Problem 14 - Basic Engineering Mathematics

Question

Differentiate the following with respect to the variable: (a) \(f(\theta)=5 \sin (100 \pi \theta-0.40)\) (b) \(f(t)=2 \cos (5 t+0.20)\)

Step-by-Step Solution

Answer

(a) 500\pi \cos(100 \pi \theta - 0.40). (b) -10 \sin(5t + 0.20).

Step 1: Understand the Problem

We need to find the derivatives of the given functions (a) \(f(\theta)=5 \sin (100 \pi \theta-0.40)\) and (b) \(f(t)=2 \cos (5 t+0.20)\) with respect to their respective variables, \(\theta\) and \(t\).

Step 2: Differentiate Part (a)

The function is \(f(\theta) = 5 \sin(100 \pi \theta - 0.40)\). Use the chain rule to differentiate: 1. Differentiate the outer function (\(\sin\)) which gives \(\cos\), leaving the inside unchanged: \(5 \cos(100 \pi \theta - 0.40)\).2. Differentiate the inner function \(100 \pi \theta - 0.40\) with respect to \(\theta\), which is \(100 \pi\).3. Combine these using the chain rule: \[f'(\theta) = 5 \cdot 100 \pi \cdot \cos(100 \pi \theta - 0.40) = 500\pi \cos(100 \pi \theta - 0.40).\]

Step 3: Differentiate Part (b)

The function is \(f(t) = 2 \cos(5t + 0.20)\). Using the chain rule again:1. Differentiate the outer function (\(\cos\)) which gives \(-\sin\), leaving the inside unchanged: \(-2 \sin(5t + 0.20)\).2. Differentiate the inner function \(5t + 0.20\) with respect to \(t\), which is \(5\).3. Combine these using the chain rule: \[f'(t) = -2 \cdot 5 \cdot \sin(5t + 0.20) = -10 \sin(5t + 0.20).\]

Step 4: Conclude the Solution

The derivatives of the given functions are:(a) For \(f(\theta) = 5 \sin(100 \pi \theta - 0.40)\), the derivative is \[f'(\theta) = 500\pi \cos(100 \pi \theta - 0.40).\](b) For \(f(t) = 2 \cos(5t + 0.20)\), the derivative is \[f'(t) = -10 \sin(5t + 0.20).\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used for differentiation. It is especially helpful when dealing with composite functions—functions composed of multiple layers. In essence, the chain rule allows us to differentiate a function inside another function.
To apply the chain rule, follow these steps:

• Differentiate the outer function: Leave the inner function unchanged and differentiate the outer function as if it were a normal, single-variable function. For example, for a function like \(\sin(100 \pi \theta - 0.40)\), the outer function is \(\sin\), which differentiates to \(\cos\).
• Differentiate the inner function: Take the derivative of the inner function separately. For instance, in \(100 \pi \theta - 0.40\), the inner function differentiates to \(100 \pi\).
• Combine the results: Multiply the derivative of the outer function by the derivative of the inner function. This gives the final derivative of the composite function.

For instance, regarding the differentiation of \(f(\theta) = 5 \sin(100 \pi \theta - 0.40)\), the outer sin function becomes cosine, and the inner function's derivative (100\(\pi\)) is multiplied, resulting in \(500 \pi \cos(100 \pi \theta - 0.40)\). Simple, right?

Trigonometric Differentiation

Trigonometric differentiation involves taking derivatives of functions with trigonometric functions like sine, cosine, and tangent. These derivatives are crucial in applied mathematics, especially in physics and engineering.
Here are fundamental derivatives:

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

These basic rules help us navigate through more complex trigonometric functions. For example, in \(f(t) = 2 \cos(5t + 0.20)\), applying this rule involves changing the cosine part to \(-\sin\) and multiplying by the derivative of the inner function, resulting in \(-10 \sin(5t + 0.20)\). This systematic approach helps us tackle any functions involving trigonometric expressions.

Calculus Techniques

The exercise highlights several calculus techniques used in derivative computation. Understanding these can significantly enhance problem-solving skills in calculus.
Here are some critical techniques:

• Recognizing Function Composition: Many functions are compositions of two or more functions. Identifying these helps in applying the chain rule correctly.
• Consistent Practice: Deriving trigonometric and other complex functions takes practice. Repeated practice helps in mastering these techniques.
• Inner Function Differentiation: Never forget to differentiate the inner function while applying the chain rule. This step is essential for accuracy.
• Verify Units and Constants: Always remember to multiply through constants, like \(5\) or \(100\pi\), as seen in these examples.By applying these techniques effectively, solving even complicated derivative problems becomes more straightforward and less intimidating.