Question

find the equation of the tangent line to the curve $x^2 - xy + y^2 = 7$ at the point (2,3). write your answer in point-slope or slope-intercept form.

1. find the equation of the tangent line to the curve $x^2y + y^3 = 4$ at the point (1,1).

Explanation:

Step1: Implicitly differentiate the curve

Differentiate $x^2y + y^3 = 4$ with respect to $x$ using product and chain rules:
$$\frac{d}{dx}(x^2y) + \frac{d}{dx}(y^3) = \frac{d}{dx}(4)$$
$$2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$$

Step2: Isolate $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ and solve for it:
$$\frac{dy}{dx}(x^2 + 3y^2) = -2xy$$
$$\frac{dy}{dx} = -\frac{2xy}{x^2 + 3y^2}$$

Step3: Calculate slope at (1,1)

Substitute $x=1, y=1$ into the slope formula:
$$\frac{dy}{dx}\bigg|_{(1,1)} = -\frac{2(1)(1)}{(1)^2 + 3(1)^2} = -\frac{2}{4} = -\frac{1}{2}$$

Step4: Write point-slope form

Use point $(1,1)$ and slope $-\frac{1}{2}$:
$$y - 1 = -\frac{1}{2}(x - 1)$$

Step5: Convert to slope-intercept form (optional)

Simplify the point-slope equation:
$$y - 1 = -\frac{1}{2}x + \frac{1}{2}$$
$$y = -\frac{1}{2}x + \frac{3}{2}$$

Answer:

Point-slope form: $y - 1 = -\frac{1}{2}(x - 1)$
Slope-intercept form: $y = -\frac{1}{2}x + \frac{3}{2}$