13. find the tangent and normal to the curve $x^{2}-xy + y^{2}=7$ at the point $(-1,2)$ (fig. 2.49). 2.49…

Explanation:

Step1: Differentiate the equation implicitly with respect to x.

Given the curve $x^2 - xy + y^2 = 7$. Differentiating both sides with respect to $x$: $$ \frac{d}{dx}(x^2) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(7) $$ Using the power rule and the product rule: $$ 2x - (1 \cdot y + x \cdot \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0 $$ $$ 2x - y - x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 $$

Step2: Solve for the derivative $\frac{dy}{dx}$.

Rearrange the terms to isolate $\frac{dy}{dx}$: $$ (2y - x) \frac{dy}{dx} = y - 2x $$ $$ \frac{dy}{dx} = \frac{y - 2x}{2y - x} $$

Step3: Evaluate the slope of the tangent line at the point (-1, 2).

Substitute $x = -1$ and $y = 2$ into the expression for $\frac{dy}{dx}$: $$ m_{tangent} = \frac{dy}{dx} \bigg|_{(-1, 2)} = \frac{2 - 2(-1)}{2(2) - (-1)} = \frac{2 + 2}{4 + 1} = \frac{4}{5} $$

Step4: Find the equation of the tangent line.

Using the point-slope form $y - y_1 = m(x - x_1)$ with the point $(-1, 2)$ and slope $m_{tangent} = \frac{4}{5}$: $$ y - 2 = \frac{4}{5}(x - (-1)) $$ $$ y - 2 = \frac{4}{5}(x + 1) $$ $$ 5(y - 2) = 4(x + 1) $$ $$ 5y - 10 = 4x + 4 $$ $$ 4x - 5y + 14 = 0 $$

Step5: Calculate the slope of the normal line.

The slope of the normal line is the negative reciprocal of the slope of the tangent line: $$ m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{4/5} = -\frac{5}{4} $$

Step6: Find the equation of the normal line.

Using the point-slope form $y - y_1 = m(x - x_1)$ with the point $(-1, 2)$ and slope $m_{normal} = -\frac{5}{4}$: $$ y - 2 = -\frac{5}{4}(x - (-1)) $$ $$ y - 2 = -\frac{5}{4}(x + 1) $$ $$ 4(y - 2) = -5(x + 1) $$ $$ 4y - 8 = -5x - 5 $$ $$ 5x + 4y - 3 = 0 $$

Answer:

The equation of the tangent line is $4x - 5y + 14 = 0$. The equation of the normal line is $5x + 4y - 3 = 0$.