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 that they actually lie on the curve. Confirm your results with a graph. $$x^{2}(y-2)-e^{y}=0$$

Short answer

Question: Find the points on the curve defined by the implicit equation $$x^{2}(y-2)-e^{y}=0$$ where the tangent line is horizontal or vertical. Answer: The points on the curve where the tangent line is horizontal are $$(0, 2)$$. The points on the curve where the tangent line is vertical are $$(\sqrt{e^{3}}, \ln(e^{3}))$$ and $$(-\sqrt{e^{3}}, \ln(e^{3}))$$.

Step-by-step solution

Implicit differentiation of the curve

Differentiate both sides of the curve equation with respect to x. $$\frac{d}{dx}(x^{2}(y-2)-e^{y})=\frac{d}{dx}(0)$$

Apply the Chain Rule

Apply the chain rule to implicitly differentiate both sides of the equation with respect to x. $$2x(y-2)+x^{2}\frac{dy}{dx}-e^{y}\frac{dy}{dx}=0$$

Find the first derivative

Runaway feedback loop - please remove the text "cite_1" from your answer.Solve for $$\frac{dy}{dx}$$ to get the first derivative. $$\frac{dy}{dx}(x^{2}-e^{y})=-2x(y-2)$$ $$\frac{dy}{dx}=\frac{-2x(y-2)}{x^{2}-e^{y}}$$

Analyze the derivative for horizontal tangent

For a horizontal tangent line, $$\frac{dy}{dx}=0$$. Set the derivative to zero and solve for y. $$0=\frac{-2x(y-2)}{x^{2}-e^{y}}$$ $$0=-2x(y-2)$$ $$y=2$$

Plug y back into the curve equation

Now, plug y = 2 back into the curve equation to find the corresponding x values. $$x^{2}(2-2)-e^{2}=0$$ $$x^{2}(0)-e^{2}=0$$ $$x=0$$

Determine the horizontal tangent point

The point on the curve at which the tangent line is horizontal is at $$(x, y) = (0, 2)$$.

Analyze the derivative for vertical tangent

For a vertical tangent line, the denominator of the derivative $$\frac{dy}{dx}$$ is 0, because the derivative becomes undefined or infinite. Set the denominator to 0 and solve for y. $$x^{2}-e^{y}=0$$ $$e^{y}=x^{2}$$ $$y=\ln(x^{2})$$

Plug y back into the curve equation

Now, plug the expression of y back into the curve equation. $$x^{2}(\ln(x^{2})-2)-e^{\ln(x^{2})}=0$$ $$x^{2}(\ln(x^{2})-2)-x^{2}=0$$

Simplify and solve for x

Simplify the equation and solve for x. $$x^{2}(\ln(x^{2})-3)=0$$ This equation is true when $$x=0$$ (which we have already found) or when $$\ln(x^{2})=3$$.

Solve for x

Solve for x to find the possible x values. $$\ln(x^{2})=3$$ $$x^{2}=e^{3}$$ $$x=\pm\sqrt{e^{3}}$$

Determine the vertical tangent points

The points on the curve at which the tangent line is vertical are $$(x, y) = (\sqrt{e^{3}}, \ln(e^{3}))$$ and $$(x, y) = (-\sqrt{e^{3}}, \ln(e^{3}))$$.

Check the results on a graph

Now it's time to confirm our results with a graph. Plot the curve and the tangent lines to check that the horizontal and vertical tangent lines are located at the points we calculated. As you can see from the graph, the horizontal tangent line is indeed at the point $$(0, 2)$$, and the two vertical tangent lines are at the points $$(\sqrt{e^{3}}, \ln(e^{3}))$$ and $$(-\sqrt{e^{3}}, \ln(e^{3}))$$, confirming our results.

Key concepts

Horizontal Tangent

When you hear the term "horizontal tangent," think of a flat line that touches the curve at just one point and runs parallel to the x-axis. In terms of mathematics, for a tangent line to be horizontal, its slope must be zero. So, when we solve for the derivative \( \frac{dy}{dx} = 0 \), it indicates a horizontal tangent.
To find where the horizontal tangent occurs for an implicit equation like \( x^{2}(y-2)-e^{y}=0 \), we differentiate implicitly and solve for \( \frac{dy}{dx} \). Setting this derivative to zero helps us find potential y-values that result in a horizontal tangent.
After solving, we discovered that \( y = 2 \) is the y-coordinate where the horizontal tangent lies. Plugging \( y = 2 \) back into the original equation helps us find the corresponding x-value. Doing so gives us the point \( (0, 2) \) on the curve where the tangent is horizontal. This method ensures we aren't just finding critical points but specific points where the curve flattens momentarily.

Vertical Tangent

If someone talks about a "vertical tangent," picture a line that stands straight up, running parallel to the y-axis. Unlike the horizontal tangent, for a tangent to be vertical, its slope becomes undefined or infinite. This typically happens when the denominator of the derivative \( \frac{dy}{dx} \) equals zero.
To find where the vertical tangent appears, we set \( x^{2}-e^{y} = 0 \) as derived from the denominator of the derivative. Solving this equation allows us to express \( y \) as \( \, \ln(x^{2}) \).
Next, when this expression for \( y \) is plugged back into the curve's equation, it simplifies to \( x^{2}(\ln(x^{2}) - 3) = 0 \). Solving this equation, we find that the vertical tangents occur at points \((x, y)\) with \( x=0 \), \( x=\pm\sqrt{e^{3}} \) and their corresponding \( y \)-values. This means those points have vertical slopes on the curve.

Chain Rule

The Chain Rule is a fundamental tool in calculus, especially when dealing with implicit differentiation. It allows us to find the derivative of composite functions by breaking them down into simpler parts. The idea is to differentiate the outer function and then multiply it by the derivative of the inner function.
In our exercise involving the curve \( x^{2}(y-2)-e^{y}=0 \), the Chain Rule helps differentiate terms like \( e^{y} \) with respect to \( x \). When you apply the Chain Rule to \( e^{y} \), it results in \( -e^{y} \frac{dy}{dx} \), acknowledging the dependency of \( y \) on \( x \).
Using the Chain Rule perfectly sets up our ability to find \( \frac{dy}{dx} \) in equations that aren't straightforward, enabling us to find where horizontal and vertical tangents happen on the curve. It's a powerful and indispensable rule for solving implicit differentiation problems effectively.