The given problem requires us to:

a. Verify that the point (3, 3) lies on the curve defined by the equation:

$x^3 + y^3 = 6xy$ We need to evaluate the expression $x^3 + y^3$ at $x = 3$ and $y = 3$ to check if the point satisfies the equation.

b. Find the equation of the tangent line to the curve at the given point (3, 3).

To do this, we will use implicit differentiation to find the slope of the tangent at that point.

Let me first tackle part (a).

Part (a) - Verify the point (3, 3) on the curve:

We need to substitute $x = 3$ and $y = 3$ into the equation:

$x^3 + y^3 = 6xy$

Substitute $x = 3$ and $y = 3$: $3^3 + 3^3 = 6(3)(3)$ $27 + 27 = 54$ $54 = 54$ Thus, the point (3, 3) does lie on the curve, as both sides of the equation are equal.

Part (b) - Find the equation of the tangent line:

To find the equation of the tangent line at the point (3, 3), we need to differentiate the equation $x^3 + y^3 = 6xy$ implicitly with respect to $x$.

Differentiating both sides with respect to $x$:

• The derivative of $x^3$ with respect to $x$ is $3x^2$.
• The derivative of $y^3$ with respect to $x$ is $3y^2 \frac{dy}{dx}$ (using the chain rule).
• The derivative of $6xy$ with respect to $x$ is $6y + 6x \frac{dy}{dx}$ (using the product rule).

So, the differentiated equation is: $3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$

Now, solve for $\frac{dy}{dx}$ (the slope of the tangent):

$3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2$ Factor out $\frac{dy}{dx}$ on the left side: $\frac{dy}{dx} (3y^2 - 6x) = 6y - 3x^2$ $\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$

Now, substitute $x = 3$ and $y = 3$ into this derivative to find the slope at the point (3, 3): $\frac{dy}{dx} = \frac{6(3) - 3(3^2)}{3(3^2) - 6(3)}$ $\frac{dy}{dx} = \frac{18 - 27}{27 - 18}$ $\frac{dy}{dx} = \frac{-9}{9} = -1$

So, the slope of the tangent line at (3, 3) is $-1$.

Finally, we can write the equation of the tangent line using the point-slope form: $y - y_1 = m(x - x_1)$ Substituting $m = -1$, $x_1 = 3$, and $y_1 = 3$: $y - 3 = -1(x - 3)$ Simplify this equation: $y - 3 = -x + 3$ $y = -x + 6$

Thus, the equation of the tangent line is: $y = -x + 6$

Would you like more details or any further explanations on these steps? Here are some related questions:

1. How do we apply implicit differentiation in more complex equations?
2. Can you explain the point-slope form of a line further?
3. How do we verify that other points lie on a curve in general?
4. Can we apply the same method to find the equation of the tangent line to a curve defined in polar coordinates?
5. What is the significance of the slope in the context of real-world problems?

Tip: Implicit differentiation is very useful when you have equations where both variables $x$ and $y$ are mixed together, and you need to find the derivative with respect to one variable.