Question: Consider the curve given by the equation $x^3 + xy - 7 = 0$.

(a) Find the equation of the tangent line to the curve at the point (-2, 3). (b) Find the equation of the normal line to the curve at the point (-2, 3).

Provide step-by-step detailed explanations to justify your answers.

Answer: (a) To find the equation of the tangent line to the curve at the point (-2, 3), we will use implicit differentiation.

Given equation: $x^3 + xy - 7 = 0$

Differentiating both sides of the equation with respect to x:

Using the product rule of differentiation, we differentiate each term:

Now, substitute the point (-2, 3) into the equation to find the slope of the tangent line:

Therefore, the slope of the tangent line to the curve at the point (-2, 3) is 7.5.

Now, using the point-slope form of the equation of a line, we can find the equation of the tangent line: [y - 3 = 7.5(x - (-2))] [y - 3 = 7.5(x + 2)] [y - 3 = 7.5x + 15] [y = 7.5x + 18]

The equation of the tangent line to the curve at the point (-2, 3) is $y = 7.5x + 18$.

(b) To find the equation of the normal line to the curve at the point (-2, 3), we utilize the fact that the slopes of the tangent and normal lines are negative reciprocals of each other.

The slope of the tangent line is 7.5, so the slope of the normal line is -1/7.5.

Using the point-slope form, the equation of the normal line is: [y - 3 = -\frac{1}{7.5}(x - (-2))] [y - 3 = -\frac{1}{7.5}(x + 2)] [y - 3 = -\frac{1}{7.5}x - \frac{1}{3.75}] [y = -\frac{1}{7.5}x + \frac{22}{7.5}] [y = -\frac{2}{15}x + \frac{22}{15}]

Therefore, the equation of the normal line to the curve at the point (-2, 3) is $y = -\frac{2}{15}x + \frac{22}{15}$.