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Use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$f(x, y)=4+2 x-3 y-x y^{2}, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(-2,1)$$

Short Answer

Expert verified
Both partial derivatives are 1 at the point (-2,1).

Step by step solution

01

Limit Definition Review

The limit definition of a partial derivative with respect to a variable gives the rate of change of a function with respect to that variable. For a function \( f(x, y) \), the partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h} \) and for \( y \) it is \( \frac{\partial f}{\partial y} = \lim_{k \to 0} \frac{f(x, y+k) - f(x, y)}{k} \).
02

Substitute for \( \frac{\partial f}{\partial x} \)

Substitute \( f(x+h, y) = 4 + 2(x+h) - 3y - (x+h)y^2 \) into the definition: \[ \frac{f(x+h, y) - f(x, y)}{h} = \frac{[4 + 2(x+h) - 3y - (x+h)y^2] - [4 + 2x - 3y - xy^2]}{h}. \]
03

Simplify \( \frac{\partial f}{\partial x} \)

Simplify the expression: \[ \frac{[2x + 2h - 3y - xy^2 - hy^2] - [2x - 3y - xy^2]}{h} = \frac{2h - hy^2}{h} = 2 - y^2. \]
04

Substitute for \( \frac{\partial f}{\partial y} \)

Substitute \( f(x, y+k) = 4 + 2x - 3(y+k) - x(y+k)^2 \) into the definition: \[ \frac{f(x, y+k) - f(x, y)}{k} = \frac{[4 + 2x - 3(y+k) - x(y+k)^2] - [4 + 2x - 3y - xy^2]}{k}. \]
05

Simplify \( \frac{\partial f}{\partial y} \)

Simplify the expression: \[ \frac{-3k - x[(y^2 + 2yk + k^2) - y^2]}{k} \] which becomes:\[ \frac{-3k - x(2yk + k^2)}{k} = -3 - 2xy. \]
06

Evaluate \( \frac{\partial f}{\partial x} \) at \((-2,1)\)

Substitute \( x = -2 \) and \( y = 1 \) into \( \frac{\partial f}{\partial x} = 2 - y^2 \): \[ 2 - 1 = 1. \]
07

Evaluate \( \frac{\partial f}{\partial y} \) at \((-2,1)\)

Substitute \( x = -2 \) and \( y = 1 \) into \( \frac{\partial f}{\partial y} = -3 - 2xy \): \[ -3 - 2(-2)(1) = -3 + 4 = 1. \]

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Key Concepts

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

Limit Definition
To understand partial derivatives, we first need to grasp the limit definition of a derivative, which provides a way to calculate how a function changes when its input variables change slightly. In the context of partial derivatives, we focus on one variable at a time while holding the others constant. This allows us to see the rate of change for one specific direction.

For a function like \( f(x, y) \), the partial derivative with respect to \( x \), written as \( \frac{\partial f}{\partial x} \), is defined using limits. It is expressed as: \[ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h} \]. This formula gives us the change in the function as \( x \) increases by a small amount \( h \) while \( y \) remains unchanged.

Similarly, the partial derivative with respect to \( y \) is: \[ \frac{\partial f}{\partial y} = \lim_{k \to 0} \frac{f(x, y+k) - f(x, y)}{k} \]. This measures how the function responds to changes in \( y \) with \( x \) being fixed. These limit processes ensure that the derivative accurately reflects the instantaneous rate of change at any point on the function.
Functions of Several Variables
Many mathematical functions we encounter involve more than one variable, like \( f(x, y) = 4 + 2x - 3y - xy^{2} \). These are called functions of several variables because they depend on two or more inputs to yield an output. In such cases, the function’s behavior is determined by the interaction between its input variables.

Partial derivatives become crucial in analyzing such functions. Each variable’s partial derivative tells us how the function changes in response to altering that specific variable. This can provide a comprehensive picture of the function's nature and how sensitive it is in different directions.
  • For \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} = 2 - y^2 \) shows how changing \( x \) affects \( f \) when \( y \) is kept constant.
  • The partial derivative \( \frac{\partial f}{\partial y} = -3 - 2xy \) reveals the effect of varying \( y \) while \( x \) remains unchanged.
In applications, these insights are invaluable in fields such as economics for optimizing resources, or physics for understanding forces acting along different axes.
Rate of Change
Partial derivatives are fundamentally about rates of change in functions involving multiple variables. Imagine you are hiking up a mountain; the steepness in different directions gives you different paths to the top. Similarly, partial derivatives indicate how steep or flat the function changes along different variables.

The concept of rate of change is central when analyzing partial derivatives. It’s akin to knowing how fast a car accelerates. Partial derivatives, like \( \frac{\partial f}{\partial x} = 2 - y^2 \), tell us the speed at which \( f \) alters as \( x \) changes, giving the 'acceleration' along the \( x \)-axis.

For \( \frac{\partial f}{\partial y} = -3 - 2xy \), it describes the 'steepness' the function experiences with variations in \( y \). This information is crucial not just for navigation, but in prediction and optimization models:
  • In meteorology, predicting how climatic variables will alter weather patterns depends on these rates.
  • In manufacturing, understanding how performance metrics change with inputs assists in improving efficiency.
By breaking down changes in any multidimensional scenario, partial derivatives provide a "compass" that points towards understanding complex variable relationships.

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Most popular questions from this chapter

Find three numbers whose sum is 9 and whose sum of squares is a minimum.

You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reasons for your answer.

Extrema on a circle of intersection \(\quad\) Find the extreme values of the function \(f(x, y, z)=x y+z^{2}\) on the circle in which the plane \(y-x=0\) intersects the sphere \(x^{2}+y^{2}+z^{2}=4.\)

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema:In Exercises \(49-54,\) use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) b. Determine all the first partial derivatives of \(h,\) including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to 0 c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z, w)=x^{2}+y^{2}+z^{2}+w^{2}\) subject to the constraints \(\quad 2 x-y+z-w-1=0 \quad\) and \(\quad x+y-z+\) \(w-1=0.\)

If you cannot make any headway with \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) in rectangular coordinates, try changing to polar coordinates. Substitute \(x=r \cos \theta, y=r \sin \theta,\) and investigate the limit of the resulting expression as \(r \rightarrow 0 .\) In other words, try to decide whether there exists a number \(L\) satisfying the following criterion: $$|r|<\delta \Rightarrow|f(r, \theta)-L|<\epsilon$$ If such an \(L\) exists, then $$\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{r \rightarrow 0} f(r \cos \theta, r \sin \theta)=L$$ For instance, $$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}}{x^{2}+y^{2}}=\lim _{r \rightarrow 0} \frac{r^{3} \cos ^{3} \theta}{r^{2}}=\lim _{r \rightarrow 0} r \cos ^{3} \theta=0$$ To verify the last of these equalities, we need to show that Equation (1) is satisfied with \(f(r, \theta)=r \cos ^{3} \theta\) and \(L=0 .\) That is, we need to show that given any \(\epsilon>0,\) there exists a \(\delta>0\) such that for all \(r\) and \(\theta\) $$|r|<\delta \Rightarrow\left|r \cos ^{3} \theta-0\right|<\epsilon$$ since $$\left|r \cos ^{3} \theta\right|=|r|\left|\cos ^{3} \theta\right| \leq|r| \cdot 1=|r|$$ the implication holds for all \(r\) and \(\theta\) if we take \(\delta=\epsilon\) In contrast, $$\frac{x^{2}}{x^{2}+y^{2}}=\frac{r^{2} \cos ^{2} \theta}{r^{2}}=\cos ^{2} \theta$$takes on all values from 0 to 1 regardless of how small \(|r|\) is, so that \(\lim _{(x, y) \rightarrow(0,0)} x^{2} /\left(x^{2}+y^{2}\right)\) does not exist. In each of these instances, the existence or nonexistence of the limit as \(r \rightarrow 0\) is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) \(\theta=\) constant and yet fail to exist in the broader sense. Example 5 illustrates this point. In polar coordinates, \(f(x, y)=\left(2 x^{2} y\right) /\left(x^{4}+y^{2}\right)\) becomes $$f(r \cos \theta, r \sin \theta)=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\sin ^{2} \theta}$$for \(r \neq 0 .\) If we hold \(\theta\) constant and let \(r \rightarrow 0,\) the limit is \(0 .\) On the path \(y=x^{2},\) however, we have \(r \sin \theta=r^{2} \cos ^{2} \theta\) and $$\begin{aligned} f(r \cos \theta, r \sin \theta) &=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\left(r \cos ^{2} \theta\right)^{2}} \\ &=\frac{2 r \cos ^{2} \theta \sin \theta}{2 r^{2} \cos ^{4} \theta}=\frac{r \sin \theta}{r^{2} \cos ^{2} \theta}=1 \end{aligned}$$ Find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. $$f(x, y)=\frac{2 x}{x^{2}+x+y^{2}}$$

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