The problem in the image asks to "Find the equation of the normal line to the curve $x^2 + 2xy = y^3$ at the point $(1, 1)$."

Let's work through this step by step.

Step 1: Implicit differentiation

The given equation is: $x^2 + 2xy = y^3$ We'll differentiate both sides of this equation with respect to $x$, using implicit differentiation. Remember that $y$ is a function of $x$ (i.e., $y = y(x)$), so when differentiating terms involving $y$, we must apply the chain rule.

Differentiating term by term:

1. The derivative of $x^2$ with respect to $x$ is $2x$.
2. The derivative of $2xy$ is: $\frac{d}{dx}(2xy) = 2 \left( \frac{d}{dx}(xy) \right) = 2 \left( x \frac{dy}{dx} + y \right)$ Here we used the product rule.
3. The derivative of $y^3$ is: $\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$

So the differentiated equation becomes: $2x + 2 \left( x \frac{dy}{dx} + y \right) = 3y^2 \frac{dy}{dx}$

Step 2: Solve for $\frac{dy}{dx}$

Now, we'll solve for $\frac{dy}{dx}$, the derivative of $y$ with respect to $x$, which gives us the slope of the tangent line at any point on the curve.

Expand and rearrange the terms: $2x + 2x \frac{dy}{dx} + 2y = 3y^2 \frac{dy}{dx}$ Group the terms involving $\frac{dy}{dx}$ on one side: $2x + 2y = 3y^2 \frac{dy}{dx} - 2x \frac{dy}{dx}$ Factor out $\frac{dy}{dx}$ on the right side: $2x + 2y = \frac{dy}{dx}(3y^2 - 2x)$ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2x + 2y}{3y^2 - 2x}$

Step 3: Find the slope of the tangent line at $(1, 1)$

Substitute $x = 1$ and $y = 1$ into the derivative: $\frac{dy}{dx} = \frac{2(1) + 2(1)}{3(1)^2 - 2(1)} = \frac{2 + 2}{3 - 2} = \frac{4}{1} = 4$ So, the slope of the tangent line at $(1, 1)$ is 4.

Step 4: Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is 4, the slope of the normal line is: $m_{\text{normal}} = -\frac{1}{4}$

Step 5: Write the equation of the normal line

The normal line passes through the point $(1, 1)$ and has slope $-\frac{1}{4}$. Using the point-slope form of the equation of a line: $y - y_1 = m (x - x_1)$ Substitute $(x_1, y_1) = (1, 1)$ and $m = -\frac{1}{4}$: $y - 1 = -\frac{1}{4}(x - 1)$ Simplifying this equation: $y - 1 = -\frac{1}{4}x + \frac{1}{4}$ $y = -\frac{1}{4}x + \frac{1}{4} + 1$ $y = -\frac{1}{4}x + \frac{5}{4}$

Final Answer:

The equation of the normal line is: $y = -\frac{1}{4}x + \frac{5}{4}$

Would you like any additional details or clarifications on this solution?

Here are some related questions:

1. How do you find the equation of a tangent line to a curve at a given point?
2. What is implicit differentiation and when is it used?
3. How do you apply the product rule in calculus?
4. What is the relationship between the slopes of tangent and normal lines?
5. How can you check if a line is perpendicular to another?

Tip: Always remember that the slope of the normal line is the negative reciprocal of the slope of the tangent line at the same point.