Problem 21 - Advanced Engineering Mathematics

Question

Find the first partial derivatives of the given function. $$ z=\cos ^{2} 5 x+\sin ^{2} 5 y $$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial x} = -10 \cos(5x) \sin(5x) \); \( \frac{\partial z}{\partial y} = 10 \sin(5y) \cos(5y) \).

Step 1: Identify the Function

The given function is \( z = \cos^{2}(5x) + \sin^{2}(5y) \). Your task is to find the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).

Step 2: Find Partial Derivative with respect to x

To find \( \frac{\partial z}{\partial x} \), treat \( y \) as a constant and differentiate the function with respect to \( x \). Use the chain rule: the derivative of \( \cos(u) \) is \( -\sin(u) \), and for \( u = 5x \), the derivative of \( u \) with respect to \( x \) is 5. Thus, \( \frac{d}{dx}(\cos^{2}(5x)) = 2 \cos(5x) \cdot (-\sin(5x)) \cdot 5 = -10 \cos(5x) \sin(5x) \). The \( \sin^{2}(5y) \) term is constant with respect to \( x \) so its derivative is 0.

Step 3: Find Partial Derivative with respect to y

To find \( \frac{\partial z}{\partial y} \), treat \( x \) as a constant and differentiate the function with respect to \( y \). Use the chain rule: the derivative of \( \sin(u) \) is \( \cos(u) \), and for \( u = 5y \), the derivative of \( u \) with respect to \( y \) is 5. Thus, \( \frac{d}{dy}(\sin^{2}(5y)) = 2 \sin(5y) \cdot \cos(5y) \cdot 5 = 10 \sin(5y) \cos(5y) \). The \( \cos^{2}(5x) \) term is constant with respect to \( y \) so its derivative is 0.

Step 4: Write the Partial Derivatives

Combine the results from the previous steps: \( \frac{\partial z}{\partial x} = -10 \cos(5x) \sin(5x) \)\( \frac{\partial z}{\partial y} = 10 \sin(5y) \cos(5y) \)

Key Concepts

Chain Rule

The chain rule is a crucial concept in calculus used for differentiating composite functions. It helps break down the complexities of dealing with multiple layers of functions. In the context of partial derivatives, the chain rule lets us handle situations where one function is nested inside another. For instance, if we have a function like \( z = f(g(x)) \), the chain rule provides the necessary steps to find the derivative of \( z \) with respect to \( x \).For the function given in the exercise, \( z = \cos^{2}(5x) + \sin^{2}(5y) \), applying the chain rule means:

• First taking the derivative of the outer functions (\( \cos^{2} \) and \( \sin^{2} \))
• Then multiplying by the derivative of the inner functions (5x and 5y)

In this way, despite their complexity, we can systematically find the rate of change with respect to different variables in multivariable functions.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions with more than one variable. Unlike single-variable calculus, where you're only dealing with how one variable changes, multivariable calculus considers how functions change when multiple inputs vary.Partial derivatives are a fundamental tool in this area. They allow us to see how a function changes as we alter one specific variable, while keeping others constant. In the exercise, finding the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), helps isolate the effect of each variable on the function \( z \).Understanding multivariable functions requires visualizing them in higher dimensions. For example, imagine a surface in 3D space, where changes in \( x \) and \( y \) position the surface differently along the \( z \)-axis. Partial derivatives tell you how steeply the surface rises or falls in the direction of each variable, offering critical insights into the behavior of multivariable systems.

Trigonometric Functions

Trigonometric functions like \( \sin \) and \( \cos \) are fundamental in calculus because they describe cyclical and oscillatory behaviors, such as waves. Calculating derivatives of these functions is an essential skill, especially when they are part of more complex, composite functions.In the provided exercise, the trigonometric identities heavily influence the partial derivatives:

• For \( \cos(u) \), the derivative is \( -\sin(u) \), indicating how the cosine function slopes negatively as it travels around the unit circle.
• For \( \sin(u) \), the slope becomes \( \cos(u) \), showing the positive incline from the sine curve.

Understanding these derivatives helps demystify how the trigonometric components of \( z = \cos^{2}(5x) + \sin^{2}(5y) \) behave. When visualizing these components, it’s like interpreting the waves generated by oscillating functions—revealing where they peak and dip, and showing dual layers of sine and cosine working in tandem.