# What is the slope of the tangent line of sinx-y^2/x= C , where C is an arbitrary constant, at (pi/3,1)?

##### 1 Answer
Mar 29, 2016

Slope of the tangent at $\left(\frac{\pi}{3} , 1\right)$ is $\frac{3}{2 \pi} + \frac{\pi}{12}$

#### Explanation:

At $\left(\frac{\pi}{3} , 1\right)$ we have $\sin \left(\frac{\pi}{3}\right) - {1}^{2} / \left(\frac{\pi}{3}\right) = C$ or $C = \frac{\sqrt{3}}{2} - \frac{1 \times 3}{\pi} = \frac{\pi \sqrt{3} - 3}{2 \pi}$

Slope of the function is given by its first derivative and differentiating the function $\sin x - {y}^{2} / x = \frac{\pi \sqrt{3} - 3}{2 \pi}$, we get

$\cos x - \frac{x \cdot 2 y \cdot \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - {y}^{2} \cdot 1}{x} ^ 2 = 0$ or

$\cos x - 2 \frac{y}{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2} / {x}^{2} = 0$ or

$2 \frac{y}{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} / {x}^{2} + \cos x$ or

$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} / {x}^{2} \cdot \frac{x}{2 y} + \frac{x}{2 y} \cos x$ or

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{2 x} + \frac{x}{2 y} \cos x$

and at $\left(\frac{\pi}{3} , 1\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \frac{\pi}{3}} + \frac{\frac{\pi}{3}}{2 \cdot 1} \cos \left(\frac{\pi}{3}\right)$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2 \pi} + \frac{\pi}{6} \cdot \frac{1}{2} = \frac{3}{2 \pi} + \frac{\pi}{12}$

Hence slope at $\left(\frac{\pi}{3} , 1\right)$ is $\frac{3}{2 \pi} + \frac{\pi}{12}$