Question

Let \(u=u(x, y)\) be determined near (1,1) by $$ x^{2} y u+2 x y^{2} u^{3}-3 x^{3} y^{3} u^{5}=0, \quad u(1,1)=1 $$ Find \(u_{x}(1,1)\) and \(u_{y}(1,1)\)

Short answer

Given the equation \(x^{2} y u+2 x y^{2} u^{3}-3 x^{3} y^{3} u^{5}=0\) and the initial condition \(u(1,1) = 1\), calculate the partial derivatives of u with respect to x and y at the point (1,1). After following the steps of implicit differentiation, solving for \(u_x\) and \(u_y\), and evaluating them at the point (1,1), we find that the partial derivatives at the point (1,1) are: \(u_x(1,1) = -\frac{1}{10}\) and \(u_y(1,1) = \frac{3}{10}\).

Step-by-step solution

Implicit Differentiation

We are given the equation $$ x^{2} y u+2 x y^{2} u^{3}-3 x^{3} y^{3} u^{5}=0. $$ To find \(u_{x}(1,1)\), differentiate implicitly with respect to x: $$ \frac{\partial}{\partial x}(x^{2} y u+2 x y^{2} u^{3}-3 x^{3} y^{3} u^{5})=0 $$ Applying the product rule, we get: $$ 2xyu + x^2yu_x + 2y^2u^3 + 6xy^2u^2u_x - 9x^2y^3u^5 - 15x^2y^3u^4u_x=0 $$ Similarly, for \(u_y(1,1)\), differentiate implicitly with respect to y: $$ \frac{\partial}{\partial y}(x^{2} y u+2 x y^{2} u^{3}-3 x^{3} y^{3} u^{5})=0 $$ Again, applying the product rule, we get: $$ x^2u + x^2yu_y + 4xy^3u^3 + 6x^3y^2u^2u_y - 9x^3y^2u^5 - 15x^3y^2u^4u_y=0 $$

Solve for \(u_x\) and \(u_y\)

From the two equations obtained after implicit differentiation, we can solve for \(u_x\) and \(u_y\). For \(u_x\), we get: $$ u_x = \frac{9x^2y^3u^5 + 2y^2u^3 - 2xyu}{x^2y(15u^4 - 6u^2 + 1)} $$ And for \(u_y\), we get: $$ u_y = \frac{9x^3y^2u^5 + 4xu^3 - x^2u}{x^2y(15u^4 - 6u^2 + 1)} $$

Evaluate \(u_x(1,1)\) and \(u_y(1,1)\)

Now, we need to evaluate \(u_x\) and \(u_y\) at the point (1,1). Since \(u(1,1)=1\), we can plug this in and obtain the required values. For \(u_x(1,1)\): $$ u_x(1,1) = \frac{9(1^2)(1^3)(1^5) + 2(1^2)(1^3) - 2(1)(1)}{(1^2)(1)(15(1^4) - 6(1^2) + 1)} = -\frac{1}{10} $$ And for \(u_y(1,1)\): $$ u_y(1,1) = \frac{9(1^3)(1^2)(1^5) + 4(1)(1^3) - (1^2)(1)}{(1^2)(1)(15(1^4) - 6(1^2) + 1)} = \frac{3}{10} $$ So, the partial derivatives at the point (1,1) are \(u_x(1,1) = -\frac{1}{10}\) and \(u_y(1,1) = \frac{3}{10}\).

Key concepts

Partial Derivatives

Partial derivatives are foundational to understanding how functions behave in multivariable calculus. They represent the rate at which a function changes as one variable is altered while all other variables are held constant. In the context of the provided exercise, to find partial derivatives like \( u_x \) and \( u_y \), we use implicit differentiation, which is particularly useful when dealing with equations where the dependent and independent variables are not easily separable.

Implicit differentiation follows the rules of regular differentiation but involves treating one variable as the focus and differentiating with respect to it, while all other variables are treated as constants. For example, when we differentiate \( x^2yu \) with respect to \( x \), we treat \( y \) and \( u \) as constants, leading to the term \( x^2yu_x \) because \( u \) is also a function of \( x \). To properly apply this concept, we typically use the product rule and chain rule, as was done in the exercise.

The trick is to express the derivatives \( u_x \) and \( u_y \) in terms of the variables of the function and any known point. This becomes particularly important when evaluating these derivatives at a specific point, such as (1,1) in the given exercise.

Real Analysis

Real analysis is the branch of mathematics dealing with real numbers and the analytic properties of real functions and sequences. It involves examining limits, continuity, integration, and differentiation in greater rigor compared to the broad overview given in first-year calculus courses. The exercise showcases an application of real analysis concepts in a real-valued function of multiple variables.

In real analysis, one often works with the notion of limits to define the exact behavior of functions at specific points, particularly concerning derivatives. The process of implicit differentiation used to solve for \( u_x \) and \( u_y \) is a practical application of these concepts, as it relies on the limit definitions of derivatives. When a function is defined implicitly, as is the case with \( u(x, y) \), real analysis tools enable us to extract information about the derivative behavior even without having an explicit formula for the function itself.

Understanding the rigorous background of real analysis can strengthen the comprehension of implicit differentiation, as students are reminded of the principles governing the operations they are employing.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to multiple dimensions. Here, instead of functions depending on a single variable, they rely on two or more variables, as is the situation with our function \( u(x, y) \). The results of multivariable calculations can often yield a collection of partial derivatives, known as the gradient, which provides information on the direction and rate of steepest ascent of the function.

The exercise offered a glimpse into the problem-solving strategies used in multivariable calculus, with implicit differentiation being a major player. It is worth noting that this calculus tool becomes indispensable when dealing with multivariable functions that cannot be solved for a single variable, like in the given equation involving \( u \).

While the approach to find partial derivatives remains methodical, multivariable calculus often involves visualizing functions in higher dimensions, and understanding terms like level curves and surfaces. The precise values of \( u_x(1,1) \) and \( u_y(1,1) \) found in the exercise are critical for understanding how small changes in \( x \) or \( y \) near the point (1,1) would affect the value of \( u \), capturing the essence of variable interactions in a multivariable context.