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Find \(f_{x}, f_{y},\) and \(f_{z}\). $$f(x, y, z)=e^{-\left(x^{2}+y^{2}+z^{2}\right)}$$

Short Answer

Expert verified
The partial derivatives are: \(f_x = -2xe^{-(x^2+y^2+z^2)}\), \(f_y = -2ye^{-(x^2+y^2+z^2)}\), \(f_z = -2ze^{-(x^2+y^2+z^2)}\).

Step by step solution

01

Identify the Function

We are given the function to find partial derivatives: \[ f(x, y, z) = e^{-(x^2 + y^2 + z^2)}. \]We need to find the partial derivatives of this function with respect to each variable: \(x\), \(y\), and \(z\).
02

Find Partial Derivative with respect to x

To find \(f_x\), take the derivative of \(f\) with respect to \(x\), treating \(y\) and \(z\) as constants.\[ f_x = \frac{\partial}{\partial x} e^{-(x^2 + y^2 + z^2)} = e^{-(x^2 + y^2 + z^2)} \cdot (-2x) = -2xe^{-(x^2 + y^2 + z^2)}. \]
03

Find Partial Derivative with respect to y

To find \(f_y\), take the derivative of \(f\) with respect to \(y\), treating \(x\) and \(z\) as constants.\[ f_y = \frac{\partial}{\partial y} e^{-(x^2 + y^2 + z^2)} = e^{-(x^2 + y^2 + z^2)} \cdot (-2y) = -2ye^{-(x^2 + y^2 + z^2)}. \]
04

Find Partial Derivative with respect to z

To find \(f_z\), take the derivative of \(f\) with respect to \(z\), treating \(x\) and \(y\) as constants.\[ f_z = \frac{\partial}{\partial z} e^{-(x^2 + y^2 + z^2)} = e^{-(x^2 + y^2 + z^2)} \cdot (-2z) = -2ze^{-(x^2 + y^2 + z^2)}. \]

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Key Concepts

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

Multivariable Calculus
Multivariable Calculus is an extension of calculus into more than one dimension. It allows us to explore how functions change when more than one variable is involved. In contrast to single-variable calculus, where functions depend on a single variable, multivariable functions can depend on two, three, or more variables. This makes it essential in fields requiring more complex mathematical modeling, like physics and engineering.

When dealing with multivariable functions, a new set of tools becomes important: partial derivatives, gradients, vectors, and more. These tools help analyze how changes in each variable influence the function's overall behavior. Understanding these concepts strengthens mathematical intuition and problem-solving skills in multivariable contexts.
  • **Partial Derivatives**: These derivatives measure how a function changes as just one of its variables changes, keeping the others constant.
  • **Gradient**: This is a vector that contains all of a function's partial derivatives, providing a direction of the steepest ascent of the function.
  • **Level Curves and Surfaces**: These are used to visualize multivariable functions by depicting sets where the function takes a constant value.
By mastering these concepts, you'll gain a deeper understanding of how multivariable equations behave and how they model the real world.
Exponential Functions
Exponential functions form the backbone of many scientific calculations and are critical in modeling growth processes. They are characterized by a constant base raised to a power, with the most common base being the natural number, denoted as 'e'. This number, approximately equal to 2.718, arises naturally in diverse mathematical contexts and has unique properties that make it especially significant.

In the context of the function given, an exponential function is combined with a negative exponent. This signifies a decrease or decay, as seen in the expression \[ e^{-(x^2 + y^2 + z^2)} \]. This formula could represent phenomena such as radioactive decay or the attenuation of sound waves.
  • The "-" sign in the exponent indicates that as the sum \( x^2 + y^2 + z^2 \) increases, the function value decreases.
  • Exponential decay is rapid and becomes smoother as the exponent grows larger, leading to the shape of the function often described as a 'bell curve' or 'Gaussian'.
Understanding how exponential functions behave, especially in a multivariable setting, provides important insights into various scientific, economic, and engineering problems.
Chain Rule
The chain rule is a fundamental technique in calculus used for finding the derivative of a composite function. When faced with functions within functions, the chain rule simplifies differentiation by breaking it down into manageable steps. In multivariable calculus, the chain rule is especially crucial when partial derivatives are involved, ensuring precision in differentiation.

The essence of the chain rule is to account for how each variable contributes to the change of the function. For example, if we consider the function \( f(x, y, z) = e^{-(x^2 + y^2 + z^2)} \), the chain rule helps determine how each constituent part of the exponent affects the exponential function by differentiating each part separately and then combining the results.
  • When applying the chain rule to partial derivatives, one must take care to only differentiate with respect to the variable of interest, keeping the other variables constant.
  • In the provided solution, the chain rule is applied by first deriving the exponent \( -(x^2 + y^2 + z^2) \), followed by deriving the outer exponential function, yielding terms like \( -2xe^{-(x^2 + y^2 + z^2)} \).
This methodical approach is crucial for tackling complex problems across domains requiring finely-tuned mathematical models.
Gradient
The gradient is a vector that captures both the direction and the rate of the steepest ascent of a multivariable function. Represented by the nabla symbol \(abla\), it consists of all the partial derivatives of a function with respect to its variables. In essence, the gradient provides a comprehensive way of understanding how a function behaves in all directions at a given point.

The importance of gradients extends across various fields, such as finding maximum or minimum values in optimization problems, or understanding force fields in physics.
  • For the function \( f(x, y, z) = e^{-(x^2 + y^2 + z^2)} \), the gradient \( abla f \) includes components \( f_x, f_y, \text{and} \ f_z \), calculated in the solution steps.
  • The vector formed by these derivatives points in the direction of the greatest rate of increase of the function. In visual terms, it's like an arrow pointing uphill on a topographic map.
  • A zero gradient indicates a flat region or an extremum point — places where a function might reach a local maximum, minimum, or saddle point.
Understanding gradients helps in exploring multivariable landscapes, ensuring you're equipped to handle diverse mathematical challenges efficiently.

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Most popular questions from this chapter

Gives a function \(f(x, y)\) and a positive number \(\epsilon\) In each exercise, show that there exists a \(\delta>0\) such that for all \((x, y)\) $$\sqrt{x^{2}+y^{2}}<\delta \Rightarrow|f(x, y)-f(0,0)|<\epsilon$$ $$f(x, y)=(x+y) /(2+\cos x), \quad \epsilon=0.02$$

Use a CAS to plot the implicitly defined level surfaces. $$x+y^{2}-3 z^{2}=1$$

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.

In economics, the usefulness or utility of amounts \(x\) and \(y\) of two capital goods \(G_{1}\) and \(G_{2}\) is sometimes measured by a function \(U(x, y) .\) For example, \(G_{1}\) and \(G_{2}\) might be two chemicals a pharmaceutical company needs to have on hand and \(U(x, y)\) the gain from manufacturing a product whose synthesis requires different amounts of the chemicals depending on the process used. If \(G_{1}\) costs \(a\) dollars per kilogram, \(G_{2}\) costs \(b\) dollars per kilogram, and the total amount allocated for the purchase of \(G_{1}\) and \(G_{2}\) together is \(c\) dollars, then the company's managers want to maximize \(U(x, y)\) given that \(a x+b y=c .\) Thus, they need to solve a typical Lagrange multiplier problem. Suppose that $$U(x, y)=x y+2 x$$ and that the equation \(a x+b y=c\) simplifies to $$2 x+y=30$$ Find the maximum value of \(U\) and the corresponding values of \(x\) and \(y\) subject to this latter constraint.

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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