What is the slope of the tangent line of sin^2y-cos^2x^2= C , where C is an arbitrary constant, at (pi/3,pi/3)?

Sep 27, 2017

$= \left(- \frac{4 \pi \sqrt{3}}{9}\right) \cdot \left(\sin \left(\frac{2 {\pi}^{2}}{9}\right)\right) \approx - 1.964$

Explanation:

Use implicit differentation to determine a formula for the derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{d}{\mathrm{dx}} \left({\sin}^{2} y\right) - \frac{d}{\mathrm{dx}} \left({\cos}^{2} {x}^{2}\right) = \frac{d}{\mathrm{dx}} C$
$2 \left(\sin y\right) \cdot \left(\cos y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - 2 \left(\cos {x}^{2}\right) \cdot \left(- \sin {x}^{2}\right) \cdot \left(2 x\right) = 0$
$\left(\sin 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + 4 x \cdot \left(\cos {x}^{2}\right) \cdot \left(\sin {x}^{2}\right) = 0$
$\left(\sin 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x \cdot \left(\sin \left(2 {x}^{2}\right)\right) = 0$
$\left(\sin 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x \left(\sin \left(2 {x}^{2}\right)\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x \left(\sin \left(2 {x}^{2}\right)\right)}{\sin 2 y}$

To get the slope, we then evaluate this at the point $\left(\frac{\pi}{3} , \frac{\pi}{3}\right)$:

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) {|}_{\frac{\pi}{3} , \frac{\pi}{3}} = \frac{- 2 \cdot \left(\frac{\pi}{3}\right) \cdot \left(\sin \left(2 \cdot {\left(\frac{\pi}{3}\right)}^{2}\right)\right)}{\sin \left(2 \cdot \frac{\pi}{3}\right)}$
$= \frac{\left(- \frac{2 \pi}{3}\right) \cdot \left(\sin \left(\frac{2 {\pi}^{2}}{9}\right)\right)}{\sin \left(\frac{2 \pi}{3}\right)} = \frac{\left(- \frac{2 \pi}{3}\right) \cdot \left(\sin \left(\frac{2 {\pi}^{2}}{9}\right)\right)}{\frac{\sqrt{3}}{2}}$
$= \frac{\left(- \frac{4 \pi}{3}\right) \cdot \left(\sin \left(\frac{2 {\pi}^{2}}{9}\right)\right)}{\sqrt{3}} = \frac{\left(- \frac{4 \pi \sqrt{3}}{3}\right) \cdot \left(\sin \left(\frac{2 {\pi}^{2}}{9}\right)\right)}{3}$
$= \left(- \frac{4 \pi \sqrt{3}}{9}\right) \cdot \left(\sin \left(\frac{2 {\pi}^{2}}{9}\right)\right)$
$\approx - 1.964$