To find the points on the curve $y^3 - xy = 2$ where the tangent is vertical, we need to identify where the derivative of $y$ with respect to $x$ becomes undefined (i.e., where $\frac{dx}{dy} = 0$).

Step-by-Step Solution:

1. Differentiate the equation implicitly:

   Given: $y^3 - xy = 2$

   Differentiate both sides with respect to $x$: $3y^2 \frac{dy}{dx} - y - x \frac{dy}{dx} = 0$

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

   Rearrange the terms to isolate $\frac{dy}{dx}$: $\frac{dy}{dx}(3y^2 - x) = y$ $\frac{dy}{dx} = \frac{y}{3y^2 - x}$

3. Condition for a vertical tangent:

   A tangent line is vertical when $\frac{dy}{dx}$ is undefined. This occurs when the denominator is zero: $3y^2 - x = 0$ Solving for $x$, we get: $x = 3y^2$

4. Substitute $x = 3y^2$ into the original equation:

   Substitute $x = 3y^2$ into $y^3 - xy = 2$ to find the corresponding $y$-values: $y^3 - (3y^2)y = 2$ $y^3 - 3y^3 = 2$ $-2y^3 = 2$ $y^3 = -1$ $y = -1$

5. Find the corresponding $x$-value:

   Substitute $y = -1$ into $x = 3y^2$: $x = 3(-1)^2 = 3$

6. Solution:

   The point at which the tangent line to the curve is vertical is: $(x, y) = (3, -1)$

Final Answer:

The coordinates of the point on the curve where the tangent is vertical are: $(3, -1)$

Would you like further details on the differentiation steps or have any questions?

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1. What is the significance of vertical tangents in curve analysis?
2. How would we find horizontal tangents for this curve?
3. Can we use implicit differentiation on other types of equations?
4. How can we verify the vertical tangent point geometrically?
5. What happens if there are multiple points with vertical tangents on a curve?

Tip: For curves given by implicit equations, look for points where the derivative’s denominator is zero to identify vertical tangents.