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Chapter 12: Problem 15

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

Short Answer

Expert verified
Answer: The gradient of the function evaluated at point \(P(-1,2)\) is \(\nabla F = \begin{bmatrix} 2 e^{-17} \\ -8 e^{-17} \end{bmatrix}\).

Step by step solution

01

Compute Partial Derivative with respect to x

To compute the partial derivative of the function with respect to x (\(\frac{\partial F}{\partial x}\)), we differentiate with respect to x keeping y constant: $$\frac{\partial F}{\partial x} = \frac{d}{dx}\left(e^{-x^{2}- 2y^{2}}\right)$$ Using the chain rule, this becomes: $$\frac{\partial F}{\partial x} = \left(-2x\right) e^{-x^{2}-2 y^{2}}$$
02

Compute Partial Derivative with respect to y

To compute the partial derivative of the function with respect to y (\(\frac{\partial F}{\partial y}\)), we differentiate with respect to y keeping x constant: $$\frac{\partial F}{\partial y} = \frac{d}{dy}\left(e^{-x^{2}- 2y^{2}}\right)$$ Using the chain rule, this becomes: $$\frac{\partial F}{\partial y} = \left(-4y\right) e^{-x^{2}-2 y^{2}}$$
03

Form the Gradient Vector

Once we have both partial derivatives, we can form the gradient vector by putting the partial derivatives with respect to x and y into the vector components: $$\nabla F = \begin{bmatrix} -2x e^{-x^{2}-2y^{2}} \\ -4y e^{-x^{2}-2y^{2}} \end{bmatrix}$$
04

Evaluate Gradient Vector at Point P(-1, 2)

Now we evaluate the gradient vector at the given point \(P (-1, 2)\): $$\nabla F = \begin{bmatrix} -2(-1) e^{-(-1)^{2}-2(2)^{2}} \\ -4(2) e^{-(-1)^{2}-2(2)^{2}} \end{bmatrix} = \begin{bmatrix} 2 e^{-17} \\ -8 e^{-17} \end{bmatrix}$$ Therefore, the gradient of the function \(F(x, y)=e^{-x^{2}-2 y^{2}}\) evaluated at point \(P(-1,2)\) is: $$\nabla F = \begin{bmatrix} 2 e^{-17} \\ -8 e^{-17} \end{bmatrix}$$

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Key Concepts

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

Partial Derivative
The concept of a partial derivative is fundamentally about observing how a function changes if only one variable changes, while others are kept constant. It's like focusing a magnifying glass on only one dimension of a multi-dimensional surface. In mathematical terms, for a function like \(F(x, y)=e^{-x^{2}-2 y^{2}}\), we find the partial derivative with respect to \(x\) by treating \(y\) as a constant. Similarly, we find the partial derivative with respect to \(y\) by treating \(x\) as a constant.

To compute the partial derivative with respect to \(x\), we apply differentiation:
  • Differentiate \(e^{-x^{2}-2 y^{2}}\) with respect to \(x\), yielding \(-2x e^{-x^{2}-2 y^{2}}\).
The negative sign and the term \(-2x\) stem from differentiating the exponent \(-x^{2}\). The same process applies to \(y\), resulting in the term \(-4y e^{-x^{2}-2 y^{2}}\).Understanding partial derivatives is crucial, as they form the foundation of gradients, which describe the direction and rate of steepest ascent of a function.
Chain Rule
The chain rule is a pivotal tool in calculus that allows us to differentiate composite functions. It's especially useful for functions where one component is nested inside another. In the context of partial derivatives, the chain rule helps us differentiate expressions that have more complex inner functions.

Consider the function \(F(x, y)=e^{-x^{2}-2 y^{2}}\). When computing the partial derivative with respect to \(x\), we use the chain rule to manage the inner function \(-x^{2}\).
Here's what happens:
  • The derivative of \(-x^{2}\) with respect to \(x\) is \(-2x\).
  • The outer function, \(e^{u}\) where \(u = -x^{2}-2 y^{2}\), differentiates to itself.
  • Combining these steps with the chain rule yields \(-2x e^{-x^{2}-2 y^{2}}\).
This systematic approach not only applies to \(x\) but also to \(y\), simplifying the handling of more intricate functions.
Gradient Vector
The gradient vector is a key concept in multivariable calculus, representing both the direction and magnitude of the greatest rate of increase of a function. It is composed of all the partial derivatives of a function, effectively summarizing how the function changes in all dimensions.

For the function \(F(x, y)=e^{-x^{2}-2 y^{2}}\), once you calculate the partial derivatives with respect to both \(x\) and \(y\), combining these gives the gradient vector:
  • The \(x\)-component is \(-2x e^{-x^{2}-2 y^{2}}\).
  • The \(y\)-component is \(-4y e^{-x^{2}-2 y^{2}}\).
Thus, the gradient vector \(abla F\) is \(\begin{bmatrix} -2x e^{-x^{2}-2y^{2}} \ -4y e^{-x^{2}-2y^{2}} \end{bmatrix}\).

When evaluated at a point like \((-1, 2)\), substitute in the coordinates to find specific directional values, giving you precise information about the function's behavior at that location.

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Most popular questions from this chapter

Given positive numbers \(x_{1}, \ldots, x_{n},\) prove that the geometric mean \(\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n}\) is no greater than the arithmetic mean \(\left(x_{1}+\cdots+x_{n}\right) / n\) in the following cases. a. Find the maximum value of \(x y z,\) subject to \(x+y+z=k\) where \(k\) is a real number and \(x>0, y>0\), and \(z>0 .\) Use the result to prove that $$(x y z)^{1 / 3} \leq \frac{x+y+z}{3}.$$ b. Generalize part (a) and show that $$\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n} \leq \frac{x_{1}+\cdots+x_{n}}{n}.$$

Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}-4 y^{2}+x y ; R=\left\\{(x, y): 4 x^{2}+9 y^{2} \leq 36\right\\}$$

Given a differentiable function \(w=f(x, y, z),\) the goal is to find its maximum and minimum values subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0\) and \(h(x, y, z)=0\).

Let \(x, y,\) and \(z\) be non-negative numbers with \(x+y+z=200\) a. Find the values of \(x, y,\) and \(z\) that minimize \(x^{2}+y^{2}+z^{2}\) b. Find the values of \(x, y,\) and \(z\) that minimize \(\sqrt{x^{2}+y^{2}+z^{2}}\). c. Find the values of \(x, y,\) and \(z\) that maximize \(x y z\) d. Find the values of \(x, y,\) and \(z\) that maximize \(x^{2} y^{2} z^{2}\).

Traveling waves (for example, water waves or electromagnetic waves) exhibit periodic motion in both time and position. In one dimension, some types of wave motion are governed by the one-dimensional wave equation $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u(x, t)\) is the height or displacement of the wave surface at position \(x\) and time \(t,\) and \(c\) is the constant speed of the wave. Show that the following functions are solutions of the wave equation. $$u(x, t)=\cos (2(x+c t))$$
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