Problem 48
Question
Find all points on the curve \(x^{2} y-x y^{2}=2\) where the tangent line is vertical, that is, where \(d x / d y=0\).
Step-by-Step Solution
Verified Answer
The point is (2, 1) where the tangent line is vertical.
1Step 1: Implicit Differentiation
We start by differentiating the given equation implicitly with respect to \[\frac{d}{dy}(x^2 y - x y^2 = 2)\].This yields the following expression:\[x^2 \frac{dy}{dy} + 2x y \frac{dx}{dy} - 2x y \frac{dy}{dy} - y^2\frac{dx}{dy}= 0\].
2Step 2: Simplify the Derivative
After simplification, we have:\[x^2 + 2xy \frac{dx}{dy} - 2xy - y^2\frac{dx}{dy} = 0\].Rearrange it to isolate \(\frac{dx}{dy}\):\[(2xy - y^2) \frac{dx}{dy} = 2xy - x^2\].
3Step 3: Finding Vertical Tangent Condition
For the tangent line to be vertical, we need \(\frac{dx}{dy} = 0\). Set the equation from the previous step to zero:\[2xy - x^2 = 0\].
4Step 4: Solve for Relationship Between x and y
Factor the expression:\[x(2y - x) = 0\].This gives two possibilities:1. \(x = 0\)2. \(2y - x = 0\), which simplifies to \(x = 2y\).
5Step 5: Find Points for x = 0
Substitute \(x = 0\) into the original equation:\[0^2y - 0y^2 = 2\implies 0 = 2\].This is a contradiction, so no solutions exist for \(x = 0\).
6Step 6: Find Points for x = 2y
Substitute \(x = 2y\) into the original equation where possible in place of \(x\):\[(2y)^2 y - (2y)y^2 = 2\]Simplify to obtain:\[4y^3 - 2y^3 = 2\]\[2y^3 = 2\]\[y^3 = 1\].
7Step 7: Solve for y and Find x Values
From \(y^3 = 1\), we find \(y = 1\).Using \(x = 2y\), we substitute back to find \(x = 2 \cdot 1 = 2\).
8Step 8: List Points with Vertical Tangents
The only solution is \((x, y) = (2, 1)\). This point has a vertical tangent on the curve.
Key Concepts
Implicit DifferentiationCalculus Problem SolvingDifferential EquationsImplicit Differentiation Steps
Implicit Differentiation
Implicit differentiation is an essential technique in calculus, particularly when dealing with equations not easily expressed explicitly as functions of a single variable. This method allows us to find the derivative of a function when it is presented in an implicit form.
For instance, in the problem at hand, we deal with the curve defined by the implicit equation \(x^2 y - x y^2 = 2\). Here, both \(x\) and \(y\) are treated as functions that depend on another variable.
By differentiating each term using the chain rule, while keeping in mind both \(x\) and \(y\) are intertwined, we arrive at a unified derivative expression. Implicit differentiation simplifies the process of finding slopes, especially in scenarios where solving for one variable explicitly isn't straightforward.
For instance, in the problem at hand, we deal with the curve defined by the implicit equation \(x^2 y - x y^2 = 2\). Here, both \(x\) and \(y\) are treated as functions that depend on another variable.
By differentiating each term using the chain rule, while keeping in mind both \(x\) and \(y\) are intertwined, we arrive at a unified derivative expression. Implicit differentiation simplifies the process of finding slopes, especially in scenarios where solving for one variable explicitly isn't straightforward.
Calculus Problem Solving
Problem-solving in calculus often involves multi-step approaches, utilizing a variety of tools and techniques to reach a solution. In this exercise, understanding the nature of the tangent line's slope was critical.
The exercise's goal was to determine where a vertical tangent line occurs on a curve. This is equivalent to finding points on the curve where the derivative \(\frac{dx}{dy}\) becomes zero.
Solving for these conditions required factoring and analyzing resulting expressions carefully, ensuring that all algebraic manipulations adhere to calculus principles. This step-by-step approach is typical in solving calculus problems, highlighting the importance of logical reasoning and analytical skills.
The exercise's goal was to determine where a vertical tangent line occurs on a curve. This is equivalent to finding points on the curve where the derivative \(\frac{dx}{dy}\) becomes zero.
Solving for these conditions required factoring and analyzing resulting expressions carefully, ensuring that all algebraic manipulations adhere to calculus principles. This step-by-step approach is typical in solving calculus problems, highlighting the importance of logical reasoning and analytical skills.
Differential Equations
Differential equations form the backbone of describing various natural phenomena in mathematics. They involve derivatives and are pivotal in the problem since they implicitly express relationships between \(x\) and \(y\).
In this case, finding \(\frac{dx}{dy}\) by differentiating the given equation is akin to solving a differential equation, albeit a simple one. Such problems teach crucial concepts of how differential equations can describe curves and paths not defined by traditional functions.
This process elucidates how specific conditions (like vertical tangents) can be extracted by analyzing the behavior of derivatives, demonstrating the power of differential equations in providing insights into geometric interpretations.
In this case, finding \(\frac{dx}{dy}\) by differentiating the given equation is akin to solving a differential equation, albeit a simple one. Such problems teach crucial concepts of how differential equations can describe curves and paths not defined by traditional functions.
This process elucidates how specific conditions (like vertical tangents) can be extracted by analyzing the behavior of derivatives, demonstrating the power of differential equations in providing insights into geometric interpretations.
Implicit Differentiation Steps
Going through the steps of implicit differentiation requires careful handling of variable interdependencies. Initially, each term of the equation \(x^2 y - x y^2 = 2\) is differentiated with respect to \(y\), keeping in mind that \(x\) is also a function of \(y\).
This results in terms involving \(\frac{dx}{dy}\), which need to be isolated to find the condition for vertical tangency.
The simplification process involves collecting like terms and solving for \(\frac{dx}{dy}\). Setting the derivative to zero unveils important relationships between \(x\) and \(y\).
This results in terms involving \(\frac{dx}{dy}\), which need to be isolated to find the condition for vertical tangency.
The simplification process involves collecting like terms and solving for \(\frac{dx}{dy}\). Setting the derivative to zero unveils important relationships between \(x\) and \(y\).
- The crucial part is recognizing terms that cancel or combine, which leads to a solvable expression.
- Step-wise factoring and substitution aid in narrowing down the specific points on the curve associated with a vertical tangent.
Other exercises in this chapter
Problem 47
Use the Product Rule to show that \(D_{x}[f(x)]^{2}=\) \(2 \cdot f(x) \cdot D_{x} f(x)\).
View solution Problem 47
In Problems 47-58, express the indicated derivative in terms of the function \(F(x)\). Assume that \(F\) is differentiable. $$ D_{x}(F(2 x)) $$
View solution Problem 48
Find the indicated derivative. \(D_{x}\left[2^{\left(e^{x}\right)}+\left(2^{e}\right)^{x}\right]\)
View solution Problem 48
$$ \text { Develop a rule for } D_{x}[f(x) g(x) h(x)] $$
View solution