Problem 36 - Thomas Calculus in SI Units

Question

Find all possible functions with the given derivative. a. \(y^{\prime}=\sec ^{2} \theta\) b. \(y^{\prime}=\sqrt{\theta} \quad\) c. \(y^{\prime}=\sqrt{\theta}-\sec ^{2} \theta\)

Step-by-Step Solution

Answer

a. \( y = \tan \theta + C \); b. \( y = \frac{2}{3} \theta^{3/2} + C \); c. \( y = \frac{2}{3} \theta^{3/2} - \tan \theta + C \).

Step 1: Identify the task for part a

We need to find a function whose derivative is given by the expression \( y' = \sec^2 \theta \).

Step 2: Integrate to find the function for part a

Recall that the derivative of the tangent function is \( \sec^2 \theta \), so integrating \( \sec^2 \theta \) with respect to \( \theta \) gives the original function \( y = \tan \theta + C \), where \( C \) is the constant of integration.

Step 3: Identify the task for part b

We need to find a function whose derivative is given by the expression \( y' = \sqrt{\theta} \).

Step 4: Integrate to find the function for part b

Integrate \( \sqrt{\theta} \), which is \( \theta^{1/2} \), with respect to \( \theta \). This gives \( y = \frac{2}{3} \theta^{3/2} + C \), where \( C \) is the constant of integration.

Step 5: Identify the task for part c

We need to find a function whose derivative is given by \( y' = \sqrt{\theta} - \sec^2 \theta \).

Step 6: Integrate to find the function for part c

To find the function, integrate the expression \( y' = \sqrt{\theta} - \sec^2 \theta \) with respect to \( \theta \). This results in combining the integrals from Steps 2 and 4: \( y = \frac{2}{3} \theta^{3/2} - \tan \theta + C \), where \( C \) is the constant of integration.

Key Concepts

Understanding Antiderivatives

Antiderivatives are like the reverse of derivatives. They help us find a function whose derivative matches a given one. This process is also known as integration. When we are given a derivative, we can find its antiderivative using integration techniques.
For instance, in the exercise, one of the derivatives provided was \(y' = \sec^2 \theta\). The antiderivative, \(y\), is the function whose derivative is \(\sec^2 \theta\). By remembering that the derivative of \(\tan \theta\) is \(\sec^2 \theta\), we can deduce that the antiderivative of \(\sec^2 \theta\) is \(\tan \theta\), up to a constant.
It's essential to recognize different forms of derivatives to apply the correct integration technique:

• Polynomial expressions are integrated using rules like \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\).


• Standard trigonometric derivatives require knowing antiderivatives, such as \(\int \sec^2 \theta \, d\theta = \tan \theta + C\).

Finding antiderivatives involves practice and familiarity with integration methods and standard derivatives.

The Role of the Constant of Integration

Whenever we find the antiderivative of a function, we add a constant \(C\). This constant represents the family of possible functions that have the given derivative. Think of it as an infinite set of parallel curves that are vertically translated versions of each other.
For example, in the solution, after calculating the antiderivative, you'll always notice a \(+ C\). This constant arises because the derivative of any constant is zero. So, given a derivative \(\sec^2 \theta\), there are infinitely many functions \(\tan \theta + C\), each with a constant shift, that share this derivative, \(\sec^2 \theta\).
Understanding the reason for this constant is crucial:

• When you only know the derivative of a function, you lack information about its vertical position on the graph. The constant fills this gap.
• If you have extra information, like a point that the function passes through, you can determine the specific value of \(C\).

This constant is an integral part of solutions, literally and figuratively!

Exploring Trigonometric Functions

Trigonometric functions often appear in integration problems due to their cyclic nature. Functions like sine, cosine, and tangent have well-known derivatives that allow us to find antiderivatives easily.
In the given exercise, the focus was on finding the antiderivative of \(\sec^2 \theta\). Knowing that \(\frac{d}{d\theta}(\tan \theta) = \sec^2 \theta\) instantly tells you that \(\tan \theta\) is an antiderivative.
Some key points about integrating trigonometric functions:

• Recognize standard derivatives such as \(\frac{d}{d\theta}(\sin \theta) = \cos \theta\) and anyone of \(\tan \theta = \sec^2 \theta\).


• Be aware of integration rules involving trig identities, which can simplify more complex integrals.
• Practice converting between different trigonometric forms, as it helps in both integration and differentiation.

Mastering trigonometric functions is key to solving a wide range of calculus problems efficiently.