Question

Vertical tangent lines a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (see Exercise 53 ). b. Does the curve have any horizontal tangent lines? Explain.

Short answer

Question: Determine the points where the curve defined by the equation \(x + y^2 - y = 1\) has a vertical tangent line and explain if the curve has any horizontal tangent lines. Answer: The curve has a vertical tangent line at the point \((\frac{3}{4}, \frac{1}{2})\). There are no horizontal tangent lines for the given curve.

Step-by-step solution

Find the derivative dy/dx

To solve this problem, we need to differentiate the given equation with respect to x implicitly. The given equation is \( x + y^2 - y = 1 \). Differentiate both sides with respect to x: \(\frac{d}{dx}(x) + \frac{d}{dx}(y^2) - \frac{d}{dx}(y) = \frac{d}{dx}(1)\) Which gives: \(1 + 2y\frac{dy}{dx} - \frac{dy}{dx} = 0\) Now solve for dy/dx: \(\frac{dy}{dx}(2y - 1) = -1\) \(\frac{dy}{dx} = \frac{-1}{2y - 1}\)

Find points with vertical tangent lines

To determine the points with vertical tangent lines, we need to find when the derivative is undefined. This occurs when the denominator, \(2y - 1\), is 0: \(2y - 1 = 0\) \(2y = 1\) \(y = \frac{1}{2}\) Now, we plug this value of y back into the original equation to find the corresponding x value: \(x + \left(\frac{1}{2}\right)^2 - \frac{1}{2}= 1\) \(x = 1 - \frac{1}{4} + \frac{1}{2}\) \(x = \frac{3}{4}\) So the point where the curve has a vertical tangent line is \((\frac{3}{4}, \frac{1}{2})\).

Check for horizontal tangent lines

A horizontal tangent line would have a slope of 0 (meaning dy/dx is 0). Let's check if there are any points on the curve for which dy/dx is 0: \(\frac{-1}{2y - 1} = 0\) To find if any such y values exist, we would need the denominator to remain non-zero while the numerator is zero. However, the numerator of the derivative, -1, is never 0 for any value of y. Therefore, there are no horizontal tangent lines for the given curve.

Key concepts

Vertical Tangent Line

A vertical tangent line on a curve is a special type of tangent where the slope is undefined. This happens because the denominator of the derivative expression becomes zero. When the derivative is in the form \(\frac{dy}{dx} = \frac{-1}{2y-1}\), a vertical tangent occurs when \(2y - 1 = 0\). Solving this equation gives \(y = \frac{1}{2}\).
To find the corresponding x-value, substitute \(y = \frac{1}{2}\) back into the original equation \(x + y^2 - y = 1\). This results in \(x = \frac{3}{4}\), leading us to conclude that the curve has a vertical tangent line at the point \(\left(\frac{3}{4}, \frac{1}{2}\right)\).
Vertical tangents are important in analyzing the behavior of functions, as they indicate points where the function stops behaving predictably as a vertical slope can imply rapid changes in the direction of a curve.

Horizontal Tangent Line

Horizontal tangent lines occur when the derivative of the function is zero, indicating a slope of zero and, thus, a completely flat tangent line. In the case of the derivative \(\frac{dy}{dx} = \frac{-1}{2y-1}\), a horizontal tangent would imply that the numerator, -1, must be zero. However, -1 is a constant and can never be zero, meaning the curve we are examining has no horizontal tangent lines.
Understanding when horizontal tangents occur helps in identifying where a function reaches a local maximum or minimum, as these points often form plateaus with horizontal tangents. Despite their nonexistence in this specific exercise, recognizing this concept is crucial for comprehensive graph analysis.

Derivative

The derivative of a function provides us with the slope of the tangent line at any point on the curve. For implicit functions, deriving is more complex but is constructive in finding tangent lines.
When we derive \(x + y^2 - y = 1\) implicitly, we find that\(\frac{dy}{dx} = \frac{-1}{2y-1}\). This process involves differentiating each term with respect to x while considering y as a function of x. The derivative equips us with critical insights, revealing how y changes concerning x, hence aiding in understanding function behavior and predicting any tangent anomalies.

Implicit Differentiation Steps

Implicit differentiation is a technique used when a function isn’t isolated as y=f(x). Instead, y and x are mixed on one side of an equation. To apply implicit differentiation:

• Differentiate every term with respect to x, applying the chain rule where necessary.
• Gather all terms containing \(\frac{dy}{dx}\) on one side of the equation.
• Factor out \(\frac{dy}{dx}\) and solve for it.

In the provided exercise, applying these steps to the equation \(x + y^2 - y = 1\) gives us\(\frac{dy}{dx} = \frac{-1}{2y - 1}\), providing information about the slope of tangent lines and aiding in finding points where tangents may be horizontal or vertical. This systematic approach using chain rule and properties of equality ensures we can handle even complex intertwined equations.