Problem 27 - Calculus with Concepts in Calculus

Question

Let \(f(x, y)=\int_{\pi}^{x^{2}+y^{2}} \sin t^{2} d t .\) Find \(f_{x}\) and \(f_{y} .\)

Step-by-Step Solution

Answer

\( f_x = 2x \cdot \sin((x^2 + y^2)^2) \); \( f_y = 2y \cdot \sin((x^2 + y^2)^2) \).

Step 1: Understand the Function Dependencies

The given function is \( f(x, y) = \int_{\pi}^{x^2 + y^2} \sin t^2 \, dt \). This suggests that \( f(x, y) \) depends on \( x \) and \( y \) through the upper limit of the integral \( x^2 + y^2 \). Thus, \( f(x, y) \) is a composition of functions where the outer function is an integral and the inner function is \( x^2 + y^2 \).

Step 2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(t) \) is an antiderivative of \( \sin t^2 \), then \( \frac{d}{dt} \left( \int_{a}^{t} \sin t^2 \, dt \right) = \sin t^2 \). Apply this to differentiate the integral with respect to its upper limit.

Step 3: Find Derivative with Respect to x (Chain Rule)

The derivative of \( f(x, y) \) with respect to \( x \) is found by using the chain rule. Compute:\[ f_x = \frac{\partial}{\partial x} \left( \int_{\pi}^{x^2 + y^2} \sin t^2 \, dt \right) = \sin((x^2 + y^2)^2) \cdot \frac{\partial}{\partial x}(x^2 + y^2) \]Now, calculate \( \frac{\partial}{\partial x}(x^2 + y^2) = 2x \). Thus,\[ f_x = 2x \cdot \sin((x^2 + y^2)^2) \]

Step 4: Find Derivative with Respect to y (Chain Rule)

Similarly, we find the derivative of \( f(x, y) \) with respect to \( y \) using the chain rule. Compute:\[ f_y = \frac{\partial}{\partial y} \left( \int_{\pi}^{x^2 + y^2} \sin t^2 \, dt \right) = \sin((x^2 + y^2)^2) \cdot \frac{\partial}{\partial y}(x^2 + y^2) \]Now, calculate \( \frac{\partial}{\partial y}(x^2 + y^2) = 2y \). Thus,\[ f_y = 2y \cdot \sin((x^2 + y^2)^2) \]

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a crucial concept in calculus that connects differentiation with integration. It states that if a function is integrable and continuous on an interval, then its integral has an antiderivative. More simply, differentiating the integral of a function within a given range brings back the original function.

For example, consider a function \( F(t) \) that acts as an antiderivative of \( \sin t^2 \). According to the theorem, the derivative of its integral from \( a \) to \( t \), which is \( \frac{d}{dt} \left( \int_{a}^{t} \sin t^2 \, dt \right) \), is equal to \( \sin t^2 \) itself.

In the given exercise, the theorem is applied to differentiate \( f(x, y) = \int_{\pi}^{x^2 + y^2} \sin t^2 \, dt \). As a result, taking the derivative concerning \( t \) returns \( \sin t^2 \), specifically evaluated at the upper boundary \( x^2 + y^2 \). This lays the foundation for finding partial derivatives with respect to \( x \) and \( y \) using the chain rule.

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating composite functions. When you have a function composed of other functions, the chain rule helps determine its derivative by multiplying the derivative of the outer function by the derivative of the inner one.

To apply the chain rule, start by identifying the outer and inner functions. In our exercise, the outer function is the integral \( \int_{\pi}^{x^2 + y^2} \sin t^2 \, dt \), and the inner function is the limit \( x^2 + y^2 \).

Let's see how it works for partial derivatives:

• For \( f_x \), differentiate the outer function concerning the upper limit, \( x^2 + y^2 \), then multiply by the derivative of the inner function, \( x^2 \), with respect to \( x \). This results in: \[ f_x = \sin((x^2 + y^2)^2) \cdot 2x \]
• For \( f_y \), apply in a similar manner by differentiating with respect to \( y \): \[ f_y = \sin((x^2 + y^2)^2) \cdot 2y \]

The chain rule connects all these calculations, helping accurately find the necessary derivatives by breaking down complex expressions into manageable parts.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of several variables, allowing us to work with tasks and problems in higher dimensions. Two essential components of multivariable calculus include partial derivatives and functions of multiple variables.

Partial derivatives measure how a function changes as one of its variables varies while the others remain constant. In the exercise, \( f(x, y) \) depends on both \( x \) and \( y \) through the composite function \( x^2 + y^2 \), requiring the use of partial derivatives to evaluate changes independently.

The notation requires consistency as you can differentiate with respect to any variable of the function of interest, not just a single dimension. When computing \( f_x \) and \( f_y \), you're essentially exploring how the function varies according to each variable while keeping the interconnected nature of \( x^2 + y^2 \) in mind.

• In \( f_x \), the change in function concerning \( x \) focuses on the component of \( x^2 \). The chain rule aids in encapsulating these dependencies.
• Similarly, for \( f_y \), it analyzes how \( f(x, y) \) varies as \( y \) changes, while \( x \) maintains its status, observing the \( y^2 \) behavior.

Mastering these techniques offers incredible insight into how variables interplay and affect outcomes in a multivariable environment, preparing students for more advanced calculus applications.