Question

If $$\left[\begin{array}{l} u \\ v \end{array}\right]=\mathbf{F}(x, y)=\left[\begin{array}{l} x^{2}+y^{2} \\ x^{2}-y^{2} \end{array}\right]$$ (Example 6.3 .1 ), find four branches \(\mathbf{G}_{1}, \mathbf{G}_{2}, \mathbf{G}_{3},\) and \(\mathbf{G}_{4}\) of \(\mathbf{F}^{-1}\) defined on $$T_{1}=\\{(u, v) \mid u+v>0, u-v>0\\}$$ and verify that \(\mathbf{G}_{i}^{\prime}(u, v)=\left(\mathbf{F}^{\prime}(x(u, v), y(u, v))\right)^{-1}, 1 \leq i \leq 4\).

Short answer

Question: Verify that the derivative of each branch of the inverse function for the given vector function is equal to the inverse of the derivative of the function itself. Given vector function: \(\mathbf{F}(x,y)=(x^2+y^2,x^2-y^2)\) Domain for inverse function: \(T_1=\{(u, v): u+v > 0, u-v > 0\}\) Branches of the inverse function: \(\mathbf{G}_i(u,v) = \left(\pm\sqrt{\frac{u+v}{2}},\pm\sqrt{\frac{u-v}{2}}\right)\) Verification: \(\mathbf{G}_i'(u, v)=\left(\mathbf{F}'(x(u, v), y(u, v))\right)^{-1}\) for \(1 \leq i \leq 4\).

Step-by-step solution

Solve for x and y in terms of u and v

Let's derive the expressions for \(x\) and \(y\) in terms of \(u\) and \(v\): We're given: $$\begin{cases} u = x^2 + y^2\\ v = x^2 - y^2 \end{cases}$$ Adding both these equations, we get: $$u + v = 2x^2$$ Now we can solve for x: $$x^2 = \frac{u+v}{2} \Rightarrow x = \pm \sqrt{\frac{u+v}{2}}$$ Subtracting both the given equations, we get: $$u-v=2y^2$$ Now we can solve for y: $$y^2=\frac{u-v}{2} \Rightarrow y=\pm\sqrt{\frac{u-v}{2}}$$

Determine the four branches of \(\mathbf{F}^{-1}\)

Now that we have both \(x\) and \(y\) expressed in terms of \(u\) and \(v\), we can define the four branches for the inverse function \(\mathbf{F}^{-1}\) as follows: $$\mathbf{G}_1(u,v) = \left(\sqrt{\frac{u+v}{2}},\sqrt{\frac{u-v}{2}}\right)$$ $$\mathbf{G}_2(u,v) = \left(-\sqrt{\frac{u+v}{2}},\sqrt{\frac{u-v}{2}}\right)$$ $$\mathbf{G}_3(u,v) = \left(\sqrt{\frac{u+v}{2}},-\sqrt{\frac{u-v}{2}}\right)$$ $$\mathbf{G}_4(u,v) = \left(-\sqrt{\frac{u+v}{2}},-\sqrt{\frac{u-v}{2}}\right)$$

Calculate the derivative of each branch, \(\mathbf{G}_i'(u, v)\)

Before we calculate the derivatives, let's define the individual components of \(\mathbf{G}_i(u,v)\) as: $$x_i(u,v)=\pm\sqrt{\frac{u+v}{2}}$$ $$y_i(u,v)=\pm\sqrt{\frac{u-v}{2}}$$ Now, we can find the derivative of each component: $$\frac{\partial x_i}{\partial u} = \pm\frac{1}{4\sqrt{\frac{u+v}{2}}}$$ $$\frac{\partial x_i}{\partial v} = \pm\frac{1}{4\sqrt{\frac{u+v}{2}}}$$ $$\frac{\partial y_i}{\partial u} = \pm\frac{1}{4\sqrt{\frac{u-v}{2}}}$$ $$\frac{\partial y_i}{\partial v} = \mp\frac{1}{4\sqrt{\frac{u-v}{2}}}$$ Hence, the derivative of each \(\mathbf{G}_i(u,v)\) is: $$\mathbf{G}_i'(u,v) = \begin{pmatrix}\frac{\partial x_i}{\partial u} & \frac{\partial x_i}{\partial v}\\ \frac{\partial y_i}{\partial u} & \frac{\partial y_i}{\partial v}\end{pmatrix} = \pm\frac{1}{4}\begin{pmatrix}\frac{1}{\sqrt{\frac{u+v}{2}}} & \frac{1}{\sqrt{\frac{u+v}{2}}}\\ \frac{1}{\sqrt{\frac{u-v}{2}}} & -\frac{1}{\sqrt{\frac{u-v}{2}}}\end{pmatrix}$$

Find the derivative of \(\mathbf{F}(x(u, v), y(u, v))\)

We need to find the derivative of \(\mathbf{F}(x(u, v), y(u, v))\): $$\mathbf{F}'(x(u,v),y(u,v)) = \begin{pmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{pmatrix} = \begin{pmatrix}2x & 2y \\ 2x & -2y\end{pmatrix}$$

Verify that \(\mathbf{G}_i'(u, v)=\left(\mathbf{F}'(x(u, v), y(u, v))\right)^{-1}\) for each branch

Now, we need to verify that \(\mathbf{G}_i'(u, v)=\left(\mathbf{F}'(x(u, v), y(u, v))\right)^{-1}\) for each branch. In other words, we need to show that: $$\mathbf{G}_i'(u, v)\mathbf{F}'(x(u, v), y(u, v)) = I$$ where \(I\) is the identity matrix. Let's calculate \(\mathbf{G}_i'(u, v)\mathbf{F}'(x(u, v), y(u, v))\): $$\mathbf{G}_i'(u, v)\mathbf{F}'(x(u, v), y(u, v)) = \pm\frac{1}{4}\begin{pmatrix}\frac{1}{\sqrt{\frac{u+v}{2}}} & \frac{1}{\sqrt{\frac{u+v}{2}}}\\ \frac{1}{\sqrt{\frac{u-v}{2}}} & -\frac{1}{\sqrt{\frac{u-v}{2}}}\end{pmatrix}\begin{pmatrix}2x & 2y \\ 2x & -2y\end{pmatrix}$$ $$= \pm\frac{1}{2}\begin{pmatrix}\frac{1}{\sqrt{\frac{u+v}{2}}} & \frac{1}{\sqrt{\frac{u+v}{2}}}\\ \frac{1}{\sqrt{\frac{u-v}{2}}} & -\frac{1}{\sqrt{\frac{u-v}{2}}}\end{pmatrix}\begin{pmatrix}x & y \\ x & -y\end{pmatrix}$$ $$= \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} = I$$ Thus, we have verified that \(\mathbf{G}_i'(u, v)=\left(\mathbf{F}'(x(u, v), y(u, v))\right)^{-1}\) for each branch \(1 \leq i \leq 4\).

Key concepts

Multivariable Calculus

Understanding Multivariable Calculus is crucial for tackling problems that involve more than one variable. As we step into this multidimensional realm, we encounter functions that map points from \f\(\textbf{R}^n\f\) to \f\(\textbf{R}^m\f\), where 'n' and 'm' indicate the number of dimensions in the domain and co-domain, respectively.

Multivariable functions often require us to think about space in three dimensions or more, which can be challenging. The function \f\(\textbf{F}(x, y)\f\) in the exercise you've been working with, maps points from a two-dimensional plane to another two-dimensional plane, transforming the variables 'x' and 'y' to 'u' and 'v'. When working with such functions, we look at how changes in the input variables affect the outputs, which brings us naturally to the concept of partial derivatives and the associated Jacobian matrix.

Derivative of Inverse Functions

In the world of single-variable calculus, the derivative of an inverse function can be found using the well-known formula \f\((f^{-1})'(y) = \frac{1}{f'(x)}\f\). However, when it comes to Multivariable Calculus, we generalize this concept using partial derivatives and the Jacobian matrix.

For a multivariable function to have an inverse, it must be locally invertible around a point. A crucial condition for invertibility is that the Jacobian determinant at that point must be non-zero. Once we establish this invertibility, we can then find the derivative of the inverse function, which will be related to the inverse of the original function's Jacobian matrix, as shown in the given exercise.

The relationship between the derivatives of the original function and its inverse is not coincidental; indeed, it's a deeply woven aspect of the function's local linear behavior, embodied by the calculation of the derivatives of the inverse branches \f\(\textbf{G}_i'(u, v)\f\).

Jacobian Matrix and Determinant

The Jacobian matrix is a powerhouse in multivariable calculus, encapsulating all the first-order partial derivatives of a vector-valued function into a single matrix. In the case of our function \f\(\textbf{F}(x, y)\f\), the Jacobian matrix contains all the possible rates of change of 'u' and 'v' with respect to 'x' and 'y'.

The determinant of the Jacobian matrix, also known as the Jacobian determinant, acts as a 'scale factor' for the transformation created by the function \f\(\textbf{F}(x, y)\f\). It tells us about the behavior of the function near a point and whether the function is locally invertible around that point - that is, if the function does not compress the input space to a lower dimension.

The nonzero determinant in our example ensures that the function \f\(\textbf{F}\f\) is locally invertible, and we are able to compute the inverse function's derivative as the inverse of the Jacobian matrix of the original function. This is crucial in the verification step, where the product of \f\(\textbf{G}_i'(u, v)\f\) and \f\(\textbf{F}'(x(u, v), y(u, v))\f\) yields the identity matrix, confirming the inverse relationship.