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Chapter 12: Problem 21

If \(F(x, y)=\frac{2 x-y}{x y}\), find \(F_{x}(3,-2)\) and \(F_{y}(3,-2) .\)

Short Answer

Expert verified
\( F_x(3, -2) = -\frac{1}{18} \) and \( F_y(3, -2) = \frac{1}{9} \)

Step by step solution

01

Understand the Problem

We need to find the partial derivatives of the function \( F(x, y) = \frac{2x - y}{xy} \). Specifically, we want to find \( F_{x}(3, -2) \) and \( F_{y}(3, -2) \). This means we must first compute the partial derivatives \( F_x(x, y) \) and \( F_y(x, y) \) and then evaluate them at the given point \((3, -2)\).
02

Compute the Partial Derivative with Respect to x

Using the quotient rule for partial differentiation with respect to \( x \), we have the numerator \( u = 2x - y \) and the denominator \( v = xy \). The derivative \( \frac{\partial}{\partial x} \frac{u}{v} \) is found using:\[ F_x = \frac{v \frac{\partial u}{\partial x} - u \frac{\partial v}{\partial x}}{v^2} \]where:- \( \frac{\partial u}{\partial x} = 2 \)- \( \frac{\partial v}{\partial x} = y \)Substitute into the formula:\[ F_x = \frac{xy \cdot 2 - (2x - y) \cdot y}{(xy)^2} = \frac{2xy - (2xy - y^2)}{x^2 y^2} = \frac{y^2}{x^2 y^2} \]Simplifying gives:\[ F_x = \frac{1}{x^2 y} \]
03

Evaluate F_x at (3, -2)

Substitute \( x = 3 \) and \( y = -2 \) into \( F_x = \frac{1}{x^2 y} \),\[ F_x(3, -2) = \frac{1}{3^2 \cdot (-2)} = \frac{1}{9 \cdot (-2)} = \frac{1}{-18} = -\frac{1}{18} \]
04

Compute the Partial Derivative with Respect to y

Using the quotient rule for partial differentiation with respect to \( y \), we have the same expressions for \( u \) and \( v \) as before. The derivative \( \frac{\partial}{\partial y} \frac{u}{v} \) is:\[ F_y = \frac{v \frac{\partial u}{\partial y} - u \frac{\partial v}{\partial y}}{v^2} \]where:- \( \frac{\partial u}{\partial y} = -1 \)- \( \frac{\partial v}{\partial y} = x \)Substitute into the formula:\[ F_y = \frac{xy \cdot (-1) - (2x - y) \cdot x}{(xy)^2} = \frac{-xy - (2x^2 - xy)}{x^2 y^2} = \frac{-2x^2}{x^2 y^2} \]Simplifying gives:\[ F_y = \frac{-2}{y \cdot x^2} \]
05

Evaluate F_y at (3, -2)

Substitute \( x = 3 \) and \( y = -2 \) into \( F_y = \frac{-2}{y \cdot x^2} \),\[ F_y(3, -2) = \frac{-2}{(-2) \cdot 3^2} = \frac{-2}{-2 \cdot 9} = \frac{-2}{-18} = \frac{1}{9} \]

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Key Concepts

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

Quotient Rule
The quotient rule is a fundamental concept in calculus, particularly useful in differentiation. This rule helps us find the derivative of a quotient of two functions. If we have a function expressed as a ratio, such as \( \frac{u}{v} \), the quotient rule becomes handy.
  • Let \( u \) be the numerator and \( v \) be the denominator.
  • The derivative of the quotient \( \frac{u}{v} \) is given by:
\[ \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]
In this formula:
  • \( \frac{du}{dx} \) is the derivative of \( u \).
  • \( \frac{dv}{dx} \) is the derivative of \( v \).
  • \( v \) and \( v^2 \) are the original and squared denominator respectively.
With partial differentiation, the procedure is similar, but we differentiate with respect to one variable, keeping others constant.
Partial Differentiation
Partial differentiation extends the idea of differentiation to functions of multiple variables. In essence, it allows us to see the effect of changing one variable while keeping others constant. For a function \( F(x, y) \), partial derivatives are denoted as \( F_x \) and \( F_y \), representing the partial derivatives with respect to the variables \( x \) and \( y \) respectively.
  • With \( F(x, y) = \frac{2x - y}{xy} \), we apply the quotient rule to find these derivatives.
  • Find \( F_x \) by treating \( y \) as constant and differentiating with respect to \( x \).
  • Find \( F_y \) by treating \( x \) as constant and differentiating with respect to \( y \).
Each partial derivative shows how the function changes as just one variable changes, proving to be a powerful tool in multivariable calculus.
Function Evaluation
Once the partial derivatives are computed, the next step is to evaluate the derivatives at a specific point. This process is called function evaluation, where we substitute given values for the variables into the function or its derivatives.
  • Given \( F(x, y) = \frac{2x - y}{xy} \), we found:
    • \( F_x(x, y) = \frac{1}{x^2 y} \)
    • \( F_y(x, y) = \frac{-2}{y \cdot x^2} \)
  • Now, evaluate \( F_x(3, -2) \) by substituting \( x = 3 \) and \( y = -2 \):
    • \( F_x(3, -2) = -\frac{1}{18} \)
  • Similarly, for \( F_y(3, -2) \):
    • \( F_y(3, -2) = \frac{1}{9} \)
Function evaluation is critical as it gives specific insights into the function's behavior at particular points in its domain.

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Most popular questions from this chapter

The period \(T\) of a pendulum of length \(L\) is given by \(T=2 \pi \sqrt{L / g}\), where \(g\) is the acceleration of gravity. Show that \(d T / T=\frac{1}{2}[d L / L-d g / g]\), and use this result to estimate the maximum percentage error in \(T\) due to an error of \(0.5 \%\) in measuring \(L\) and \(0.3 \%\) in measuring \(g\).

The elevation of a mountain above sea level at the point \((x, y)\) is \(f(x, y) .\) A mountain climber at \(\mathbf{p}\) notes that the slope in the easterly direction is \(-\frac{1}{2}\) and the slope in the northerly direction is \(-\frac{1}{4}\). In what direction should he move for fastest descent?

For the function \(f(x, y)=\tan \left(\left(x^{2}+y^{2}\right) / 64\right)\), find the second-order Taylor approximation based at \(\left(x_{0}, y_{0}\right)=(0,0)\). Then estimate \(f(0.2,-0.3)\) using (a) the first-order approximation, (b) the second-order approximation, and (c) your calculator directly.

Find a unit vector in the direction in which \(f\) increases most rapidly at \(\mathbf{p} .\) What is the rate of change in this direction? \(f(x, y, z)=x e^{y z} ; \mathbf{p}=(2,0,-4)\)

If the temperature of a plate at the point \((x, y)\) is \(T(x, y)=10+x^{2}-y^{2}\), find the path a heat-seeking particle (which always moves in the direction of greatest increase in temperature) would follow if it starts at \((-2,1) .\) Hint: The particle moves in the direction of the gradient $$ \nabla T=2 x \mathbf{i}-2 y \mathbf{j} $$ We may write the path in parametric form as $$ \mathbf{r}(t)=x(t) \mathbf{i}+y(t) \mathbf{j} $$ and we want \(x(0)=-2\) and \(y(0)=1 .\) To move in the required direction means that \(\mathbf{r}^{\prime}(t)\) should be parallel to \(\nabla T\). This will be satisfied if $$ \frac{x^{\prime}(t)}{2 x(t)}=-\frac{y^{\prime}(t)}{2 y(t)} $$ together with the conditions \(x(0)=-2\) and \(y(0)=1\). Now solve this differential equation and evaluate the arbitrary constant of integration.
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