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Chapter 3: Problem 86

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. $$y^{2}-3 x y=2$$

Short Answer

Expert verified
Question: Determine the points, if any, on the curve $$y^{2}-3xy=2$$ where the tangent line is horizontal or vertical. Answer: There are no points on the curve $$y^{2}-3xy=2$$ where the tangent line is horizontal or vertical.

Step by step solution

01

Implicit Differentiation

Differentiate both sides of the equation with respect to x. Remember to apply the chain rule when necessary. $$\frac{d}{dx}(y^2) - \frac{d}{dx}(3xy)=\frac{d}{dx}(2)$$
02

Compute the Derivative

After applying the chain rule and derivative, we have: $$2y\frac{dy}{dx}-3x\frac{dy}{dx}-3y=0$$ Now, solve for dy/dx, $$\frac{dy}{dx}(2y-3x)=-3y$$ Now, compute the final derivative: $$\frac{dy}{dx}=\frac{-3y}{2y-3x}$$
03

Determine Horizontal Tangent Points

To find the points of horizontal tangent, we need to find where $$\frac{dy}{dx}=0$$: $$\frac{-3y}{2y-3x}=0$$ The only way this fraction equals 0 is when $$y=0$$. After plugging y=0 in the original equation, $$0-3(0)x=2$$ There is no solution, the equation is not consistent. Therefore, there are no points on the curve with a horizontal tangent.
04

Determine Vertical Tangent Points

To find points of vertical tangent, we need to find where $$\frac{dx}{dy}=0$$ or equivalently, $$\frac{dy}{dx}=\frac{-3y}{2y-3x}$$ is undefined. This can happen when the denominator is 0, $$2y-3x=0$$ Solve for x: $$x=\frac{2}{3}y$$ Now plug this expression for x in the original equation: $$y^2-3\left(\frac{2}{3}y\right)y=2$$ which simplifies to $$y^2-2y^2=2$$ $$y^2=-2$$ Since there is no real solution for y, there are no points on the curve with a vertical tangent.
05

Confirm Results with a Graph

After graphing the function $$y^2-3xy=2$$, you will see that there are no points on the curve where the tangent line is horizontal or vertical. In conclusion, there are no points on the curve $$y^2-3xy=2$$ where the tangent line is horizontal or vertical.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Tangents
When analyzing curves, horizontal tangents occur where the slope of the tangent line is zero. To find these points using implicit differentiation, you'll set the derivative \( \frac{dy}{dx} \)to zero.
In the given exercise's solution, we derived the equation for \( \frac{dy}{dx} \):\[\frac{dy}{dx} = \frac{-3y}{2y-3x}\]This tells us that the slope is zero when the numerator \(-3y\) is zero.
However, when substituting \( y = 0 \) back into the original equation \( y^2 - 3xy = 2 \), we found the equation becomes inconsistent, indicating there are no real points where the tangent line is horizontal.
  • Horizontal tangents imply flat sections of the curve, but not always easy solutions.
  • Always verify on the original equation.
  • If the tangent slope can't be zero with real numbers, no horizontal tangent exists.
Vertical Tangents
Vertical tangents are present when the slope becomes undefined. In implicit differentiation, this occurs when the denominator of the slope \(\frac{dy}{dx}\) is zero.
For this problem:\[\frac{dy}{dx} = \frac{-3y}{2y-3x}\]The denominator \(2y - 3x\)must be zero for the slope to be undefined, i.e.:\[2y - 3x = 0\]Solving for \(x\) gives:\[x = \frac{2}{3}y\]After substituting this back into the original equation, we discovered that the resulting equation \(y^2 = -2\)does not yield real solutions.
This means that although the process should lead us to vertical tangents, there are no actual vertical tangent points on this curve.
  • Vertical tangents can signal undefined behavior or abrupt shifts in the curve.
  • Successfully substitute findings back to ensure solutions fit within the context of the original equation.
  • Real solutions are key; non-real results usually mean there are no tangents fulfilling the criteria.
Chain Rule
The chain rule is a fundamental tool in differentiation, especially necessary when differentiating terms in implicit equations involving products or powers involving more than one variable. When applying implicit differentiation, like in this problem, utilize the chain rule to handle terms where both variables are involved.
For example, differentiate \(y^2\) and \(-3xy\)with respect to \(x\):- For \(y^2\), differentiate to get \(2y\frac{dy}{dx}\). The chain rule is applied considering \(y\) as a function of \(x\).- For \(-3xy\), use both the product rule and chain rule to result in \(-3(x\frac{dy}{dx} + y)\).Here you need to account for both variables being present, using differentiation respectively.Properly applying the chain rule helps solve implicit differentiation correctly.
  • Implicit differentiation involves treating one variable as a function of another.
  • The chain rule is crucial for handling multi-variable relationships.
  • Be attentive to differentiate every element contributing to implicit terms.

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Most popular questions from this chapter

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=m x ; x^{2}+y^{2}=a^{2},\) where \(m\) and \(a\) are constants

Horizontal tangents The graph of \(y=\cos x \cdot \ln \cos ^{2} x\) has seven horizontal tangent lines on the interval \([0,2 \pi] .\) Find the approximate \(x\) -coordinates of all points at which these tangent lines occur.

a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. b. Graph the tangent and normal lines on the given graph. $$\begin{aligned} &3 x^{3}+7 y^{3}=10 y\\\ &\left(x_{0}, y_{0}\right)=(1,1) \end{aligned}$$

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).

Identifying functions from an equation. The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. \(y^{3}=a x^{2}(\text { Neile's semicubical parabola })\)
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