Question

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$

Short answer

Based on the given equation of the curve and the steps we've followed, we have found that: 1. At the point (-1, 1), the curve has a horizontal tangent. 2. At the point (1, -1), the curve has a horizontal tangent. 3. At the point (0, 0), the curve has a vertical tangent. These results can be confirmed by graphing the original curve and observing that the tangent lines are horizontal or vertical at these points.

Step-by-step solution

Use implicit differentiation to find dy/dx

Let's take the derivative of the given equation with respect to x, using implicit differentiation. This process involves differentiating both sides of the equation with respect to x and solving for dy/dx. $$\frac{d}{dx} (x(1-y^2) + y^3) = 0$$ Now let's differentiate term by term: $$(1 - y^2) + x(-2y) \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0$$ Let's group the terms with dy/dx on the left-hand side and the other terms on the right-hand side: $$\frac{dy}{dx}(x(-2y)+3y^2) = (y^2-1)$$ Now we can solve for dy/dx: $$\frac{dy}{dx} = \frac{y^2 -1}{-2yx+3y^2}$$

Find points where the tangent line is horizontal or vertical

Horizontal tangent line occurs when dy/dx = 0, and a vertical tangent line occurs when the denominator of dy/dx is equal to 0. So let's solve these two cases separately. a) Horizontal tangent (dy/dx = 0): $$0 = \frac{y^2 -1}{-2yx+3y^2}$$ Since the numerator is equal to 0, we have: $$y^2-1=0$$ Solving this equation for y, we get two critical points: $$y = \pm1$$ b) Vertical tangent (-2yx + 3y^2 = 0): $$-2yx+3y^2=0$$ Factoring out the y, we get: $$y(-2x+3y)=0$$ Solving this equation for y, we get two critical points: $$y=0$$ and $$y=\frac{2}{3}x$$

Find the corresponding x values and verify that the points lie on the curve

Now we need to find the x values for the critical points we found in Step 2. a) For y = 1: Plug in y = 1 in the original equation: $$x(1-1^2) + 1^3 = 0$$ This gives us x = -1. b) For y = -1: Plug in y = -1 in the original equation: $$x(1-(-1)^2) + (-1)^3 = 0$$ This gives us x = 1. c) For y = 0: Plug in y = 0 in the original equation: $$x(1-0^2) + 0^3 = 0$$ This gives us x = 0. Now we will check if the point satisfying the second equation of vertical tangent (y = 2/3x) lies on the curve. Plugging y = 2/3x into the curve equation: $$x\left(1-\left(\frac{2}{3}x\right)^{2}\right)+\left(\frac{2}{3}x\right)^{3}=0$$ Unfortunately, this equation does not give a specific x value. Hence, we cannot confirm that the point lies on the curve. From these calculations, we have found that the following points lie on the curve and have horizontal/vertical tangent lines: 1. The point (-1, 1) has a horizontal tangent. 2. The point (1, -1) has a horizontal tangent. 3. The point (0, 0) has a vertical tangent. Now you can confirm these results with a graph of the original equation, making sure that the curve has horizontal and vertical tangents at these points.

Key concepts

Tangent Line

A tangent line is a line that touches a curve at a single point, and its slope is the derivative of the function at that specific point. In simpler terms, the tangent line represents the instantaneous rate of change or the 'direction' of the curve at that point. Calculating the tangent line for a curve not only helps in determining the slope but also aids in sketching and understanding the overall behavior of the curve.

When using implicit differentiation, the slope of the tangent line is found by differentiating the given equation concerning both variables and solving for \(\frac{dy}{dx}\). This form tells us the slope of the tangent at any given point on the curve.

For example, in our exercise, implicit differentiation of the equation gives the derivative: \[\frac{dy}{dx} = \frac{y^2 -1}{-2yx+3y^2}\] This formula allows us to analyze when the tangent line is horizontal, vertical, or has any slope depending on the values of \(x\) and \(y\). Such tangents can be useful in various applications such as physics, engineering, and economics for predictions and optimizations.

Horizontal Tangent

A horizontal tangent is a special case where the tangent line to the curve is completely horizontal, meaning its slope is zero. To determine where a curve has a horizontal tangent, we need to set the derivative \(\frac{dy}{dx}\) equal to zero.

In our problem, this condition leads to setting the numerator of the derivative equal to zero: \[y^2 - 1 = 0\] Solving this, we find the critical points \(y = \pm 1\). At these points, the slope of the tangent line is 0, indicating a flat tangent.

It's essential to verify these \(y\)-values in the original equation to ensure whether such points \((x, \pm 1)\) truly lie on the curve. Solving for \(x\) reveals the points \((-1, 1)\) and \((1, -1)\), where the curve has horizontal tangents. Horizontal tangents often indicate local maxima, minima, or points of inflection on the graph.

Vertical Tangent

A vertical tangent occurs when the tangent line of the curve is straight up and down, meaning the slope is undefined or infinite. We find a vertical tangent by setting the denominator of \(\frac{dy}{dx}\) to zero.

In the provided problem, this involves setting:\[-2yx + 3y^2 = 0\]Factoring gives\[y(-2x + 3y) = 0\] This solution presents two important points:

• \(y = 0\)
• \(y = \frac{2}{3}x\)

The first solution, \(y = 0\), quickly leads us to \(x = 0\) upon substitution back into the curve equation, validating the point \((0, 0)\) as having a vertical tangent line.

Although the second solution \(y = \frac{2}{3}x\) is valid algebraically, it does not yield specific points on the curve after substitution, indicating that not all algebraically derived points necessarily manifest on the curve. This highlights the importance of verification.

Critical Points

Critical points are crucial in calculus as they reveal important features of a function's graph, such as where it changes direction or has extrema. These are found where the derivative is zero (horizontal tangent) or undefined (vertical tangent).

In this exercise, the critical points are determined through investigating where the derivative \(\frac{dy}{dx}\) equals zero or its denominator is zero. Solving both conditions help locate

• Horizontal tangents: which occur when \(y = \pm 1\), giving points \((-1, 1)\) and \((1, -1)\).
• Vertical tangents: which occur when \(x = 0\), resulting in the point \((0, 0)\).

Plotting these critical points on the graph of a curve shows significant changes in direction and behavior around these points.

Understanding critical points allows us to deeply analyze the function's behavior, offering insights into its increasing or decreasing trends and potential local maxima or minima.