Let C be the curve y³ - 3xy + 2 = 0. If H is the set of points on the curve C where the tangent is h

Q: Wait, why do we reject the negative root here? Or why is the set H empty?

Great question! When we set the numerator of our derivative to zero, we get $y = 0$. If we plug $y = 0$ into the original equation $y^3 - 3xy + 2 = 0$, we get $0 - 0 + 2 = 0$, which simplifies to $2 = 0$. This is a mathematical impossibility! It means there is no point on the curve where $y = 0$. Thus, no horizontal tangent can exist.

Q: Why can't we just use energy conservation or some other method?

This is a calculus problem, not a physics one! We are dealing with the geometry of a curve. The only way to find the slope of a tangent line at any point on an implicit curve is through implicit differentiation. It is the most direct and powerful tool in your arsenal for this type of problem.

Q: How do I know if I should set the numerator or denominator to zero?

Think about the slope $\frac{dy}{dx}$. A horizontal line has a slope of $0$, so the fraction $\frac{dy}{dx}$ must be $0$, which happens when the numerator is $0$. A vertical line has an undefined slope, which happens when the denominator of the fraction is $0$. It is all about the behavior of the fraction!

Question

Let $C$ be the curve $y^3 - 3xy + 2 = 0$. If $H$ is the set of points on the curve $C$ where the tangent is horizontal and $V$ is the set of the point on the curve $C$ where the tangent is vertical then $H = .........$ and $V = .........$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before the curve defined by the equation $y^3 - 3xy + 2 = 0$. It is not a simple parabola or a circle; it is an implicit landscape where $x$ and $y$ are locked in a complex dance.

Our goal is to find the points where the tangent line is horizontal (Set $H$) and where it is vertical (Set $V$). This is a detective story where we use calculus to uncover the hidden geometry of the curve.

### The Calculus Engine: Implicit Differentiation

Because we cannot easily isolate $y$ in $y^3 - 3xy + 2 = 0$, we must use the power of implicit differentiation. We treat $y$ as a function of $x$ and differentiate every term with respect to $x$.

Applying the derivative operator $\frac{d}{dx}$ to the entire equation, we get:

$$\frac{d}{dx}(y^3) - \frac{d}{dx}(3xy) + \frac{d}{dx}(2) = \frac{d}{dx}(0)$$

Using the chain rule on $y^3$, we get $3y^2 \frac{dy}{dx}$. For the term $3xy$, we must use the product rule: $3(x \frac{dy}{dx} + y)$. The derivative of the constant $2$ is $0$.

Putting it all together, we have:

$$3y^2 \frac{dy}{dx} - 3\left(x \frac{dy}{dx} + y\right) = 0$$

Dividing by $3$ simplifies this to $y^2 \frac{dy}{dx} - x \frac{dy}{dx} - y = 0$. Finally, isolating $\frac{dy}{dx}$, we find our golden formula:

$$\frac{dy}{dx} = \frac{y}{y^2 - x}$$

### The Hunt for Horizontal Tangents ($H$)

A horizontal tangent means the slope is zero. For the fraction $\frac{y}{y^2 - x}$ to be zero, the numerator must be zero, so $y = 0$.

Does this point exist on our curve? Let us substitute $y = 0$ into the original equation $y^3 - 3xy + 2 = 0$.

We get $0^3 - 3x(0) + 2 = 0$, which simplifies to $2 = 0$. This is a blatant contradiction!

It means there is no point on the curve where $y = 0$. Therefore, the set $H$ is the empty set, denoted as $\phi$.

### The Hunt for Vertical Tangents ($V$)

A vertical tangent occurs when the slope is undefined, which happens when the denominator of our derivative is zero: $y^2 - x = 0$, or $x = y^2$.

Now, we substitute $x = y^2$ into the original equation to find the coordinates. The equation becomes:

$$y^3 - 3(y^2)y + 2 = 0$$

This simplifies to $y^3 - 3y^3 + 2 = 0$, or $-2y^3 + 2 = 0$. This gives $2y^3 = 2$, so $y^3 = 1$.

The only real solution is $y = 1$. Using our condition $x = y^2$, we find $x = (1)^2 = 1$. Thus, the only point with a vertical tangent is $(1, 1)$.

### Conclusion

We have successfully navigated the curve. We found that **$H = \phi$** and **$V = \{(1, 1)\}$**.

This problem beautifully demonstrates how calculus allows us to probe the secrets of implicit equations, turning a daunting algebraic expression into a clear geometric reality. Keep practicing, and you will find that every curve has a story to tell.

Animated Solution for Mathematics - Differentiation: Let $C$ be the curve $y^{3} - 3 x y + 2 = 0$ . If $H$ is the set of points on the curve $C$ where the tangent is horizontal and $V$ is the set of the point on the curve $C$ where the tangent is vertical then $H = .........$ and $V = .........$

Visualized Solution

The Curve Equation

Conditions for $H$ and $V$

Implicit Differentiation Setup

Applying Derivative Rules

Assembling the Differentiated Equation

Solving for $\frac{d y}{d x}$

Set $H$ : Horizontal Tangents

Evaluating Set $H$

Set $V$ : Vertical Tangents

Solving for Points in $V$

Calculating the $y$ -coordinate

Calculating the $x$ -coordinate

Final Answer

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Calculus Engine

The Hunt for Horizontal Tangents ( $H$ )

The Hunt for Vertical Tangents ( $V$ )

Conclusion

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Visualized Solution

The Curve Equation

Conditions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V

Implicit Differentiation Setup

Applying Derivative Rules

Assembling the Differentiated Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H: Horizontal Tangents

Evaluating Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V: Vertical Tangents

Solving for Points in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V

Calculating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate

Calculating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate