Problem 40 - Thomas Calculus in SI Units

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: Identify the given derivative

The given derivative function is \( r'(t) = \sec t \tan t - 1 \). Our task is to find the original function \( r(t) \).

Step 2: Integrate the derivative

To find the function \( r(t) \), we need to integrate \( r'(t) = \sec t \tan t - 1 \).First, integrate \( \sec t \tan t \):\[ \int \sec t \tan t \, dt = \sec t + C \].Then, integrate \(-1\):\[ \int -1 \, dt = -t + C \].Thus, the integral of the entire expression becomes:\[ r(t) = \sec t - t + C \].

Step 3: Determine the constant of integration

Use the given point \( P(0,0) \) to find the constant \( C \).Substitute \( t = 0 \) and \( r(t) = 0 \) into the equation:\[ 0 = \sec(0) - 0 + C \].Since \( \sec(0) = 1 \), we have:\[ 0 = 1 + C \].Solving for \( C \):\[ C = -1 \].

Step 4: Write the final function

Substitute \( C = -1 \) back into the function:\[ r(t) = \sec t - t - 1 \].This is the function with the given derivative whose graph passes through the point \( P(0,0) \).

Key Concepts

Integration

Integration is an essential concept in calculus. It involves reversing the process of differentiation. When we have a derivative, integrating it helps us find the original function. This is why integration is also known as an "antiderivative."

In our exercise, we integrated the derivative function \( r'(t) = \sec t \tan t - 1 \). This was done step by step:

• First, we integrated \( \sec t \tan t \), which gave us \( \sec t + C \).
• We then integrated \(-1\), resulting in \(-t + C \).

Combining these results led us to the integral of the original function \( r(t) = \sec t - t + C \). Essentially, integration helps us recover a function from its rate of change, similar to assembling a full picture from its pieces.

Derivative

A derivative represents the rate of change of a function. In simpler terms, it tells us how a function is changing at any given point. It's a fundamental tool in calculus for understanding the dynamics of how things change.

In our exercise, we started with a given derivative function: \( r'(t) = \sec t \tan t - 1 \). Here:

• \( \sec t \tan t \) is part of the derivative that indicates a trigonometric rate of change.
• The constant \(-1\) suggests a uniform change in the opposite direction.

Finding the original function required integrating this derivative, essentially reversing the process to understand how the function develops from its derivative.

Constant of Integration

The constant of integration, often denoted as \( C \), emerges when we perform indefinite integrals. This constant is crucial because every derivative of a constant is zero, meaning there are infinite "families" of functions that could yield the same derivative.

In our example, after integrating \( r'(t) \) to find \( r(t) \), we had \( r(t) = \sec t - t + C \). However, because derivatives are indifferent to constant values, identifying this particular constant requires additional information.

Using the point \( P(0,0) \), we substituted \( t = 0 \) into \( r(t) \), leading us to solve for \( C \) as \(-1\). This kind of problem illustrates why the constant of integration is vital: it helps pinpoint the specific solution out of infinitely possible ones. The function ultimately becomes \( r(t) = \sec t - t - 1 \).

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are key elements in calculus. They frequently appear in problems involving waves, oscillations, or anything periodic. These functions have specific derivatives, which play prominently in calculus tasks.

In this exercise:

• The derivative \( r'(t) = \sec t \tan t \) uses the secant function, related to cosine, and the tangent function, derived from sine and cosine.
• Integrating this required knowledge of trigonometric identities: \( \int \sec t \tan t \, dt \) equals \( \sec t + C \).

Understanding these relationships is crucial. Trigonometric functions often present unique integration challenges, which can be mastered through practice and familiarity with these identities.