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. $$x^{2}(y-2)-e^{y}=0$$

Short answer

Answer: There are no horizontal tangent lines for the given curve. However, there are two vertical tangent lines at the points \((-\sqrt{e^{3}},3)\) and \((\sqrt{e^{3}},3)\).

Step-by-step solution

Take the derivative of the equation implicitly with respect to x

Differentiate both sides of the equation with respect to x, using implicit differentiation. Apply the product rule where necessary and remember that the derivative of \(y\) with respect to \(x\) would be denoted as \(\frac{dy}{dx}\). $$\frac{d}{dx}(x^{2}(y-2)) - \frac{d}{dx}(e^{y}) = \frac{d}{dx}(0)$$ $$2x(y-2) + x^{2}\frac{dy}{dx} - e^{y}\frac{dy}{dx} = 0$$

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

Now, solve the equation for \(\frac{dy}{dx}\) to get the slope of the tangent line. $$x^{2}\frac{dy}{dx} - e^{y}\frac{dy}{dx} = -2x(y-2)$$ $$\frac{dy}{dx}(x^{2} - e^{y}) = -2x(y-2)$$ $$\frac{dy}{dx} = \frac{-2x(y-2)}{(x^{2} - e^{y})}$$

Identify conditions for horizontal and vertical tangents

Now that we have the derivative, let's identify the conditions for horizontal and vertical tangent lines: - Horizontal tangent: The slope of the tangent line will be 0. So, set \(\frac{dy}{dx} = 0\) and solve for \(x\) and \(y\) in that case. - Vertical tangent: The slope of the tangent line will be undefined. So, set the denominator of the derivative to 0 and solve for \(x\) and \(y\). (i.e. \(x^{2} - e^{y} = 0\))

Solve for horizontal tangents

Setting \(\frac{dy}{dx} = 0\), we get $$-2x(y-2) = 0$$ This implies that either \(x = 0\) or \(y = 2\). For each case, substitute back into the original equation and solve for the other variable. Case 1: \(x=0\) $$0^{2}(y-2)-e^{y}=0 \Rightarrow -e^{y}=0$$ There is no real solution for \(y\) in this case, so there is no horizontal tangent line at \(x=0\). Case 2: \(y=2\) $$x^{2}(2-2)-e^{2}=0 \Rightarrow x^{2}(0)-e^{2}=0$$ The equation simplifies to \(-e^{2}=0\). There's no real solution for \(x\), so there is no horizontal tangent line at \(y=2\). Thus, there are no horizontal tangent lines for the given curve.

Solve for vertical tangents

For vertical tangents, we set \(x^{2} - e^{y} = 0\), which implies \(x^{2} = e^{y}\). Replacing \(x^{2}\) in the original equation with \(e^{y}\), we get: $$e^{y}(y-2)-e^{y} = 0$$ Now factor out \(e^{y}\): $$e^{y}(y-2-1)=0$$ $$e^{y}(y-3)=0$$ Since \(e^{y}\) is never equal to 0, we have only one solution, \(y=3\). Plugging \(y=3\) back into \(x^{2}=e^{y}\), we get \(x^{2}=e^{3}\). Thus, \(x=\pm\sqrt{e^{3}}\). So, we have two vertical tangent lines at the points \((-\sqrt{e^{3}},3)\) and \((\sqrt{e^{3}},3)\).

Confirm results with a graph

Graph the curve \(x^{2}(y-2)-e^{y}=0\) and observe the tangent lines at the points \((-\sqrt{e^{3}},3)\) and \((\sqrt{e^{3}},3)\). You should see vertical tangent lines at these points, confirming our results. In conclusion, there are no horizontal tangent lines, but there are two vertical tangent lines at the points \((-\sqrt{e^{3}},3)\) and \((\sqrt{e^{3}},3)\).

Key concepts

Calculus

Calculus is a vast branch of mathematics that studies how things change. It's a powerful tool that allows us to understand and describe the behavior of functions and their curves, exploring rates of change and accumulations. Calculus can be divided into two main branches:
- **Differential Calculus:** This focuses on the concept of a derivative, which examines how a function changes as its input changes. It helps us understand the slope of curves and the behavior of dynamic systems. - **Integral Calculus:** This involves the accumulation of quantities and the areas under and between curves. It's useful for calculating totals and quantities over an interval.
In this problem, we use differential calculus, particularly implicit differentiation, to explore how a curve behaves and determine where its tangent lines are horizontal or vertical. This technique is crucial in understanding the relationship between variables in equations not solved explicitly for one variable.

Tangent Lines

Tangent lines are straight lines that touch a curve at a particular point without crossing it at that location. They represent the instantaneous direction of the curve and the slope of the curve at that exact point. Understanding tangent lines is crucial because:
- They give us the slope at a specific point, telling us how steep the curve is there. - They are used in approximation, where a function is represented by its tangent line for small intervals due to their local linearity.
In this exercise, we are particularly focused on finding points where the tangent line is either horizontal or vertical. A horizontal tangent line implies that the slope is zero, meaning the curve is flat at that point. Conversely, a vertical tangent implies an undefined slope, showing a steep rise or fall in the curve at that point. Identifying these characteristics helps us deeply understand the nature and behavior of the curve at different points.

Derivative

The derivative is a fundamental concept in calculus, conveying the rate of change of a function with respect to a variable. In simple terms, it tells us how a function behaves as we make small changes to its input. The derivative of a function at any given point is the slope of the tangent line at that point.
- **Implicit Differentiation:** This is a technique used when differentiating equations not presented in the classic "y = f(x)" form. It allows us to find the derivative of one variable with respect to another, even when they are intertwined.
In this exercise, implicit differentiation enlightens us on how to find derivatives when multiple variables are involved. We wanted to find where the slope of the tangent, given by the derivative \( \frac{dy}{dx} \), is either zero or undefined to locate horizontal and vertical tangents respectively. This involves considering both the numerator and the denominator of the derivative separately.

Product Rule

The product rule is an essential differentiation technique used when dealing with products of two functions. If you have two differentiable functions, say \( u(x) \) and \( v(x) \), their product can be differentiated using:
\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]
This means you derive each function separately and sum the resulting terms, taking into account the order and association of the functions.
In the exercise, the product rule is utilized in implicit differentiation. The expression \( x^2(y-2) \) is a product of \( x^2 \) and \( (y-2) \). We applied the product rule to differentiate this product with respect to \( x \). This step is crucial because it simplifies the problem of differentiating complex expressions, making it manageable to solve for critical values of \( x \) and \( y \) that define the nature of the tangent lines.