Question

find the slope of the tangent line to the curve (a lemniscate)
$2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2})$
at the point $(-3,1)$
slope $=$
question help: video message instructor

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Using the chain rule and product rule.
For the left - hand side:
Let \(u=x^{2}+y^{2}\), then \(2u^{2}\). The derivative of \(2u^{2}\) with respect to \(x\) is \(4u\cdot(2x + 2y\frac{dy}{dx})\) (by the chain rule \(\frac{d}{dx}(u^{n})=nu^{n - 1}\frac{du}{dx}\) and \(\frac{du}{dx}=2x + 2y\frac{dy}{dx}\)). Substituting \(u=x^{2}+y^{2}\) back, we get \(4(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})\).
For the right - hand side:
The derivative of \(25(x^{2}-y^{2})\) with respect to \(x\) is \(25(2x-2y\frac{dy}{dx})\) (by the sum/difference rule \(\frac{d}{dx}(f(x)-g(x))=\frac{df}{dx}-\frac{dg}{dx}\) and \(\frac{d}{dx}(y^{2}) = 2y\frac{dy}{dx}\) by the chain rule).
So, \(4(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=25(2x - 2y\frac{dy}{dx})\).

Step2: Substitute the point \((-3,1)\)

First, calculate \(x=-3\) and \(y = 1\). Then \(x^{2}+y^{2}=(-3)^{2}+1^{2}=9 + 1=10\).
Substitute into \(4(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=25(2x - 2y\frac{dy}{dx})\):
\(4\times10\times(-6 + 2\frac{dy}{dx})=25\times(-6-2\frac{dy}{dx})\).
Expand: \(-240+80\frac{dy}{dx}=-150 - 50\frac{dy}{dx}\).

Step3: Solve for \(\frac{dy}{dx}\)

Add \(50\frac{dy}{dx}\) to both sides: \(-240+(80\frac{dy}{dx}+50\frac{dy}{dx})=-150\).
\(-240 + 130\frac{dy}{dx}=-150\).
Add \(240\) to both sides: \(130\frac{dy}{dx}=240 - 150=90\).
Then \(\frac{dy}{dx}=\frac{90}{130}=\frac{9}{13}\).

Answer:

\(\frac{9}{13}\)