Question

a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (Exercise 61 ). b. Does the curve have any horizontal tangent lines? Explain.

Short answer

Question: Find the points on the curve \(x+y^2-y=1\) where the curve has vertical and horizontal tangent lines. Answer: The point where the curve has a vertical tangent line is \((\frac{3}{4}, \frac{1}{2})\). There are no points on the curve where the tangent line is horizontal.

Step-by-step solution

Express y as a function of x or x as a function of y

We can express \(x\) as a function of \(y\) by solving for \(x\) in the equation: \(x+y^2-y=1\) \(x = 1 - y^2 + y\) Now we have \(x\) in terms of \(y\): \(x(y) = 1 - y^2 + y\).

Find the derivative dx/dy

To find the points where the tangent lines are vertical or horizontal, we need to take the derivative of \(x\) with respect to \(y\). Differentiating \(x(y)\) with respect to \(y\): \(\frac{dx}{dy} = \frac{d}{dy}(1 - y^2 + y) = -2y + 1\)

Find the points where the curve has a vertical tangent line

A vertical tangent line occurs when the derivative \(\frac{dx}{dy}\) is equal to zero. So, we need to solve for \(y\) when \(\frac{dx}{dy} = 0\): \(-2y + 1 = 0\) \(2y = 1\) \(y = \frac{1}{2}\) Now we need to find the corresponding \(x\) value for this \(y\) value from the function \(x(y)\): \(x(\frac{1}{2}) = 1 - (\frac{1}{2})^2 + \frac{1}{2} = 1 - \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\) So, the point where the curve has a vertical tangent line is \((\frac{3}{4}, \frac{1}{2})\).

Determine if the curve has any horizontal tangent lines

A horizontal tangent line occurs when the derivative is undefined. The derivative \(\frac{dx}{dy} = -2y + 1\) is defined for all values of \(y\), so there are no points on the curve where the tangent line is horizontal.

Key concepts

Tangent Lines in Calculus

In calculus, tangent lines represent a concept that is fundamental to understanding the behavior of functions and curves. A tangent line to a curve at a given point is a straight line that just touches the curve at that point and has the same slope as the curve at that point. This line effectively approximates the curve near the point of tangency.
Tangent lines can be either vertical or horizontal, or they can have other slopes. A vertical tangent occurs at a point where the derivative of the curve, when considered as a function of either variable in a two-dimensional plane, approaches infinity, or in more technical terms, when the horizontal change is 0. Conversely, a horizontal tangent line occurs when the derivative of the curve is 0, indicating that there is no vertical change at that point.
In the context of the exercise, the curve represented by the equation \(x+y^{2}-y=1\) exhibits a vertical tangent line at a specific point. To locate this point, the derivative is set to zero, reflecting the condition for a horizontal tangent, thus indicating the presence of a vertical tangent due to the reciprocal nature of the relationship between the variables in the implicit equation.

Derivative of a Function

The derivative of a function at a point is the rate at which the function value changes with respect to a change in the input value. In more familiar terms, it's the slope of the tangent line to the curve of the function at that point. For a function \(y = f(x)\), the derivative with respect to x is usually denoted as \(f'(x)\) or \(\frac{dy}{dx}\).
In the exercise given, we've expressed x as a function of y and then found the derivative \(\frac{dx}{dy}\). This is a slightly less common scenario as most functions are expressed as y in terms of x. Nevertheless, the principle is the same, and finding the derivative still tells us the slope of the tangent line — but this time, we are considering how x changes as y changes, which informs us about vertical tangents as opposed to horizontal ones.
Understanding the derivative is crucial for solving many types of problems in calculus, including optimization and motion, and it's foundational for more advanced topics like differential equations.

Implicit Differentiation

Sometimes functions are not provided in a simple #y = f(x)# form, but rather in an equation involving both x and y that cannot be easily solved for one variable in terms of the other. This is where implicit differentiation comes into play. It is a method used to find the derivative of a function implicitly defined by an equation involving both variables.
To apply implicit differentiation to the equation \(x+y^{2}-y=1\), we treat y as a function of x (or vice versa) and differentiate both sides of the equation with respect to x. In doing so, we must apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function. This rule is pivotal when y is itself a function of x.
Implicit differentiation allows us to work with the derivative \(\frac{dx}{dy}\) without isolating y as a function of x, and vice versa. It’s an elegant technique that comes in handy especially when dealing with curves described by more complex equations in calculus.