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Handout Tuesday 10/14
Today we continue exploring implicit functions and their derivatives.
Using desmos, graph a few of the following:
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\(\displaystyle x^3 + y^3 = 7xy\)
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\(\displaystyle y^2 = x^3 - 4x + 10\)
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\(\displaystyle (x^2+y^2)^2 = 16xy\)
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\(\displaystyle x^2 + y^2 = (x^2 + y^2 - 2x)^2\)
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\(\displaystyle (x^2 + y^2)^2 = 7xy^2 + x^2\)
A very useful way to understand these problems is to find their local maxima. This will be a spot where the tangent line is horizontal, or where \(\frac{dy}{dx} = 0\text{.}\)
To find \(\frac{dy}{dx}\) we evaluate the derivative of both sides, with respect to \(x\) (i.e., take \(\frac{d}{dx}\) of both sides). Any term with a \(y\) will require the chain rule, since \(y\) is a function of \(x\text{.}\)
Let’s do this for one of the examples, say \(x^3 + y^3 = 7xy\text{.}\)
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\(\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(7xy)\text{.}\) This becomes\begin{equation*} 3x^2 + 3y^2\frac{dy}{dx} = 7y + 7x\frac{dy}{dx}\text{.} \end{equation*}
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Now we solve for \(frac{dy}{dx}\text{:}\)\begin{align*} 3y^2\frac{dy}{dx} - 7x\frac{dy}{dx} \amp = 7y - 3x^2\\ (3y^2 - 7x)\frac{dy}{dx} \amp = 7y - 3x^2\\ \frac{dy}{dx} \amp = \frac{7y - 3x^2}{3y^2 - 7x} \end{align*}
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The maximum value here will be where this derivative is 0. The only way for the fraction to be zero is if its numerator is 0. So we want to find \(x\) such that \(7y - 3x^2 = 0\text{.}\)
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First, find \(y = \frac{3}{7}x^2\text{.}\) Now substitute this in to the original implicit function:\begin{equation*} x^3 + \left(\frac{3}{7}x^2\right)^3 = 6x\frac{3}{7}x^2 \end{equation*}Now we can combine like terms and divide both sides by \(x^3\) (although note what happens at \(x = 0\text{,}\) this is also a place where the tangent line is horizontal.)
