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Solve the following relations for \(x\) and \(y,\) and compute the Jacobian \(J(u, v)\) $$u=2 x-3 y, v=y-x$$

Short Answer

Expert verified
Question: Compute the Jacobian \(J(u,v)\) for the given relations \(u=2x-3y\) and \(v=y-x\). Answer: The Jacobian \(J(u,v)\) is \(-\frac{1}{5}\).

Step by step solution

01

Solve the given relations for x and y

We are given the relations $$u=2 x-3 y, \qquad v=y-x$$ First, rearrange and solve the second equation for \(x\): $$x = y - v$$ Now substitute this expression for \(x\) into the first equation: $$u = 2(y - v) - 3y$$ Now solve for \(y\): $$y = \frac{2v + u}{5}$$ And finally, by substituting this expression for \(y\) back into the expression for \(x\), we get: $$x = \frac{3v-u}{5}$$ Now we have the expressions for \(x\) and \(y\) in terms of \(u\) and \(v\): $$x = \frac{3v-u}{5}, \qquad y = \frac{2v + u}{5}$$
02

Find the partial derivatives

To compute the Jacobian, we need to find the following partial derivatives: $$\frac{\partial x}{\partial u}, \quad \frac{\partial x}{\partial v}, \quad \frac{\partial y}{\partial u}, \quad \frac{\partial y}{\partial v}$$ Using the expressions for \(x\) and \(y\) in terms of \(u\) and \(v\), compute the partial derivatives: $$\frac{\partial x}{\partial u} = -\frac{1}{5}$$ $$\frac{\partial x}{\partial v} = \frac{3}{5}$$ $$\frac{\partial y}{\partial u} = \frac{1}{5}$$ $$\frac{\partial y}{\partial v} = \frac{2}{5}$$
03

Compute the Jacobian

Now, we can compute the Jacobian \(J(u,v)\) using the partial derivatives found above. The Jacobian is the determinant of the matrix of partial derivatives: $$J(u, v) = \det \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}$$ Substitute in the values of the partial derivatives: $$J(u, v) = \det \begin{bmatrix} -\frac{1}{5} & \frac{3}{5} \\ \frac{1}{5} & \frac{2}{5} \end{bmatrix}$$ Compute the determinant: $$J(u, v) = \left(-\frac{1}{5}\right)\left(\frac{2}{5}\right) - \left(\frac{3}{5}\right)\left(\frac{1}{5}\right)$$ Simplify: $$J(u, v) = -\frac{2}{25} - \frac{3}{25} = -\frac{5}{25}$$ Therefore, the Jacobian \(J(u,v)\) is: $$J(u, v) = -\frac{1}{5}$$

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Key Concepts

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

Partial Derivatives
When we progress from single-variable calculus to multivariable calculus, we encounter functions with more than one input. For example, a function might depend on both x and y, like z = f(x, y). In order to understand how this function changes with respect to each input variable, we compute partial derivatives. These derivatives are computed with respect to one variable at a time, while treating all other variables as constants.

To find a partial derivative of a function with respect to a particular variable, we use the same differentiation rules we are familiar with from single-variable calculus. For instance, if we have a function f(x, y), the partial derivative of f with respect to x is denoted as \( \frac{\partial f}{\partial x} \) and is calculated by differentiating f with respect to x while holding y constant. Similarly, the partial derivative of f with respect to y is denoted as \( \frac{\partial f}{\partial y} \) and focuses on the change in f as y varies, with x held constant.

Understanding partial derivatives is crucial when analyzing the behavior of multivariable functions. It tells us how a small change in one input can affect the output, assuming we do not change the other inputs. This concept is integral to solving practical problems in physics, engineering, economics, and many other fields.
Multivariable Calculus
Multivariable calculus extends the principles and methods of calculus to functions of several variables. Unlike single-variable calculus, where we deal with functions that depend on one variable, in multivariable calculus, we explore functions that have two or more variables, such as f(x, y) or g(x, y, z).

The beauty of multivariable calculus lies in its ability to model and solve problems in higher dimensions. Imagine trying to predict the temperature at a particular location, which could depend on both the elevation (x) and time of year (y); a multivariable function would allow us to express this relationship. Key topics in this field include partial derivatives, multiple integrals, line and surface integrals, and vector analysis.

One of the central applications of multivariable calculus is optimization—finding the maximum or minimum values of functions with several variables. This involves concepts like gradients, which generalize the derivative to higher dimensions, and techniques like Lagrange multipliers for constrained optimization problems. The discipline not only broadens our comprehension of mathematical landscapes but also equips us with tools to navigate the complexities of the physical world.
Determinants
In linear algebra, the determinant is a scalar value associated with a square matrix. It is denoted as 'det(M)' or |M|, where 'M' is a matrix. The determinant provides important insights into the matrix it is associated with, such as whether the matrix is invertible, and it is used to solve systems of linear equations.

The determinant can be thought of as a measure of how a linear transformation associated with a matrix scales the area or volume when applied to a geometric region. If a 2×2 matrix, for example, transforms a unit square into a parallelogram, the absolute value of the determinant gives the area of that parallelogram. For higher-dimensional matrices, the determinant gives the signed volume of the transformed region.

A key fact about determinants is that a matrix is invertible if and only if its determinant is non-zero. In the context of multivariable calculus, the determinant plays an essential role in finding the Jacobian, which is a determinant consisting of partial derivatives. The Jacobian determinant helps us to understand how volume changes under a transformation represented by a function of several variables. A non-zero Jacobian implies that the function is locally invertible near that point.

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Most popular questions from this chapter

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Meaning of the Jacobian The Jacobian is a magnification (or reduction) factor that relates the area of a small region near the point \((u, v)\) to the area of the image of that region near the point \((x, y)\) a. Suppose \(S\) is a rectangle in the \(u v\) -plane with vertices \(O(0,0)\) \(P(\Delta u, 0),(\Delta u, \Delta v),\) and \(Q(0, \Delta v)\) (see figure). The image of \(S\) under the transformation \(x=g(u, v), y=h(u, v)\) is a region \(R\) in the \(x y\) -plane. Let \(O^{\prime}, P^{\prime},\) and \(Q^{\prime}\) be the images of O, \(P,\) and \(Q,\) respectively, in the \(x y\) -plane, where \(O^{\prime}, P^{\prime},\) and \(Q^{\prime}\) do not all lie on the same line. Explain why the coordinates of \(\boldsymbol{O}^{\prime}, \boldsymbol{P}^{\prime},\) and \(Q^{\prime}\) are \((g(0,0), h(0,0)),(g(\Delta u, 0), h(\Delta u, 0))\) and \((g(0, \Delta v), h(0, \Delta v)),\) respectively. b. Use a Taylor series in both variables to show that $$\begin{array}{l} g(\Delta u, 0) \approx g(0,0)+g_{u}(0,0) \Delta u \\ g(0, \Delta v) \approx g(0,0)+g_{v}(0,0) \Delta v \\ h(\Delta u, 0) \approx h(0,0)+h_{u}(0,0) \Delta u \\ h(0, \Delta v) \approx h(0,0)+h_{v}(0,0) \Delta v \end{array}$$ where \(g_{u}(0,0)\) is \(\frac{\partial x}{\partial u}\) evaluated at \((0,0),\) with similar meanings for \(g_{v}, h_{u},\) and \(h_{v}\) c. Consider the vectors \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\) and the parallelogram, two of whose sides are \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\). Use the cross product to show that the area of the parallelogram is approximately \(|J(u, v)| \Delta u \Delta v\) d. Explain why the ratio of the area of \(R\) to the area of \(S\) is approximately \(|J(u, v)|\)

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