Problem 36 - Thomas Calculus

Question

Find the function with the given derivative whose graph passes through the point \(P .\) \(r^{\prime}(t)=\sec t \tan t-1, \quad P(0,0)\)

Step-by-Step Solution

Answer

The function is \( r(t) = \sec t - t - 1 \).

Step 1: Set Up the Integral

To find the original function from its derivative, integrate the given derivative function. The derivative given is \( r'(t) = \sec t \tan t - 1 \). We need to find \( r(t) \) which satisfies this.

Step 2: Integrate Each Term

Integrate each component of the derivative separately: For the first term, the integral of \( \sec t \tan t \) is \( \sec t \), and for the second term, the integral of \(-1\) is \( -t \).Therefore, the indefinite integral of \( r'(t) \) is \[ r(t) = \sec t - t + C \] where \( C \) is the constant of integration.

Step 3: Use Given Point to Solve for Constant

Substitute point \( P(0, 0) \) into the equation to solve for \( C \) :\( 0 = \sec(0) - 0 + C \). Since \( \sec(0) = 1 \), we have:\( 0 = 1 + C \)Thus, \( C = -1 \).

Step 4: Write the Specific Function

Write the specific form of the function \( r(t) \) using the calculated constant:\[ r(t) = \sec t - t - 1 \]

Key Concepts

Indefinite Integral

In calculus, an indefinite integral refers to the most general form of the antiderivative of a function. When we have a derivative, like in this exercise, we are interested in reversing the differentiation process to find the original function. This is achieved through integration. The symbol for an indefinite integral is \( \int \), followed by the function we want to integrate and \( dt \) when our variable is \( t \). It is called indefinite because it includes a constant term, which we call the constant of integration. This form represents not just one, but an entire family of functions.

• To compute the indefinite integral of a function \( f(t) \), we find the function \( F(t) \) such that \( F'(t) = f(t) \).
• For the exercise given, \( r'(t) \) is \( \sec t \tan t - 1 \), and its indefinite integral is \( \int (\sec t \tan t - 1)\, dt \).
• This results in \( \sec t - t + C \), where \( C \) is the constant of integration.

Constant of Integration

The constant of integration, symbolized as \( C \), comes into play every time we evaluate an indefinite integral. This constant is essential because it encompasses all possible vertical shifts of the antiderivative on a graph. When differentiating, constants vanish, meaning we cannot retrieve unique information about the constant from just the derivative.

• In our exercise, after integrating \( \sec t \tan t - 1 \), we add \( C \) to account for this freedom in vertical shifts.
• To determine the exact value of \( C \), we utilize an additional piece of information, such as a specific point on the graph. This is often given in problems as a point \( P \), through which the function passes.
• For instance, using the point \( P(0, 0) \), we substitute into the integrated function because \( 0 = \sec(0) - 0 + C \). Solving gives us \( C = -1 \).

Derivative Integration

Derivative integration is the inverse operation of taking a derivative. "To integrate a derivative" means to find an original function whose derivative was given initially. This happens frequently in calculus. We transform a derivative back into its antiderivative or original function through the integration process.

• The process involves finding the antiderivative and adjusting it to meet given conditions.
• For our exercise, to transform the given derivative \( r'(t) = \sec t \tan t - 1 \) back to its original function \( r(t) \), we integrated to get \( r(t) = \sec t - t + C \).
• Given \( P(0,0) \) helps adjust this general outcome into a specific function \( r(t) = \sec t - t - 1 \), identifying the exact function related to the initial condition.