Problem 37 - University Calculus: Early Transcendentals

Question

Find all possible functions with the given derivative. a. \(y^{\prime}=\sin 2 t\) b. \(y^{\prime}=\cos \frac{t}{2} \quad\) c. \(y^{\prime}=\sin 2 t+\cos \frac{t}{2}\)

Step-by-Step Solution

Answer

a. \(-\frac{1}{2}\cos 2t + C\), b. \(2\sin\frac{t}{2} + C\), c. \(-\frac{1}{2}\cos 2t + 2\sin \frac{t}{2} + C\).

Step 1: Understanding the Problem

We are given three derivative functions and need to find the original functions (antiderivatives) for each. This involves using integration to reverse the process of differentiation.

Step 2: Solve Part (a)

For the derivative function \(y' = \sin 2t\), find the antiderivative. Integrate \(\sin 2t\) with respect to \(t\). The antiderivative of \(\sin ax\) is \(-\frac{1}{a}\cos ax\). Therefore, the function is \(-\frac{1}{2}\cos 2t + C\), where \(C\) is the integration constant.

Step 3: Solve Part (b)

For the derivative function \(y' = \cos \frac{t}{2}\), integrate \(\cos \frac{t}{2}\) with respect to \(t\). The antiderivative of \(\cos ax\) is \(\frac{1}{a}\sin ax\). Therefore, the function is \(2\sin \frac{t}{2} + C\).

Step 4: Solve Part (c)

For the derivative function \(y' = \sin 2t + \cos \frac{t}{2}\), integrate each term separately. We have already determined that the antiderivatives for \(\sin 2t\) and \(\cos \frac{t}{2}\) are \(-\frac{1}{2}\cos 2t + C_1\) and \(2\sin \frac{t}{2} + C_2\), respectively. Thus, the antiderivative for the whole function is \(-\frac{1}{2}\cos 2t + 2\sin \frac{t}{2} + C\), where \(C = C_1 + C_2\).

Key Concepts

Antiderivatives

Finding the antiderivative of a function is like working backward from differentiation. Essentially, this process involves identifying a function whose derivative matches the given function. When finding antiderivatives, you integrate the given function with respect to a variable, such as \(t\) in this exercise. Let’s see how this looks for different types of functions:

• Power functions: The antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1}\), provided \(n eq -1\).
• Exponential functions: The antiderivative of \(e^x\) is \(e^x\) itself.
• Trigonometric functions: As showcased in the problem, for functions such as \(y' = \sin ax\), the antiderivative would be \(-\frac{1}{a}\cos ax\). For \(y' = \cos ax\), the antiderivative is \(\frac{1}{a}\sin ax\).

When you find an antiderivative, you have a general solution that includes the constant of integration. More on that in a bit.

Trigonometric Functions

Integrating trigonometric functions often involves formulas adapted from their derivatives. Remembering these relationships can be quite handy. Let’s revisit how this applies to our exercise:* For \(y' = \sin 2t\), the antiderivative is \(-\frac{1}{2}\cos 2t + C\). Here, \(\sin ax\) integrates to \(-\frac{1}{a}\cos ax\).* For \(y' = \cos \frac{t}{2}\), the antiderivative becomes \(2\sin \frac{t}{2} + C\). This stems from the fact that \(\cos ax\) integrates to \(\frac{1}{a}\sin ax\).When you have a function like \(y' = \sin 2t + \cos \frac{t}{2}\), you can split the integration process:

• Integrate each term separately.
• Combine the results, keeping in mind to add a constant of integration only once in the final expression.

This approach makes it easier to handle complex expressions or combinations of different trigonometric functions.

Constant of Integration

Whenever you find an antiderivative, it’s crucial to include a constant of integration, usually denoted as \(C\). This constant represents an entire family of functions. Here’s why the constant of integration is important:

• Differentiation removes any constant term (because the derivative of a constant is zero), meaning you lose unique information when you differentiate.
• Integrating brings back a whole family of potential original functions differing by just that constant value.

Given the functions derived in our exercise, each comes with a \(C\):
- For part (a): \(-\frac{1}{2}\cos 2t + C\).- For part (b): \(2\sin \frac{t}{2} + C\).- For part (c): \(-\frac{1}{2}\cos 2t + 2\sin \frac{t}{2} + C\).
These constants allow for the full spectrum of solutions that map back to the given derivative function, providing flexibility and completeness to the solutions. Knowing when and why to include \(C\) is key to mastering integration.