Use implicit differentiation to find the equation of the tangent line to the curve 𝑥2 + 𝑥𝑦 + 𝑦2 = 3 at the point (1,1).

1. Break Down the Problem

We need to find the slope of the tangent line to the curve defined by the equation $x^2 + xy + y^2 = 3$ at the point $(1,1)$ using implicit differentiation. After finding the slope, we can use the point-slope form of the equation of a line to determine the tangent line.

2. Relevant Concepts

• Implicit differentiation: This technique involves taking derivatives of both sides of an equation with respect to $x$, treating $y$ as a function of $x$.
• The point-slope form of a line is given by: $y - y_1 = m(x - x_1)$ where $m$ is the slope and $(x_1, y_1)$ is the point on the line.

3. Analysis and Detail

1. Differentiate both sides of the equation $x^2 + xy + y^2 = 3$:

   • The derivative of $x^2$ with respect to $x$ is $2x$.
   • For $xy$, use the product rule: $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$.
   • The derivative of $y^2$ with respect to $x$ is $2y\frac{dy}{dx}$.

   Combining these, we get: $2x + (x\frac{dy}{dx} + y) + 2y\frac{dy}{dx} = 0$

2. Rearrange this equation to solve for $\frac{dy}{dx}$: $(x + 2y)\frac{dy}{dx} = -2x - y$ $\frac{dy}{dx} = \frac{-2x - y}{x + 2y}$

3. Evaluate $\frac{dy}{dx}$ at the point $(1,1)$: $\frac{dy}{dx}\bigg|_{(1,1)} = \frac{-2(1) - 1}{1 + 2(1)} = \frac{-2 - 1}{1 + 2} = \frac{-3}{3} = -1$

4. Verify and Summarize

The slope of the tangent line at the point $(1,1)$ is $-1$.

Using the point-slope form of the line: $y - 1 = -1(x - 1)$ Simplifying this: $y - 1 = -x + 1$ $y = -x + 2$

Final Answer

The equation of the tangent line to the curve at the point $(1,1)$ is: $y = -x + 2$