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Chapter 12: Problem 1

Show that the equation \(x^{2} y^{2}+2 e^{x y}-4-2 e^{2}=0\) can be solved for \(y\) in terms of \(x\) in a neighborhood of the point \(x=1\) with \(y(1)=2\). Calculate \(\frac{d y}{\text { t }}\) when \(x=1\).

Short Answer

Expert verified
\( y(x) \) can be solved in a neighborhood of \( x=1 \) with \( y = 2 \), and \( \frac{dy}{dx} = \frac{-8 - 4e^2}{4 + 2e^2} \) at \( x=1 \).

Step by step solution

01

Verify conditions for Implicit Function Theorem

To determine if the equation can be solved for \( y \) in terms of \( x \), apply the Implicit Function Theorem. We have the equation \( F(x, y) = x^{2} y^{2} + 2 e^{xy} - 4 - 2 e^{2} = 0 \). Check if \( F \) is continuous at \( (x, y) = (1, 2) \) and both functions \( F_x \) and \( F_y \) are partially differentiable at this point. The point \( (x, y) = (1, 2) \) gives \( F(1, 2) = 1^2 \times 2^2 + 2e^{1 \times 2} - 4 - 2e^2 = 0 \), indicating that the function is satisfied at the point.
02

Compute partial derivatives

Calculate the partial derivatives of \( F \) with respect to \( x \) and \( y \).For \( F_x \): Use the product rule and chain rule,\[ F_x = 2xy^2 + 2ye^{xy} \]For \( F_y \): Also use the product rule and chain rule,\[ F_y = 2x^2y + 2xe^{xy} \] We need \( F_y(1, 2) eq 0 \) to apply the Implicit Function Theorem.
03

Evaluate \( F_y(1, 2) \)

Substitute \( x = 1 \), \( y = 2 \) into \( F_y \):\[ F_y(1, 2) = 2(1)^2(2) + 2(1)e^{1 \cdot 2} = 4 + 2e^2 \]This is not equal to zero, \( 4 + 2e^2 eq 0 \), confirming that \( y \) can locally be expressed as a function of \( x \) near \( (1, 2) \) using the Implicit Function Theorem.
04

Applying the Implicit Function Theorem

Since \( F(x, y) = 0 \) can be solved for \( y \) as a function of \( x \) and \( F_y(1, 2) eq 0 \), the Implicit Function Theorem assures us of the existence of \( y(x) \) near \( x=1 \).
05

Differentiate implicitly to find \( \frac{dy}{dx} \)

Differentiate the equation implicitly with respect to \( x \) to find \( \frac{dy}{dx} \):1. Differentiate \( x^2 y^2 \) to get \( 2xy^2 + 2x^2y \frac{dy}{dx} \).2. Differentiate \( 2e^{xy} \) using the chain rule: \( 2e^{xy}(y + x\frac{dy}{dx}) \).3. Set the differential of the constant \( -4-2e^2 \) to zero.The full implicit differentiation gives\[ 2xy^2 + 2x^2y \frac{dy}{dx} + 2e^{xy}(y + x\frac{dy}{dx}) = 0 \].
06

Solve for \( \frac{dy}{dx} \) when \( x=1 \)

Substitute \( x = 1 \), \( y = 2 \) into the differentiated equation:\[ 2(1)(2)^2 + 2(1)^2(2) \frac{dy}{dx} + 2e^2(2 + 1\frac{dy}{dx}) = 0 \].This simplifies to:\[ 8 + 4\frac{dy}{dx} + 4e^2 + 2e^2\frac{dy}{dx} = 0 \].Reorganize to isolate \( \frac{dy}{dx} \):\[ 4\frac{dy}{dx} + 2e^2\frac{dy}{dx} = -8 - 4e^2 \].\[ \frac{dy}{dx}(4 + 2e^2) = -8 - 4e^2 \].\[ \frac{dy}{dx} = \frac{-8 - 4e^2}{4 + 2e^2} \].

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Key Concepts

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

Partial Derivatives
Partial derivatives are essential in understanding how a multivariable function changes as you vary one of the variables, holding the others constant. For a function of two variables, say \( F(x,y) \), a partial derivative with respect to \( x \) is written as \( F_x \), and it indicates how \( F \) changes as \( x \) changes while \( y \) remains fixed.

In the given problem, the function \( F(x, y) = x^2 y^2 + 2e^{xy} - 4 - 2e^2 \) can be dissected by computing its partial derivatives. This means finding \( F_x \) and \( F_y \). We consider the terms separately:
  • \( x^2 y^2 \) becomes \( 2xy^2 \) when differentiated with respect to \( x \).
  • \( 2e^{xy} \) turns into \( 2ye^{xy} \) using the chain rule since \( x \) is involved in the exponent.
Similarly, finding \( F_y \) gives insight about changes in \( F \) with respect to \( y \), providing these derivatives
  • \( 2x^2y \) as a result from \( x^2 y^2 \), and
  • \( 2xe^{xy} \) from the chain rule applied to \( 2e^{xy} \).
Partial derivatives are fundamental in calculus, showing how each variable in a multivariable function influences the result.
Implicit Differentiation
Implicit differentiation is a technique used when a function is not given in a specific form, like \( y = f(x) \). The equation from the exercise is provided implicitly, meaning it connects \( x \) and \( y \) without explicitly solving for one or the other.

To use implicit differentiation, you treat \( y \) as a function of \( x \) and differentiate both sides of the equation with respect to \( x \). It also requires you to use the chain rule when differentiating terms like \( 2e^{xy} \), where both \( x \) and \( y \) appear in the expression.
  • For \( x^2 y^2 \), differentiating gives \( 2xy^2 + 2x^2 y \frac{dy}{dx} \).
  • For \( 2e^{xy} \), it results in \( 2e^{xy}(y + x\frac{dy}{dx}) \).
By setting the differentiated equation equal to zero, you solve for \( \frac{dy}{dx} \), acknowledging how changes in \( x \) affect \( y \). Implicit differentiation allows you to find the derivative without needing to solve the equation for \( y \) first. This technique is crucial when dealing with relationships that are naturally implicit.
Local Existence of Functions
The concept of local existence deals with whether a function, such as \( y \) in terms of \( x \), is well-defined around a particular point. For this, the Implicit Function Theorem comes into play. It reassures us that under certain conditions, we can locally express one variable as a function of the other.

For the theorem to apply, certain key conditions must be met:
  • The function must be continuous and differentiable in a region around the point of interest.
  • The partial derivative \( F_y \) should not be zero at the given point. In this exercise, \( F_y(1, 2) eq 0 \) was verified, meaning that the required condition holds true.
Once these prerequisites are satisfied, the Implicit Function Theorem confirms the existence of \( y(x) \) around the point \( x = 1, \ y = 2 \). This means, locally, you can express \( y \) as a function of \( x \), solving the equation near this specific point.

Understanding the local existence is important in calculus, especially where direct solutions are complex or impossible to compute.

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Most popular questions from this chapter

Consider a rectangle with horizontal side \(x\) and vertical side \(y .\) Its area is given by the formula \(A(x, y)=x y .\) Its perimeter is \(P(x, y)=2 x+2 y\). We readily compute $$ \frac{\partial A}{\partial x}=y $$ Now let us view \(A\) as a function of \(x\) and \(P\). $$ A(x, P)=x\left(\frac{P}{2}-x\right)=\frac{P x}{2}-x^{2} $$ Here \(\frac{\partial A}{\partial x}=\frac{P}{2}-2 x\), which in terms of \(x\) and \(y\) gives $$ \frac{\partial A}{\partial x}=y-x . $$ We see that \((50)\) and \((51)\) don't agree. (a) Explain the apparent discrepancy. (b) Obtain (50) by viewing \(A\) as a function of \(x\) and \(y\) via the intermediate variable \(P\).

Let \(F\) be differentiable on \(\mathbb{R}^{2}\). Let \(x=r \cos \theta, y=r \sin \theta\) and let $$ G(r, \theta)=F(x, y)=F(r \cos \theta, r \sin \theta) . $$ Thus \(F\) is transformed into \(G\) when we transform rectangular coordinates into polar coordinates. Show that $$ \frac{\partial G}{\partial r}=\frac{\partial F}{\partial x} \cos \theta+\frac{\partial F}{\partial y} \sin \theta $$ and $$ \frac{\partial G}{\partial \theta}=-\frac{\partial F}{\partial x} r \sin \theta+\frac{\partial F}{\partial y} r \cos \theta $$

Show that the Jacobian determinant of the mapping $$ x=e^{u} \cos v, y=e^{u} \sin v $$ is never zero, yet the mapping is not one-to-one on all of \(\mathbb{R}^{2}\).

Does the pair of equations \(x=u^{2}-v^{2}, y=2 u v\), have an inverse on a neighborhood of \((0,0)\) ?

Calculate \(F^{\prime}(y)\) for each of the following functions \(F\). (a) \(F(y)=\int_{0}^{1} \mathrm{e}^{-x^{2} y^{2}} d x\) (b) \(F(y)=\int_{0}^{y} \mathrm{e}^{-x^{2} y^{2}} d x\) (c) \(F(y)=\int_{0}^{y^{3}} \mathrm{e}^{-x^{2} y^{2}} d x\)
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