Problem 12 - Basic Engineering Mathematics

Question

Differentiate the following with respect to the variable: (a) \(y=2 \sin 5 \theta\) (b) \(f(t)=3 \cos 2 t\)

Step-by-Step Solution

Answer

(a) \( \frac{dy}{d\theta} = 10 \cos 5\theta \), (b) \( f'(t) = -6 \sin 2t \).

Step 1: Differentiate 2sin5θ with respect to θ

To differentiate the function \(y = 2 \sin 5\theta\), we use the chain rule. The derivative of \( \sin u \) with respect to \( u\) is \( \cos u \). Let \(u = 5\theta\), so the derivative of \(u\) with respect to \(\theta\) is 5. The derivative of \(2\sin 5\theta\) is \(2 \cdot (\cos 5\theta) \cdot 5\). Thus, the derivative is \(10 \cos 5\theta\).

Step 2: Differentiate 3cos2t with respect to t

To find the derivative of \(f(t) = 3 \cos 2t\), apply the chain rule. The derivative of \( \cos u \) with respect to \( u \) is \(-\sin u\). Let \(u = 2t\), so the derivative of \(u\) with respect to \(t\) is 2. Hence, the derivative of \(3 \cos 2t\) is \(3 \cdot (-\sin 2t) \cdot 2\), which simplifies to \(-6 \sin 2t\).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It allows us to find the derivative of a function that is composed of two or more functions. Essentially, if you have a function nested inside another function, the chain rule helps us to understand how changes in one variable affect another.

Here's how the chain rule works in a simplified manner:

• Identify the "outer" and "inner" functions.
• Differentiate the "outer" function with respect to the "inner" function.
• Differentiate the "inner" function with respect to the variable of interest.
• Multiply these derivatives together.

For example, in the function given in the problem (a) "2sin5θ", the outer function is "2sin(u)" and the inner function is "u = 5θ". By applying the chain rule, you first find the derivative of the sine function, then multiply it by the derivative of the inner function, 5θ, giving the result.

The chain rule is crucial for differentiating compositions of functions, making it invaluable in more advanced calculus applications.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are essential in both mathematics and various applications, including physics and engineering. When differentiating trigonometric functions, each has specific rules that simplify the process.

For differentiation, here are some basic derivatives:

• The derivative of \(\sin(u)\) is \(\cos(u)\).
• The derivative of \(\cos(u)\) is \(-\sin(u)\).
• The derivative of \(\tan(u)\) is \(\sec^2(u)\).

These rules are used as foundational blocks for applying more complex instances like using the chain rule with functions inside sine or cosine. As seen in our exercise, one must take into account these rules when differentiating functions like "2sin5θ" and "3cos2t".

Mastering these derivatives allows mathematicians and scientists to analyze waves, oscillations, and circular movements more easily.

Calculus

Calculus is a branch of mathematics concerned with change. It involves two main operations: differentiation and integration. Differentiation, our focus here, deals with the rate at which things change. It's like finding a slope of a curve at any point.

In the context of the given problems, we are differentiating trigonometric functions within calculus. This involves:

• Recognizing the form of the function (trigonometric in this case).
• Applying the chain rule to find the derivative of these functions with respect to the given variable.
• Simplifying the results to obtain clean expressions, like \(10 \cos 5\theta\) and \(-6 \sin 2t\).

Differentiation is particularly beneficial in fields requiring the calculation of velocities, accelerations, and understanding the behavior of dynamic systems.

As a foundation of modern mathematics, gaining a firm grasp of calculus concepts is crucial for solving real-world problems with efficiency and precision.