Problem 15 - Applied Calculus

Question

Differentiate each function. a. \(f(t)=4 \cos \frac{t}{2}\) b. Find \(f^{\prime}\left(\frac{\pi}{3}\right)\)

Step-by-Step Solution

Answer

The derivative at \(t = \frac{\pi}{3}\) is \(-1\).

Step 1: Differentiate Part a

Let's begin by differentiating the function \(f(t) = 4 \cos \frac{t}{2}\). First, use the chain rule. The derivative of \(\cos x\) is \(-\sin x\), and we have an inner function \(\frac{t}{2}\). Thus:\[ f'(t) = 4 \times (-\sin \frac{t}{2}) \times \frac{d}{dt}(\frac{t}{2}) \]The derivative of \(\frac{t}{2}\) with respect to \(t\) is \(\frac{1}{2}\), so we have:\[ f'(t) = -2 \sin \frac{t}{2} \]

Step 2: Evaluate at \(t = \frac{\pi}{3}\)

Now we need to find the derivative at \(t = \frac{\pi}{3}\). Using the derivative we found in Step 1, substitute \(t = \frac{\pi}{3}\):\[ f'\left(\frac{\pi}{3}\right) = -2 \sin \left(\frac{\pi}{6}\right) \]The value of \(\sin \left(\frac{\pi}{6}\right)\) is \(\frac{1}{2}\), so:\[ f'\left(\frac{\pi}{3}\right) = -2 \times \frac{1}{2} = -1 \]

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus for finding the derivative of composite functions. When you have one function nested inside another, the chain rule helps in differentiating the outer and inner functions systematically.
Let's break down how the Chain Rule works. It states that if you have a function \(y = g(h(x))\), where \(g\) is the outer function and \(h\) is the inner function, the derivative \(y'\) is given by:

• Derive the outer function \(g\) with respect to the inner function \(h\)
• Multiply it by the derivative of the inner function \(h\) with respect to \(x\)

This can be expressed as: \(\frac{dy}{dx} = g'(h(x)) \times h'(x)\).
In our problem, the outer function is cos (\(\cos x\)) and the inner function is \(\frac{t}{2}\). We compute the derivative:

• Outer derivative: Derivative of \(\cos x\) is \(-\sin x\).
• Inner derivative: \(\frac{d}{dt}\left(\frac{t}{2}\right) = \frac{1}{2}\).

Thus, multiplying these derivatives gives us \(-\sin\left(\frac{t}{2}\right) \times \frac{1}{2}\), leading to the result of \(-2\sin\left(\frac{t}{2}\right)\) for \(f'(t)\).

Trigonometric Functions

Trigonometric functions are core building blocks in mathematics, dealing with the relationships between angles and lengths in right triangles. Among them, the cosine and sine functions are especially important for calculus.
In this context, we used the cosine function \(\cos(x)\). Its derivative, \(-\sin(x)\), is critical when applying the Chain Rule. The cosine is an even function, meaning \(\cos(-x) = \cos(x)\), and its range is between -1 and 1.
Sine and cosine functions have a key property: their derivatives are cyclic and directly relate to one another:

• Derivative of \(\cos(x)\): \(-\sin(x)\)
• Derivative of \(\sin(x)\): \(\cos(x)\)

This cycle makes them easy to interchange, especially when differentiating composite functions involving trigonometric functions. In our case, understanding that \(f'(t) = -2\sin\left(\frac{t}{2}\right)\) provided a seamless calculation into the next steps.

Derivative Evaluation

Derivative Evaluation involves substituting a specific value into a derivative to find the rate of change at that point. In the given problem, after deriving the function, we needed to evaluate it at a specific \(t\) value.
For \(f'(t) = -2\sin\left(\frac{t}{2}\right)\), we needed to find the derivative at \(t = \frac{\pi}{3}\). Substitution into our derived formula gives:

• Calculate \(\frac{t}{2}\) which becomes \(\frac{\pi}{6}\)
• Find \(\sin\left(\frac{\pi}{6}\right)\) known to be \(\frac{1}{2}\)

Plug these values into the equation, resulting in \(-2 \times \frac{1}{2} = -1\).
This simple substitution process gives us the precise rate of change at the specific input. Evaluating derivatives at specific points is essential to understanding a function's behavior instantaneously, not just generally.