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Magazine Chapter 4: Problem 42
Find the function with the given derivative whose graph passes through the point \(P\). $$r^{\prime}(t)=\sec t \tan t-1, \quad P(0,0)$$
Short Answer
Expert verified
The function is \( r(t) = \sec t - t - 1 \).
Step by step solution
01
Identify the Given Derivative
The problem provides the derivative of the function, which is \( r'(t) = \sec t \tan t - 1 \). We need to find the function \( r(t) \) whose derivative matches this expression.
02
Integrate the Derivative
To find the function \( r(t) \), we need to integrate its derivative: \[ r(t) = \int (\sec t \tan t - 1) \, dt.\] We integrate \( \sec t \tan t \) separately from \( -1 \).
03
Integration of \( \sec t \tan t \)
The integral of \( \sec t \tan t \) is straightforward as it is a standard result: \[ \int \sec t \tan t \, dt = \sec t + C_1,\] where \( C_1 \) is an integration constant.
04
Integration of \(-1\)
Now, integrate \(-1\) with respect to \( t \):\[\int -1 \, dt = -t + C_2,\] where \( C_2 \) is another integration constant.
05
Combine the Integrals
Combining the results from steps 3 and 4, we have: \[ r(t) = \sec t - t + C.\] Here, \( C \) is the combined constant of integration, \( C = C_1 + C_2 \).
06
Apply the Initial Condition
Using the point \( P(0, 0) \), substitute \( t = 0 \) and \( r(t) = 0 \) into the function we found:\[0 = \sec(0) - (0) + C.\] Since \( \sec(0) = 1 \), this equation simplifies to:\[0 = 1 + C \Rightarrow C = -1.\]
07
Write the Final Function
Substitute \( C = -1 \) back into the function: \[ r(t) = \sec t - t - 1.\] This is the function whose derivative is \( \sec t \tan t - 1 \) and passes through the point \( P(0,0) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Derivatives
In calculus, a derivative represents the rate of change of a function concerning its variable. It tells us how a function changes as its input changes. The derivative can be thought of as the "slope" or "steepness" of the function at any given point.
- The derivative is often denoted using the prime symbol. For example, for a function \( r(t) \), it's derivative is \( r'(t) \).
- In this exercise, the given derivative is \( r'(t) = \sec t \tan t - 1 \). This tells us how the function \( r(t) \) changes as \( t \) changes.
- The goal is to determine the original function \( r(t) \) by integrating its derivative.
Role of the Constant of Integration
When integrating a function, you'll encounter something called the constant of integration, often represented as \( C \). This constant arises because integration is, in a sense, the "reverse" process of taking a derivative.
- The derivative of a constant is zero, so when we work backwards by integrating, any constant value could potentially match the original function, leading to the "+C" term in our integral solution.
- In our example, when finding \( r(t) \), we separately integrated \( \sec t \tan t \) and \(-1\). Each part generated its own constant of integration.
- The constants from each part combine into a single constant \( C \) when we compile our function: \[ r(t) = \sec t - t + C \].
The Importance of Initial Conditions
Initial conditions are crucial for pinning down the exact form of a function when integrating a derivative. They provide specific details that help us determine the constant of integration. In simpler terms, an initial condition is a known value that a function must satisfy.
- In the problem exercise, we were given the initial condition \( P(0,0) \), which means at \( t = 0 \), \( r(t) \) equals zero.
- We used this condition to solve for \( C \) in \( r(t) = \sec t - t + C \) by substituting \( r(0) = 0 \) and \( t = 0 \).
- This let us solve \( 0 = 1 + C \), leading to \( C = -1 \), thus determining the precise function: \( r(t) = \sec t - t - 1 \).
