Question

Find \(u\left(x_{0}-y_{0}\right), u_{x}\left(x_{0}, y_{0}\right),\) and \(u_{y}\left(x_{0}, y_{0}\right)\) for all continuously differentiable functions \(u\) that satisfy the given equation near \(\left(x_{0}-y_{0}\right) .\) (a) \(2 x^{2} y^{4}-3 u x y^{3}+u^{2} x^{4} y^{3}=0 ; \quad\left(x_{0}-y_{0}\right)=(1,1)\) (b) \(\cos u \cos x+\sin u \sin y=0 ; \quad\left(x_{0}, y_{0}\right)=(0, \pi)\)

Short answer

In summary, for the given equation \(\cos u \cos x + \sin u \sin y = 0\) and \(\left(x_{0}, y_{0}\right)=(0, \pi)\), we have: \(u(x_0, y_0) = u(0, \pi) = \pi\) \(\frac{\partial u}{\partial x}(x_0, y_0) = u_x(0, \pi) = 0\) \(\frac{\partial u}{\partial y}(x_0, y_0) = u_y(0, \pi) = 0\)

Step-by-step solution

Differentiating with respect to x

Differentiate the given equation with respect to \(x\): $$\frac{\partial}{\partial x}\left(2x^2y^4 - 3uxy^3 + u^2x^4y^3\right) = 0$$

Differentiating with respect to y

End the equation with respect to \(y\): $$\frac{\partial}{\partial y}\left(2x^2y^4 - 3uxy^3 + u^2x^4y^3\right) = 0$$

Solving for u(x, y) and its partial derivatives

From the above equations, find the function \(u(x, y)\) and its partial derivatives \(u_x(x, y)\) and \(u_y(x, y)\).

Evaluate the expression at the given point and find u, u_x, and u_y

Find \(u(1,1)\), \(u_x(1,1)\), and \(u_y(1,1)\). Unfortunately, the equation in part (a) cannot be rewritten in an explicit form \(u(x, y)\). Without an explicit expression, we cannot find \(u(1,1)\), \(u_x(1,1)\) and \(u_y(1,1)\). (b) For \(\cos u \cos x + \sin u \sin y = 0\) and \(\left(x_{0}, y_{0}\right)=(0, \pi)\):

Differentiating with respect to x

Differentiate the given equation with respect to \(x\): $$\frac{\partial}{\partial x}\left(\cos u \cos x + \sin u \sin y\right) = 0$$

Differentiating with respect to y

Differentiate the equation with respect to \(y\): $$\frac{\partial}{\partial y}\left(\cos u \cos x + \sin u \sin y\right) = 0$$

Solving for u(x, y) and its partial derivatives

From the above partial derivative equations, find the function \(u(x, y)\) and its partial derivatives \(u_x(x, y)\) and \(u_y(x, y)\). Here, we can rewrite the equation as: $$u(x, y) = \arccos(-\cos x \cos y)$$ Now, we can find the partial derivatives as follows: $$u_x(x, y) =\frac{\partial u}{\partial x}(x, y) = \frac{-\sin x\cos y}{\sqrt{1-\cos^2x\cos^2y}}$$ $$u_y(x, y)=\frac{\partial u}{\partial y}(x, y)=\frac{-\cos x\sin y}{\sqrt{1-\cos^2x\cos^2y}}$$

Evaluate the expression at the given point and find u, u_x, and u_y

Find \(u(0, \pi)\), \(u_x(0, \pi)\), and \(u_y(0, \pi)\): $$u(0, \pi) = \arccos(-\cos0\cos\pi) = \pi$$ $$u_x(0, \pi) = \frac{-\sin0\cos\pi}{\sqrt{1-\cos^20\cos^2\pi}}=0$$ $$u_y(0, \pi)=\frac{-\cos0\sin\pi}{\sqrt{1-\cos^20\cos^2\pi}}=0$$

Key concepts

Continuously Differentiable Functions

In the world of real analysis, continuously differentiable functions are a cornerstone. These are functions that not only have derivatives at every point in a given domain, but their derivatives are also continuous functions. This implies that the behavior of the function around any point is stable and predictable, avoiding sharp turns or gaps that non-continuous functions might exhibit. It is important for students to understand that continuity and differentiability together provide a powerful tool when dealing with complex equations, such as those involving multiple variables.

To visualize this, think of a smooth hill without any abrupt cliffs or breaks. Just like you could walk up this hill without any unexpected drops, you can navigate the function's graph without encountering sudden changes in slope. This quality is especially vital when dealing with functions that will be integrated or used in models and simulations, as it ensures the absence of erratic behavior in the function's derivative.

Partial Derivative Equations

Moving onto the realm of functions with multiple variables, partial derivative equations become the heroes of the story. Partial derivatives involve taking the derivative of a function with respect to one variable at a time, treating other variables as constants. This may sound simple, but when applied to equations that model real-world scenarios, partial derivatives become essential in understanding the direction and rate of change.

For example, imagine a hill that rises and falls differently depending on whether you walk north or east. By computing the partial derivatives, you can gauge the incline in each direction separately. In solving the provided textbook exercise, the partial derivatives with respect to both x and y are calculated to discover the gradient of the function in each direction. This helps in understanding the behavior of the function locally, near the point of interest \(x_{0}, y_{0}\), providing insights necessary to solve for the function and its gradients.

Implicit Differentiation

Finally, when functions are intertwined in an equation without an obvious single-variable relationship, implicit differentiation comes to the rescue. This technique allows us to find the derivative of one variable with respect to another without isolating one side of the equation. Implicit differentiation is particularly useful when dealing with complex relationships where setting one variable as a function of another is not straightforward, such as in the case of an ellipse equation x^2/a^2 + y^2/b^2 = 1.

In the context of the exercise, the equation for part (b) could not be solved explicitly for u. This is a classic situation where implicit differentiation shines. Taking the derivative implicitly with respect to both x and y enabled us to determine the partial derivatives of u, even though we did not have an explicit formula for u itself. Understanding this concept allows students to approach such equations with confidence, knowing they have the tools to find derivatives and explore the trends of functions that might initially seem daunting.