Find the equation of the tangent and normal to the curve $4x^3 + 4xy + y^2 = 4$ at the point (0, 1).

Steps to Solve

1. Implicit Differentiation

Differentiate both sides of the equation $4x^3 + 4xy + y^2 = 4$ with respect to $x$. Remember that $y$ is a function of $x$, so we'll need to use the chain rule when differentiating terms involving $y$. Also, we have a product $4xy$, so we'll need the product rule.

$$ \frac{d}{dx}(4x^3) + \frac{d}{dx}(4xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4) $$

2. Applying differentiation rules

Apply the power rule, product rule, and chain rule to get

$$ 12x^2 + (4x \frac{dy}{dx} + 4y) + 2y \frac{dy}{dx} = 0 $$

3. Rearrange to solve for $\frac{dy}{dx}$

Group terms with $\frac{dy}{dx}$ and move other terms to the other side:

$$ 4x \frac{dy}{dx} + 2y \frac{dy}{dx} = -12x^2 - 4y $$

Factor out $\frac{dy}{dx}$:

$$ \frac{dy}{dx}(4x + 2y) = -12x^2 - 4y $$

Divide to isolate $\frac{dy}{dx}$:

$$ \frac{dy}{dx} = \frac{-12x^2 - 4y}{4x + 2y} = \frac{-6x^2 - 2y}{2x + y} $$

4. Evaluate $\frac{dy}{dx}$ at the point (0, 1)

Substitute $x = 0$ and $y = 1$ into the expression for $\frac{dy}{dx}$:

$$ \frac{dy}{dx} \Big|_{(0,1)} = \frac{-6(0)^2 - 2(1)}{2(0) + 1} = \frac{-2}{1} = -2 $$

This is the slope of the tangent at the point (0, 1).

5. Find the equation of the Tangent Line

Use the point-slope form of a line $y - y_1 = m(x - x_1)$, where $(x_1, y_1) = (0, 1)$ and $m = -2$:

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

6. Find the slope of the Normal Line

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent's slope. Since the slope of the tangent is $-2$, the slope of the normal is $\frac{1}{2}$.

7. Find the equation of the Normal Line

Use the point-slope form of a line again, but with the slope of the normal, $m = \frac{1}{2}$, and the point $(0, 1)$:

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