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Chapter 13: Problem 57

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

Short Answer

Expert verified
The partial derivatives at (1,2) are -13 for \( \frac{\partial f}{\partial x} \) and -2 for \( \frac{\partial f}{\partial y} \).

Step by step solution

01

Identify the Function and Point

We need to find the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) of the function \( f(x, y) = 1 - x + y - 3x^2y \) at the point \( (1, 2) \).
02

Limit Definition for Partial Derivative with Respect to x

Using the limit definition, the partial derivative of \( f \) with respect to \( x \) is:\[ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h,y) - f(x,y)}{h}. \] Substitute function \( f \) into this definition.
03

Substitute the Function into the Limit (x-direction)

Replace \( x \) with \( x+h \) in the function: \[ f(x+h, y) = 1 - (x+h) + y - 3(x+h)^2y. \] So, \[ f(x+h, y) = 1 - x - h + y - 3(x^2 + 2xh + h^2)y. \] Simplify our expression: \[ 1 - x - h + y - 3x^2y - 6xhy - 3h^2y. \]
04

Compute the Difference and Simplify (x-direction)

Now, use this result in the limit equation: \[ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{[1 - x - h + y - 3x^2y - 6xhy - 3h^2y] - [1 - x + y - 3x^2y]}{h}. \] This simplifies to: \[ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{-h - 6xhy - 3h^2y}{h}. \] Further simplification gives: \[ \frac{\partial f}{\partial x} = \lim_{h \to 0} (-1 - 6xy - 3hy). \] As \( h \to 0 \), this becomes \( -1 - 6xy \).
05

Evaluate Partial Derivative at (1,2) (x-direction)

Substitute \( x = 1 \) and \( y = 2 \) into the simplified expression: \(-1 - 6xy\). Thus, \(-1 - 6(1)(2) = -13\).
06

Limit Definition for Partial Derivative with Respect to y

Similarly, use the limit definition for \( y \): \[ \frac{\partial f}{\partial y} = \lim_{h \to 0} \frac{f(x,y+h) - f(x,y)}{h}. \] Substitute the function into this definition.
07

Substitute the Function into the Limit (y-direction)

Replace \( y \) with \( y+h \) in the function: \[ f(x, y+h) = 1 - x + (y+h) - 3x^2(y+h). \] This expands to: \[ 1 - x + y + h - 3x^2y - 3x^2h. \]
08

Compute the Difference and Simplify (y-direction)

Use this result in the limit equation: \[ \frac{\partial f}{\partial y} = \lim_{h \to 0} \frac{[1 - x + y + h - 3x^2y - 3x^2h] - [1 - x + y - 3x^2y]}{h}. \] This simplifies to: \[ \frac{\partial f}{\partial y} = \lim_{h \to 0} \frac{h - 3x^2h}{h}. \] Further simplification gives: \[ \frac{\partial f}{\partial y} = \lim_{h \to 0} (1 - 3x^2). \] As \( h \to 0 \), this becomes \( 1 - 3x^2 \).
09

Evaluate Partial Derivative at (1,2) (y-direction)

Substitute \( x = 1 \) into the simplified expression: \(1 - 3x^2\). So: \(1 - 3(1)^2 = -2\).

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Key Concepts

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

limit definition
The concept of a limit is crucial in calculus and helps us understand how a function behaves as its inputs approach a certain value. The limit definition forms the backbone of deriving derivatives, including partial derivatives. In essence, a limit evaluates the behavior of a function as the change in its input approaches zero.

For partial derivatives, we take the limit as we vary one variable at a time while keeping others constant. Partially differentiating a function like \( f(x,y) \) with respect to \( x \) involves finding \[ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h,y) - f(x,y)}{h}. \] This expression tells us how the function \( f \) changes as we make a small change \( h \) in \( x \).
  • The variable you're differentiating with respect to changes slightly (\( h \) moves to 0).
  • The other variables remain constant, showing the function's sensitivity to just one variable at a time.
Understanding the limit is about seeing the instantaneous rate of change, which, when carried out for all variables individually, deepens our understanding of multivariable functions.
calculus
Calculus often stands as the foundation of mathematics, allowing us to study changes and motion. It's divided into two main branches: differential calculus and integral calculus.

Differential calculus focuses on the concept of a derivative, which represents the rate of change of a function. When applied to functions of multiple variables, we get partial derivatives. These show how a function changes with respect to each variable independently, providing insight into how joint variable changes affect the function as a whole.
  • A derivative provides a precise measurement of how a function responds to a modification in its input.
  • Partial derivatives disassemble the impact into parts that allow detailed exploration.
While integral calculus concerns the accumulation of quantities, it is differential calculus and particularly derivatives that enable predictions and problem-solving in physics, engineering, and beyond.
partial differentiation
Partial differentiation expresses how a multivariable function changes as one variable changes, holding others constant. For the function \( f(x,y) \), this means examining how \( f \) changes as either \( x \) or \( y \) changes independently.

The introduction of partial derivatives transforms our understanding of functions from static to dynamic, showing the effects each variable independently contributes. Partial differentiation is denoted by \( \frac{\partial f}{\partial x} \) or \( \frac{\partial f}{\partial y} \), representing the function's derivative in terms of \( x \) or \( y \), respectively.
  • Compute \( \frac{\partial f}{\partial x} \) by considering changes only in \( x \), fixing \( y \).
  • Compute \( \frac{\partial f}{\partial y} \) by varying only \( y \), maintaining \( x \).
Partial differentiation is a key tool in evaluating systems where multiple factors change yet you need to isolate the effect of one, explaining phenomena in environments like thermodynamics and economics.

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Most popular questions from this chapter

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. $$f(x, y)=\tan ^{-1}\left(\frac{|x|+|y|}{x^{2}+y^{2}}\right)$$

Can you conclude anything about \(f(a, b)\) if \(f\) and its first and second partial derivatives are continuous throughout a disk centered at the critical point \((a, b)\) and \(f_{x x}(a, b)\) and \(f_{y y}(a, b)\) differ in sign? Give reasons for your answer.

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. $$f(x, y)=\frac{x^{3}-x y^{2}}{x^{2}+y^{2}}$$

Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I\), you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$\begin{array}{l} x=2 \cos u \cos v, \quad y=2 \cos u \sin v, \quad z=2 \sin u \\ 0 \leq u \leq 2 \pi, \quad 0 \leq v \leq \pi \end{array}$$

If a function \(f(x, y)\) has continuous second partial derivatives throughout an open region \(R,\) must the first-order partial derivatives of \(f\) be continuous on \(R ?\) Give reasons for your answer.
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