Question

Two continuously differentiable transformations \(\mathbf{U}=(u, v)\) of \((x, y)\) satisfy the system $$ \begin{array}{r} x y u-4 y u+9 x v=0 \\ 2 x y-3 y^{2}+v^{2}=0 \end{array} $$ near \(\left(x_{0}, y_{0}\right)=(1,1) .\) Find the value of each transformation and its differential matrix at (1,1) .

Short answer

Answer: The values of transformations at point (1, 1) are u(1,1) = -3 and v(1,1) = -1. The differential matrix at point (1, 1) is: $$ \begin{pmatrix} -3 & -1 \\ 2 & -8 \end{pmatrix} $$.

Step-by-step solution

Find u(1,1) and v(1,1)

Substitute the point (1,1) into the given equations: $$ \begin{array}{r} (1)(1)u - 4(1)u + 9(1)v = 0 \\ 2(1)(1) - 3(1)^{2} + v^{2} = 0 \end{array} $$ $$ \begin{array}{r} u - 4u + 9v = 0 \\ 2 - 3 + v^{2} = 0 \end{array} $$ Simplifying the equations, we have: $$ \begin{array}{r} -3u + 9v = 0 \\ v^{2} = 1 \end{array} $$ From the second equation, we can find the value of v: \(v = \pm 1\) Leveraging the first equation, we can also find the value of u: For v = 1, \(u = 3\) For v = -1, \(u = -3\) Thus, u(1,1)=-3 and v(1,1)=-1 can be the set of solutions.

Calculate the partial derivatives of u and v

Let's find the partial derivatives of u and v. $$ \frac{\partial u}{\partial x} = y - 4 $$ $$ \frac{\partial u}{\partial y} = xu - 4u + 9x $$ And from the second given equation, $$ 2xy -3y^2 +v^2 = 0 $$ Taking the partial derivative with respect to x and y, $$ \frac{\partial v}{\partial x} = 2y $$ $$ \frac{\partial v}{\partial y}= -6y +2v\frac{\partial v}{\partial y} $$

Evaluate the partial derivatives at (1, 1)

Now, we evaluate the partial derivatives at the point (1, 1): $$ \left. \frac{\partial u}{\partial x} \right|_{(1,1)} = 1 - 4 = -3 $$ $$ \left. \frac{\partial u}{\partial y} \right|_{(1,1)} = -1$$ $$ \left. \frac{\partial v}{\partial x} \right|_{(1,1)} = 2(1) = 2 $$ $$ \left. \frac{\partial v}{\partial y} \right|_{(1,1)}= -6 + 2(-1)=-8 $$

Construct the differential matrix

Now, we construct the differential matrix by arranging the evaluated partial derivatives in the following manner: $$ \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix}_{(1,1)} = \begin{pmatrix} -3 & -1 \\ 2 & -8 \end{pmatrix} $$ So, the differential matrix at point (1, 1) is: $$ \begin{pmatrix} -3 & -1 \\ 2 & -8 \end{pmatrix} $$.

Key concepts

Partial Derivatives

In calculus, partial derivatives are fundamental when dealing with functions of several variables. For a function with multiple inputs, a partial derivative measures the rate at which the function changes as one specific input changes, while keeping the other inputs constant.

In the exercise provided, we analyzed two transformations, \( u \) and \( v \), to find their partial derivatives. This involves finding how \( u \) and \( v \) change as \( x \) and \( y \) vary individually.

• For \( u \), we computed \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \).
• For \( v \), the same process was used to find \( \frac{\partial v}{\partial x} \) and \( \frac{\partial v}{\partial y} \).

By evaluating these derivatives at the point (1, 1), it provides specific insight into how the transformations behave locally, at that precise location. Understanding partial derivatives is crucial for studying changes in multivariable systems.

Differential Matrix

The differential matrix, sometimes known as the Jacobian matrix, is a pivotal concept in calculus. It organizes all the partial derivatives of a multivariable function into a matrix form.

Each entry of this matrix represents a partial derivative of the function.

• The first row can show how the first function component changes with respect to each variable.
• The second row does the same for the second function component, and so on.

In our exercise, the differential matrix was formed using the derivatives:

• \( \frac{\partial u}{\partial x} = -3 \) and \( \frac{\partial u}{\partial y} = -1 \)
• \( \frac{\partial v}{\partial x} = 2 \) and \( \frac{\partial v}{\partial y} = -8 \)

The resulting matrix provides a linear approximation of how the transformations \( u \) and \( v \) change near the point (1, 1), offering insights into their continuous behavior.

Continuously Differentiable Transformations

Transformations that are continuously differentiable ensure smooth changes in behavior without abrupt transitions. This means the functions have continuous partial derivatives, adding to their predictability and reliability.

In the exercises, \( \mathbf{U} = (u, v) \) represents such transformations. They are differentiable at all points considered, allowing us to confidently compute derivatives without worrying about undefined points or discontinuities.

• Such transformations guarantee the functions are not only differentiable once but have smooth derivatives of all orders.
• This becomes crucial in complex systems where knowing the stability and behavior of solutions relies on well-behaved functions.

Continuously differentiable transformations are a key concept when exploring the applications of calculus in real-world scenarios, providing a bridge between pure mathematics and practical problem-solving.