! How can a derivative of this equation be found?
This is an example of an implicit equation, no variable is solved for. Some algebra could be applied and rewrite the equation explicitly into \(y=\) format, giving two separate equations: \(y=\sqrt{1-x^2}\) and \(y=-\sqrt{1-x^2}\). Now normal derivative techniques can help find two derivative equations, but then which equation would be used depends on the point where the tangent line occurs. In this case, it would be the positive square root.
To avoid dealing with multiple derivative equations, there is a simplier technique that will allow a single derivative equation to be found, it is called implicit differentiation. The main point to this method is to understand that the variable \(y\) is considered to be a function explicitly defined in terms of \(x\), in other words the OUTSIDE function, so an application of the Chain Rule must be applied.
Find an equation of a line tangent to the unit circle at the point \(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\).
| Take the derivative at each term: | \(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}(1)\) |
| Use Chain Rule on \(y^2\) term: | \(2x+2y\cdot\frac{dy}{dx}=0\) |
| Apply algebra to solve for \(\frac{dy}{dx}\): | \(\begin{array}{l} \implies 2y\cdot\frac{dy}{dx}=-2x \\ \implies \large{\frac{2y\cdot\frac{dy}{dx}}{2y}}\normalsize{=}\large{\frac{-2x}{2y}} \\ \implies \large{\frac{dy}{dx}}\normalsize{=}\large{\frac{-2x}{2y}} \\ \implies \large{\frac{dy}{dx}}\normalsize{=}\large{\frac{-x}{y}}\;\color{green}{\checkmark} \end{array}\) |
| Plug in the point to find the slope: | \(\large{\frac{dy}{dx}}\normalsize{=}\large{\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}} \implies \large{\frac{dy}{dx}}\normalsize{=}\large{\frac{-1}{\sqrt{3}}}\;\color{green}{\checkmark}\) |
| Use Point-Slope form equation: | \(y-\large{\frac{\sqrt{3}}{2}}\normalsize{=}\large{\frac{-1}{\sqrt{3}}}\normalsize{\left(x-\large{\frac{1}{2}}\right)\; \color{green}{\checkmark}}\) |