/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Computing gradients Compute the ... [FREE SOLUTION] | 91Ó°ÊÓ

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Computing gradients Compute the gradient of the following functions and evaluate it at the given point \(P\). $$f(x, y)=x e^{2 \pi y} ; P(1,0)$$

Short Answer

Expert verified
Answer: The gradient at point P(1,0) is (1, 2Ï€).

Step by step solution

01

Recall the Gradient Definition

The gradient of a scalar function \(f(x,y)\) is defined as the vector formed by its partial derivatives with respect to x and y, denoted by \(\nabla f\): $$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$$
02

Compute the Partial Derivative with respect to x

Differentiate the given function \(f(x,y) = xe^{2\pi y}\) with respect to x: $$\frac{\partial f}{\partial x} = e^{2\pi y}$$
03

Compute the Partial Derivative with respect to y

Differentiate the given function \(f(x,y) = xe^{2\pi y}\) with respect to y: $$\frac{\partial f}{\partial y} = x\cdot(2\pi e^{2\pi y})$$
04

Write down the gradient

Combine the partial derivatives in Step 2 and Step 3 to obtain the gradient vector: $$\nabla f = \left(e^{2\pi y}, 2\pi x e^{2\pi y}\right)$$
05

Evaluate the gradient at point P(1,0)

Substitute the x and y values of point P(1,0) into the gradient vector: $$\nabla f(1,0) = \left(e^{2\pi \cdot 0}, 2\pi \cdot 1 \cdot e^{2\pi \cdot 0}\right) = \left(e^0, 2\pi \cdot e^0\right) = (1, 2\pi)$$ The gradient of the function \(f(x, y)=x e^{2 \pi y}\) at point P(1,0) is \((1, 2\pi)\).

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Key Concepts

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

Partial Derivatives
The concept of partial derivatives is fundamental in the field of multivariable calculus. It involves the differentiation of a function with respect to one variable, while keeping all other variables constant. Imagine you're exploring a hill's terrain, and you're interested in just how steep it is in the north direction while you are not concerned about the east or west. That's what partial derivatives help with—they measure the rate of change of a function in one direction.

In the given problem, the function of interest is a two-variable function, specifically, the scalar function \( f(x, y) = x e^{2\textbackslash pi y} \). To find the partial derivative of this function with respect to \( x \), we treat \( y \) as a constant, resulting in \( \frac{\partial f}{\partial x} = e^{2\textbackslash pi y} \). Similarly, to find the partial derivative with respect to \( y \), we treat \( x \) as a constant, thus obtaining \( \frac{\partial f}{\partial y} = x\cdot(2\textbackslash pi e^{2\textbackslash pi y}) \). These steps are crucial to calculate the gradient vector of the function, which is our next concept to explore.
Gradient Vector
The gradient vector represents the direction and rate of the fastest increase of a scalar function. In our function \( f(x, y) \), the gradient vector is determined by combining the partial derivatives calculated previously. Symbolically, it is denoted as \( abla f \), where \( abla \) is called the 'nabla' or 'del' operator.

By putting together the partial derivatives of \( f(x, y) \) with respect to both \( x \) and \( y \), we can construct the gradient vector \( abla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \). For our function, this translates to a mathematical representation \( abla f = (e^{2\textbackslash pi y}, 2\textbackslash pi x e^{2\textbackslash pi y}) \). This vector not only points in the direction of greatest increase of our function but also gives the slope of the function in that direction. When evaluated at a particular point, such as \( P(1,0) \), the gradient vector can provide concrete values which indicate how steeply the function ascends from that point.
Multivariable Calculus
Multivariable calculus extends single-variable calculus concepts like limits, differentiation, and integration to functions of several variables. It provides powerful tools for analyzing physical systems where variables interact and are dependent on one another.

While in single-variable calculus we deal with functions like \( y = f(x) \) and their single derivatives, in multivariable calculus, we study functions like \( f(x, y, z) \) and their partial derivatives, gradient vectors, and even higher multidimensional analogs. Understanding such concepts is vital in fields ranging from physics and engineering to economics, where multiple factors affect outcomes. For instance, in optimizing a surface's design or finding a minimal path, multivariable calculus is indispensable.
Scalar Function
A scalar function is a function that produces a single real number output (scalar) given one or more input values. This is in contrast to vector functions that output vectors. Scalar functions are often visualized as surfaces or contours in space.

For example, the function \( f(x, y) = x e^{2\textbackslash pi y} \) involved in our exercise is a scalar function because for each point \((x, y)\) in the function's domain, there corresponds a single real number value of \( f \). In the context of gradients, knowing a scalar function's output relative to varying inputs lets us investigate how it changes, slopes, and ultimately defines the landscape of possibilities when analyzing multivariable scenarios.

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Most popular questions from this chapter

Looking ahead- tangent planes Consider the following surfaces \(f(x, y, z)=0,\) which may be regarded as a level surface of the function \(w=f(x, y, z) .\) A point \(P(a, b, c)\) on the surface is also given. a. Find the (three-dimensional) gradient of \(f\) and evaluate it at \(P\). b. The set of all vectors orthogonal to the gradient with their tails at \(P\) form a plane. Find an equation of that plane (soon to be called the tangent plane). $$f(x, y, z)=x^{2}+y^{2}+z^{2}-3=0 ; P(1,1,1)$$

Check assumptions Consider the function \(f(x, y)=x y+x+y+100\) subject to the constraint \(x y=4\) a. Use the method of Lagrange multipliers to write a system of three equations with three variables \(x, y,\) and \(\lambda\) b. Solve the system in part (a) to verify that \((x, y)=(-2,-2)\) and \((x, y)=(2,2)\) are solutions. c. Let the curve \(C_{1}\) be the branch of the constraint curve corresponding to \(x>0 .\) Calculate \(f(2,2)\) and determine whether this value is an absolute maximum or minimum value of \(f\) over \(C_{1} \cdot(\text {Hint}: \text { Let } h_{1}(x), \text { for } x>0, \text { equal the values of } f\) over the \right. curve \(C_{1}\) and determine whether \(h_{1}\) attains an absolute maximum or minimum value at \(x=2 .\) ) d. Let the curve \(C_{2}\) be the branch of the constraint curve corresponding to \(x<0 .\) Calculate \(f(-2,-2)\) and determine whether this value is an absolute maximum or minimum value of \(f\) over \(C_{2} .\) (Hint: Let \(h_{2}(x),\) for \(x<0,\) equal the values of \(f\) over the curve \(C_{2}\) and determine whether \(h_{2}\) attains an absolute maximum or minimum value at \(x=-2 .\) ) e. Show that the method of Lagrange multipliers fails to find the absolute maximum and minimum values of \(f\) over the constraint curve \(x y=4 .\) Reconcile your explanation with the method of Lagrange multipliers.

Challenge domains Find the domain of the following functions. Specify the domain mathematically, and then describe it in words or with a sketch. $$f(x, y, z)=\ln \left(z-x^{2}-y^{2}+2 x+3\right)$$

Find the values of \(K\) and \(L\) that maximize the following production functions subject to the given constraint, assuming \(K \geq 0\) and \(L \geq 0\) $$P=f(K, L)=10 K^{1 / 3} L^{2 / 3} \text { for } 30 K+60 L=360$$

Suppose you make a one-time deposit of \(P\) dollars into a savings account that earns interest at an annual rate of \(p \%\) compounded continuously. The balance in the account after \(t\) years is \(B(P, r, t)=P e^{r^{n}},\) where \(r=p / 100\) (for example, if the annual interest rate is \(4 \%,\) then \(r=0.04\) ). Let the interest rate be fixed at \(r=0.04\) a. With a target balance of \(\$ 2000\), find the set of all points \((P, t)\) that satisfy \(B=2000 .\) This curve gives all deposits \(P\) and times \(t\) that result in a balance of \(\$ 2000\). b. Repeat part (a) with \(B=\$ 500, \$ 1000, \$ 1500,\) and \(\$ 2500,\) and draw the resulting level curves of the balance function. c. In general, on one level curve, if \(t\) increases, does \(P\) increase or decrease?

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