Question

Find all points on the curve \(x^{2} y-x y^{2}=2\) where the tangent line is vertical, that is, where \(d x / d y=0\).

Short answer

The point on the curve where the tangent is vertical is \((2, 1)\).

Step-by-step solution

Rearrange the Equation

Start by working with the given implicit equation of the curve: \( x^2 y - xy^2 = 2 \). Simplify the equation to make it easier to differentiate later.

Differentiate Implicitly

Differentiate the equation \( x^2 y - xy^2 = 2 \) implicitly with respect to \( x \). This will involve using the product rule on both \( x^2 y \) and \(-xy^2\).

Apply the Product Rule

For the term \( x^2 y \), the product rule gives \( 2xy + x^2 \frac{dy}{dx} \). For the term \(-xy^2\), the product rule gives \( -y^2 - 2xy \frac{dy}{dx} \).

Set Up the Equation

Combine the differentiated terms: \( 2xy + x^2 \frac{dy}{dx} - y^2 - 2xy \frac{dy}{dx} = 0 \). Reorder to form \( (x^2 - 2xy) \frac{dy}{dx} = y^2 - 2xy \).

Find \( \frac{dx}{dy} \)

If the tangent line is vertical, then \( \frac{dx}{dy} = 0 \). Rewriting \( \frac{dx}{dy} \) involves reciprocating \( \frac{dy}{dx} \), leading to the condition when \( x^2 - 2xy = 0 \).

Solve for x and y

Solve the equation \( x^2 - 2xy = 0 \) to find relationships between \( x \) and \( y \). Factor the equation as \( x(x - 2y) = 0 \), giving \( x = 0 \) or \( x = 2y \).

Substitute Back for Points

Substitute \( x = 0 \) and \( x = 2y \) into the original curve equation to find any valid points.

Check Valid Points

For \( x = 0 \): Substitute into \( x^2 y - xy^2 = 2 \), which becomes \( 0 = 2 \), so no solution. For \( x = 2y \): Substitute \( x = 2y \) into \( (2y)^2 y - (2y)y^2 = 2 \) which simplifies to \( 4y^3 - 2y^3 = 2 \rightarrow 2y^3 = 2 \), giving \( y = 1 \). Therefore, \( x = 2\), and the point is \((2, 1)\).

Final Answer

With \( x = 2 \) and \( y = 1 \), the solution provides the point on the curve where the tangent line is vertical.

Key concepts

Implicit Differentiation

Implicit differentiation is a critical technique in calculus, especially when dealing with equations where variables are intertwined, and it's difficult to solve for one variable in terms of another. Unlike explicit functions where you have a clear relationship like \( y = f(x) \), implicit functions are mixed, such as \( x^2y - xy^2 = 2 \) in this exercise.

To differentiate such an equation, we treat all variables equally and differentiate both sides of the equation with respect to a chosen variable, typically \( x \). This method requires us to apply the chain rule because each variable is treated as a function that might depend on \( x \).

Remember: When differentiating \( y \) terms, you must multiply by \( \frac{dy}{dx} \), as this accounts for the derivative of \( y \) with respect to \( x \). This step maintains the integrity of the relationship between the variables.

Product Rule

The product rule is essential when differentiating products of two functions, which appears often in implicit differentiation. The product rule states that the derivative of a product \( u(x) \cdot v(x) \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). This rule helps differentiate products systematically.

In our exercise, we use the product rule to differentiate both terms in the implicit equation. For term \( x^2y \), we consider \( u=x^2 \) and \( v=y \). Differentiating gives \( 2xy + x^2 \frac{dy}{dx} \). Similarly, for \(-xy^2\), with \( u=-x \) and \( v=y^2 \), we derive \( -y^2 - 2xy \frac{dy}{dx} \).

Understanding and applying the product rule is crucial when working with complex functions expressed as products, ensuring each component's change is accurately represented.

Curve Analysis

Curve analysis in calculus involves examining the properties and behaviors of a curve, which helps us understand its geometry and features like tangents and slopes. For vertical tangents, we are particularly interested in where the derivative \( \frac{dx}{dy} \) becomes zero, indicating a vertical slope.

The equation \( (x^2 - 2xy) \frac{dy}{dx} = y^2 - 2xy \) gives insights into the slope. A vertical tangent occurs when "\( (x^2 - 2xy) \)" is zero. By solving this, we see that vertical tangents occur when \( x(x - 2y) = 0 \), leading to potential solutions for \( x = 0 \) and \( x = 2y \).

After substituting these solutions back into the original equation, we validate which points actually satisfy the curve's equation. This analytic step is fundamental in confirming where the curve's geometry leads to vertical tangents.

Calculus Problem Solving

Calculus problem solving combines various skills, such as logical reasoning, critical thinking, and a solid understanding of calculus techniques. It requires interpreting problems, applying appropriate methods, and systematically solving, as seen in this example.

Every step - from implicit differentiation and using the product rule, to analyzing curves - is a part of a comprehensive problem-solving process.

• Start by interpreting the problem: identify what the question is asking.
• Select the correct methods: here, implicit differentiation and product rule.
• Work through the algebra: ensure each manipulation maintains the equation's integrity.
• Verify solutions: confirm that solutions satisfy all original conditions.

By following these steps, calculus not only solves problems but also enhances understanding of why solutions work, building a deeper appreciation for mathematical analysis.