### Math Notes

Subjects

#### Differential Calculus Solutions

##### Topics || Problems

Find the tangent and normal to the curve $$x^2 -2xy +2y^2 -x = 0$$ at $$x=1$$.

At $$x=1$$, $$y = 0$$ and $$y = 1$$

At $$(1, 0)$$

$$x^2 -2xy +2y^2 -x = 0$$

$$2xdx - 2xdy - 2ydx + 4ydy -dx =0$$

$$2(1)dx - 2(1)dy - 2(0)dx + 4(0)dy -dx =0$$

$$2dx - 2dy -dx =0$$

$$dx - 2dy =0$$

$$\frac{dy}{dx} = \frac{1}{2}$$

The tangent line; $$y-y_1 = \frac{1}{2}(x-x_1)$$

$$y-0 = \frac{1}{2}(x-1)$$

$$2y = x - 1$$

The normal line: A normal line is a line perpendicular to the tangent line.

$$y-0 = -2(x-1)$$

$$y = -2x +2$$

At $$(1, 1)$$

$$x^2 -2xy +2y^2 -x = 0$$

$$2xdx - 2xdy - 2ydx + 4ydy -dx =0$$

$$2(1)dx - 2(1)dy - 2(1)dx + 4(1)dy -dx =0$$

$$-dx + 2dy =0$$

$$\frac{dy}{dx} = \frac{1}{2}$$

The tangent line; $$y-y_1 = \frac{1}{2}(x-x_1)$$

$$y-1 = \frac{1}{2}(x-1)$$

$$2y -2 = x - 1$$

$$2y = x +1$$

The normal line:

$$y-1 = -2(x-1)$$

$$y = -2x +3$$