To find the gradient of the curve $x^2 + 3xy + y^2 = x + 3y$ at the point $(2, -1)$, we'll use implicit differentiation. Here's the step-by-step process:

1. Differentiate the equation implicitly with respect to $x$:

Given: $x^2 + 3xy + y^2 = x + 3y$ Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^2) + \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(x) + \frac{d}{dx}(3y)$

2. Apply the differentiation rules:

• For $x^2$: $\frac{d}{dx}(x^2) = 2x$
• For $3xy$: Use the product rule: $\frac{d}{dx}(3xy) = 3\left(\frac{d}{dx}(x)y + x\frac{d}{dx}(y)\right) = 3(y + x\frac{dy}{dx})$
• For $y^2$: Use the chain rule: $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$
• For $x$: $\frac{d}{dx}(x) = 1$
• For $3y$: $\frac{d}{dx}(3y) = 3\frac{dy}{dx}$

So the equation becomes: $2x + 3(y + x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 1 + 3\frac{dy}{dx}$

3. Collect terms involving $\frac{dy}{dx}$ on one side:

$2x + 3y + 3x\frac{dy}{dx} + 2y\frac{dy}{dx} = 1 + 3\frac{dy}{dx}$ $3x\frac{dy}{dx} + 2y\frac{dy}{dx} - 3\frac{dy}{dx} = 1 - 2x - 3y$ Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(3x + 2y - 3) = 1 - 2x - 3y$

4. Solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1 - 2x - 3y}{3x + 2y - 3}$

5. Substitute the point $(2, -1)$ into the derivative:

$\frac{dy}{dx}\Bigg|_{(2, -1)} = \frac{1 - 2(2) - 3(-1)}{3(2) + 2(-1) - 3}$ Simplify the expression: $\frac{dy}{dx}\Bigg|_{(2, -1)} = \frac{1 - 4 + 3}{6 - 2 - 3} = \frac{0}{1} = 0$

Final Answer:

The gradient of the curve at the point $(2, -1)$ is $\boxed{0}$.

Would you like more details or have any questions?

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Here are 8 related questions:

1. How do you find the gradient of 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 differentiation?
4. What is the chain rule, and how is it used in differentiation?
5. How would you solve a problem where the gradient is not zero?
6. Can you find the equation of the tangent line at the point $(2, -1)$?
7. How would the solution change if the curve was different?
8. What are the common errors to avoid when using implicit differentiation?

Tip: Always simplify your expressions as much as possible before substituting numerical values to reduce the chance of errors.