Question

a. Determine the points where the curve \(x+y^{3}-y=1\) has a vertical tangent line (see Exercise 60 ). b. Does the curve have any horizontal tangent lines? Explain.

Short answer

To find the points on the curve defined by \(x+y^3-y=1\) with vertical tangent lines, we first rearrange the equation for \(y\) and find its derivative with respect to \(x\). After solving for \(\frac{dy}{dx}\), we identify the points where \(\frac{dy}{dx}\) is infinite or undefined. The resulting points where the curve has vertical tangent lines are \((\frac{2}{\sqrt{3}} - 1, \frac{1}{\sqrt{3}})\) and \((-\frac{2}{\sqrt{3}} - 1, -\frac{1}{\sqrt{3}})\). To check for horizontal tangent lines, we look for points where \(\frac{dy}{dx}=0\), but in this case, there are no such points. Therefore, there are no horizontal tangent lines for this curve.

Step-by-step solution

Solve for \(y\) and differentiate

First, rearrange the given equation for \(y\): $$x+y^3-y=1 \Rightarrow y = y^3 + x - 1$$ Now, find the derivative of \(y\) with respect to \(x\) using implicit differentiation: $$\frac{dy}{dx} = 3y^2 \frac{dy}{dx} + 1$$

Solve for \(\frac{dy}{dx}\)

Rearrange the equation to isolate \(\frac{dy}{dx}\): $$\frac{dy}{dx}(3y^2 - 1) = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{3y^2 - 1}$$

Find points for vertical tangent line

To find the points where the curve has a vertical tangent line, we need the points where \(\frac{dy}{dx}\) is infinite or undefined: $$3y^2 - 1 = 0 \Rightarrow y^2 = \frac{1}{3} \Rightarrow y = \pm\frac{1}{\sqrt{3}}$$ Now, find the corresponding x-coordinates by substituting the values of \(y\) back into the original equation: For \(y = \frac{1}{\sqrt{3}}\), $$x = \frac{1}{\sqrt{3}^3} + \frac{1}{\sqrt{3}} - 1 = \frac{2}{\sqrt{3}} - 1$$ For \(y = -\frac{1}{\sqrt{3}}\), $$x = -\frac{1}{\sqrt{3}^3} -\frac{1}{\sqrt{3}} - 1 = -\frac{2}{\sqrt{3}} - 1$$ Hence, the points where the curve has vertical tangent lines are \((\frac{2}{\sqrt{3}} - 1, \frac{1}{\sqrt{3}})\) and \((-\frac{2}{\sqrt{3}} - 1, -\frac{1}{\sqrt{3}})\).

Check for points with horizontal tangent lines

To check if the curve has a horizontal tangent line, we need to find the points where \(\frac{dy}{dx}=0\): $$\frac{1}{3y^2 - 1} = 0$$ However, a fraction can only be zero if its numerator is zero. In this case, the numerator is 1, which is nonzero. Therefore, the curve doesn't have any horizontal tangent lines.

Key concepts

Vertical Tangent Lines

A vertical tangent line occurs at points on a curve where the slope becomes undefined or infinite. This typically happens when the denominator of the derivative is zero. In our exercise, after computing the derivative using implicit differentiation, we found out that the slope \( \frac{dy}{dx} = \frac{1}{3y^2 - 1} \) becomes infinite or undefined when \( 3y^2 - 1 = 0 \). Solving this equation, we deduced that \( y = \pm \frac{1}{\sqrt{3}} \).

To find the corresponding \( x \)-values, we substituted these \( y \)-values back into the original equation. This gave us the points:

• \( \left(\frac{2}{\sqrt{3}} - 1, \frac{1}{\sqrt{3}}\right) \)
• \( \left(-\frac{2}{\sqrt{3}} - 1, -\frac{1}{\sqrt{3}}\right) \)

These are the locations on the graph where vertical tangent lines exist, indicating sharp turns or vertical asymptotes on the curve.

Horizontal Tangent Lines

Horizontal tangent lines are seen where the slope of the curve becomes zero. A zero slope means that the curve is perfectly flat at that point. Mathematically, a horizontal tangent line occurs where \( \frac{dy}{dx} = 0 \).

In the solved exercise, the derivative of the curve equation was \( \frac{dy}{dx} = \frac{1}{3y^2 - 1} \). For this fraction to equal zero, the numerator must be zero. However, the numerator here is 1, a constant and non-zero number. Therefore, there are no points on the curve where the slope equals zero, which indicates there are no horizontal tangent lines.

Recognizing the absence of horizontal tangent lines can also hint at the nature of the curve. In this case, it means that the curve never becomes flat or perfectly horizontal at any point.

Derivatives

Derivatives are fundamental in calculus and help understand the rate of change of a function. In simpler terms, they tell us how a function's output changes as its input changes. For the function given in the exercise, we needed to find the derivative to determine where the curve had vertical or horizontal tangent lines.

Using implicit differentiation, we derived \( y \) with respect to \( x \) even though \( y \) was not isolated. Implicit differentiation allows solving for the derivative without explicitly solving for one variable in terms of another, which is particularly useful in equations involving multiple variables like this one. The resulting derivative \( \frac{dy}{dx} = \frac{1}{3y^2 - 1} \) provided crucial insights into the behavior of the curve, helping identify vertical and horizontal tangents.

Calculus

Calculus is a branch of mathematics focused on change and motion and is fundamental for understanding complex systems involving dynamic changes. It consists mainly of two areas: differentiation and integration. In this exercise, we concentrated on differentiation.

Differentiation deals with the concept of derivatives and how they describe rates of change. This is precisely what we did when we differentiated the equation \( x + y^3 - y = 1 \) implicitly to find \( \frac{dy}{dx} \). By engaging with the derivative, we looked at the tangent lines—the instant rates of change of the curve.

Implicit differentiation further highlights a powerful tool in calculus, where differentiation does not require explicit relationships between variables. This method allowed us to explore the curve's behavior and identify where tangent lines would become vertical or identify the non-existence of horizontal tangent lines.