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Chapter 12: Problem 14

Compute the gradient of the following functions and evaluate it at the given point \(P\). $$f(x, y)=\sin (3 x+2 y) ; P(\pi, 3 \pi / 2)$$

Short Answer

Expert verified
Question: Compute the gradient of the function \(f(x, y) = \sin(3x + 2y)\) at the point \(P(\pi, \frac{3\pi}{2})\). Answer: The gradient of the function \(f(x, y) = \sin(3x + 2y)\) at the point \(P(\pi, \frac{3\pi}{2})\) is \(\nabla f(\pi, \frac{3\pi}{2}) = \left\langle 3, 2 \right\rangle\).

Step by step solution

01

Find the Partial Derivative with respect to x

Let's compute the partial derivative of the function \(f(x, y)\) with respect to \(x\): $$\frac{\partial}{\partial x}f(x, y) = \frac{\partial}{\partial x}(\sin(3x + 2y))$$ Using the chain rule, we get: $$\frac{\partial}{\partial x}f(x, y) = \cos(3x + 2y) \cdot \frac{\partial}{\partial x}(3x + 2y)$$ Now, differentiate \((3x + 2y)\) with respect to \(x\) to obtain: $$\frac{\partial}{\partial x}f(x, y) = \cos(3x + 2y) \cdot 3$$
02

Find the Partial Derivative with respect to y

Next, we'll compute the partial derivative of the function \(f(x, y)\) with respect to \(y\): $$\frac{\partial}{\partial y}f(x, y) = \frac{\partial}{\partial y}(\sin(3x + 2y))$$ Using the chain rule again, we obtain: $$\frac{\partial}{\partial y}f(x, y) = \cos(3x + 2y) \cdot \frac{\partial}{\partial y}(3x + 2y)$$ Now, differentiate \((3x + 2y)\) with respect to \(y\) to get: $$\frac{\partial}{\partial y}f(x, y) = \cos(3x + 2y) \cdot 2$$
03

Compute the Gradient of the function f(x, y)

Now that we have the partial derivatives with respect to \(x\) and \(y\), we can compute the gradient vector of the function \(f(x, y)\): $$\nabla f(x, y) = \left\langle \frac{\partial}{\partial x}f(x, y), \frac{\partial}{\partial y}f(x, y) \right\rangle$$ Substitute the partial derivatives obtained in Steps 1 and 2: $$\nabla f(x, y) = \left\langle 3\cos(3x + 2y), 2\cos(3x + 2y) \right\rangle$$
04

Evaluate the Gradient at the given point P

Now, we will evaluate the gradient at the given point \(P(\pi, \frac{3\pi}{2})\). Substitute the values of \(x\) and \(y\) into the gradient vector: $$\nabla f(\pi, \frac{3\pi}{2}) = \left\langle 3\cos(3\pi + 3\pi), 2\cos(3\pi + 3\pi) \right\rangle$$ Simplify and compute the trigonometric functions: $$\nabla f(\pi, \frac{3\pi}{2}) = \left\langle 3\cos(6\pi), 2\cos(6\pi) \right\rangle$$ $$\nabla f(\pi, \frac{3\pi}{2}) = \left\langle 3(1), 2(1) \right\rangle$$ Finally, we have the gradient evaluated at the given point \(P\): $$\nabla f(\pi, \frac{3\pi}{2}) = \left\langle 3, 2 \right\rangle$$

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus. They represent the rate at which a function changes with respect to one of its variables while keeping the other variables constant. In our exercise, the function given is a two-variable function, which we denote as f(x, y).

To find the partial derivative of f with respect to x, we treat y as a constant and differentiate accordingly. This gives us insight into how f changes in the direction of the x-axis. Similarly, taking the partial derivative with respect to y, we can understand how f changes along the y-axis. For functions that involve trigonometric expressions, as in our exercise, it's crucial to apply the chain rule carefully to ensure the derivatives are computed correctly.
Chain Rule
The chain rule is a pivotal rule in differentiation that deals with the derivative of composite functions. In essence, if a given function f can be expressed as a composition of two functions, say u(g(x)), where g(x) is a function of x and u is a function of g, the derivative of f with respect to x is found by multiplying the derivative of u with respect to g by the derivative of g with respect to x. In the context of our exercise, when finding the partial derivatives of f(x, y) = sin(3x + 2y), the chain rule helps us determine the influence of both x and y on the sine function. By applying the chain rule, we were able to correctly find the partial derivatives, which form the components of the gradient vector.
Vector Calculus
Vector calculus expands the rules and techniques of calculus to vector fields, which are functions that assign a vector to every point in a subset of space. One of the key operations in vector calculus is finding the gradient of a scalar-valued function, like the f(x, y) in our exercise. The gradient is a vector field that points in the direction of the greatest rate of increase of the function, and its magnitude represents the rate of the increase.

The components of the gradient are the partial derivatives of the function with respect to each variable. As we calculated, the gradient of f at any point (x, y) is a vector composed of these partial derivatives, providing important directional information about f's behavior at that point.
Trigonometric Functions
Trigonometric functions鈥攕uch as sine, cosine, and tangent鈥攁re fundamental in mathematics, representing the relationship between the angles and sides of a triangle projected onto a unit circle. They are periodic and oscillate between fixed values, making them useful in describing wave patterns and circular motion.

In our exercise, we deal with the sine function, which varies between -1 and 1. Understanding the properties of trigonometric functions is crucial when taking derivatives, especially within more complex functions involving multiple variables. As seen in the solution, knowing that cos(2n蟺) = 1 for any integer n allows us to evaluate the gradient at specific points efficiently. This knowledge simplifies the process of evaluating the gradient at the given point P(蟺, 3蟺/2) in the solution.

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Most popular questions from this chapter

Suppose \(n\) houses are located at the distinct points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right) .\) A power substation must be located at a point such that the sum of the squares of the distances between the houses and the substation is minimized. a. Find the optimal location of the substation in the case that \(n=3\) and the houses are located at \((0,0),(2,0),\) and (1,1) b. Find the optimal location of the substation in the case that \(n=3\) and the houses are located at distinct points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right),\) and \(\left(x_{3}, y_{3}\right)\) c. Find the optimal location of the substation in the general case of \(n\) houses located at distinct points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots\) \(\left(x_{n}, y_{n}\right)\) d. You might argue that the locations found in parts (a), (b), and (c) are not optimal because they result from minimizing the sum of the squares of the distances, not the sum of the distances themselves. Use the locations in part (a) and write the function that gives the sum of the distances. Note that minimizing this function is much more difficult than in part (a).

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Generalize the procedure in Exercise 70 by assuming that \(n\) data points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) are given. Write the function \(E(m, b)\) (summation notation allows for a more compact calculation). Show that the coefficients of the best-fit line are $$ \begin{aligned} m &=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}} \text { and } \\ b &=\frac{1}{n}\left(\sum y_{k}-m \Sigma x_{k}\right) \end{aligned}, $$ where all sums run from \(k=1\) to \(k=n\).

(Adapted from 1938 Putnam Exam) Consider the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) and the plane \(P\) given by \(A x+B y+C z+1=0\). Let \(h=\left(A^{2}+B^{2}+C^{2}\right)^{-1 / 2}\) and \(m=\left(a^{2} A^{2}+b^{2} B^{2}+c^{2} C^{2}\right)^{1 / 2}\) a. Find the equation of the plane tangent to the ellipsoid at the point \((p, q, r)\) b. Find the two points on the ellipsoid at which the tangent plane is parallel to \(P\) and find equations of the tangent planes. c. Show that the distance between the origin and the plane \(P\) is \(h\) d. Show that the distance between the origin and the tangent planes is \(h m\) e. Find a condition that guarantees that the plane \(P\) does not intersect the ellipsoid.

a. Show that the point in the plane \(a x+b y+c z=d\) nearest the origin is \(P\left(a d / D^{2}, b d / D^{2}, c d / D^{2}\right),\) where \(D^{2}=a^{2}+b^{2}+c^{2} .\) Conclude that the least distance from the plane to the origin is \(|d| / D\). (Hint: The least distance is along a normal to the plane.) b. Show that the least distance from the point \(P_{0}\left(x_{0}, y_{0}, z_{0}\right)\) to the plane \(a x+b y+c z=d\) is \(\left|a x_{0}+b y_{0}+c z_{0}-d\right| / D\) (Hint: Find the point \(P\) on the plane closest to \(P_{0}\).)

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=e^{a x} \cos a y, \text { for any real number } a$$
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