Problem 36 - Thomas Calculus

Question

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

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: Solve for a. Derivative of sec²θ

The derivative is given as \( y' = \sec^2 \theta \). We know that the derivative of \( \tan \theta \) is \( \sec^2 \theta \). Therefore, the function \( y \) such that \( y' = \sec^2 \theta \) is \( y = \tan \theta + C \), where \( C \) is the constant of integration.

Step 2: Solve for b. Derivative of sqrt(θ)

We need to determine an antiderivative of \( \sqrt{\theta} \). Rewrite \( \sqrt{\theta} \) as \( \theta^{1/2} \). The antiderivative is calculated by adding 1 to the exponent and dividing by the new exponent: \( \int \theta^{1/2} \, d\theta = \frac{\theta^{3/2}}{3/2} = \frac{2}{3} \theta^{3/2} + C \). Therefore, \( y = \frac{2}{3}\theta^{3/2} + C \), where \( C \) is the constant of integration.

Step 3: Solve for c. Sum of Two Functions

The derivative is \( y' = \sqrt{\theta} - \sec^2 \theta \). Split this into two separate antiderivatives: \( \int \sqrt{\theta} \) and \( \int -\sec^2 \theta \). As calculated in Step 2, \( \int \sqrt{\theta} = \frac{2}{3}\theta^{3/2} \). For \( \int -\sec^2 \theta \), we have \( -\tan \theta \) from Step 1. Combine these to find the function: \( y = \frac{2}{3}\theta^{3/2} - \tan \theta + C \), where \( C \) is the constant of integration.

Key Concepts

Constant of Integration

When we work with antiderivatives, there is an important element that we must always include in our solution: the constant of integration, denoted usually by \( C \). Whenever we find the antiderivative of a function, we're essentially reversing differentiation, which means there could be an infinite number of possible functions that have the same derivative. This happens because the derivative of a constant is zero, so adding or subtracting any constant does not affect the derivative.

As a result, the general solution to an integral (i.e., finding an antiderivative) includes this arbitrary constant. This is why, when calculating the antiderivative of a function, we add \( C \) at the end of our result. For example, if the derivative is \( y' = \sec^2 \theta \), the antiderivative could be \( y = \tan \theta + C \), where \( C \) represents all possible constants.

Trigonometric Derivatives

Trigonometric functions and their derivatives are a key part of calculus. One of the trigonometric derivatives that often appear in exercises is the derivative of \( \tan \theta \), which is \( \sec^2 \theta \).

This means if you are asked to find a function whose derivative is \( \sec^2 \theta \), you should immediately think of the function \( \tan \theta \). This is a result of the differentiation process of the tangent function and can be a helpful piece of information when working on calculus problems related to trigonometric functions.

Knowing these standard derivatives ensures that you can quickly spot the right antiderivative, as seen in the original step-by-step solution for problem (a) in the exercise.

Power Rule

The power rule is a straightforward and immensely useful technique for finding antiderivatives, especially for polynomial functions. It states that if you have a function \( x^n \), the antiderivative is \( \frac{x^{n+1}}{n+1} \), as long as \( n eq -1 \).

To apply this to our exercise, consider the function \( \sqrt{\theta} \) which can be rewritten as \( \theta^{1/2} \). Using the power rule, the antiderivative is \( \frac{\theta^{3/2}}{3/2} \), simplifying to \( \frac{2}{3} \theta^{3/2} \).

Practicing applying the power rule allows you to tackle a wide variety of integral problems, clearing the path to solving more complex integrals that often combine multiple rules.

Integration Techniques

Integration techniques refer to various methods used to solve integrals, which come in handy when you need to integrate more complicated functions.

• The direct antiderivative method, as seen in the examples, is straightforward for functions like \( \sec^2 \theta \) and \( \sqrt{\theta} \).
• For composite functions, like \( \sqrt{\theta} - \sec^2 \theta \) in the exercise, you can find the antiderivative by addressing each term separately. Integrate each term and then add (or subtract) them together.

These techniques allow you to break down more complex derivatives into parts that are easier to handle. This is especially useful in applied mathematics, where longer expressions are quite common, and efficiency in solving these expressions becomes crucial.