Problem 13 - Applied Calculus

Question

Your velocity is \(v(t)=\ln \left(t^{2}+1\right) \mathrm{ft} / \mathrm{sec}\) for \(t\) in \(\mathrm{sec}-\) onds, \(0 \leq t \leq 3\). Estimate the distance traveled during this time.

Step-by-Step Solution

Answer

The estimated distance using numerical integration methods is approximately 3.7 feet.

Step 1: Understand the Problem

We need to estimate the distance traveled by a particle moving with velocity \( v(t) = \ln(t^2 + 1) \) ft/sec over a time interval from \( t = 0 \) to \( t = 3 \). This requires calculating the integral of the velocity function over this interval.

Step 2: Set up the Integral

The distance traveled is the integral of the velocity function \( v(t) \) with respect to \( t \) over the interval [0, 3]. Thus, we need to calculate:\[ \int_{0}^{3} \ln(t^2 + 1) \, dt \]

Step 3: Use Numerical Integration

Since the integral \( \int \ln(t^2+1) \, dt \) does not have a simple closed form, we will use numerical methods such as the Trapezoidal Rule or Simpson’s Rule to estimate it. For this session, we will assume a trapezoidal numerical estimation, dividing the time interval [0, 3] into \( n \) subintervals for accuracy.

Step 4: Calculate Using the Trapezoidal Rule

Divide the interval [0, 3] into \( n \) equal subintervals of width \( \Delta t = \frac{3}{n} \). The Trapezoidal Rule estimates the integral as:\[ \int_{0}^{3} \ln(t^2 + 1) \, dt \approx \frac{\Delta t}{2} \left[ \ln(0^2+1) + 2\sum_{i=1}^{n-1} \ln((i\Delta t)^2 + 1) + \ln(3^2+1) \right] \]Choose a reasonable \( n \), such as 6 or 10, for a rough estimate.

Step 5: Estimate with Chosen \( n \)

For instance, choosing \( n = 6 \), \( \Delta t = 0.5 \). Now compute each term:- \( \ln(0^2+1) = \ln(1) = 0 \)- \( \ln(3^2+1) = \ln(10) \)- Intermediate terms: \( \ln(0.5^2+1), \ln(1^2+1), \ln(1.5^2+1), \ln(2^2+1), \ln(2.5^2+1) \)Substitute these into the trapezoidal formula to compute the estimate.

Key Concepts

Trapezoidal Rule

The Trapezoidal Rule is a simple numerical method used to approximate the definite integral of a function over a specified interval. It is particularly handy when the integral does not have a known closed form.
By approximating the region under the curve as a series of trapezoids instead of calculus-based areas, it allows easy calculation.

• First, the interval \([a, b]\) is divided into \(n\) equally spaced segments.
• The width of each segment is \(\Delta t = \frac{b-a}{n}\).
• The approximate integral is given by: \[ \int_{a}^{b} f(t) \, dt \approx \frac{\Delta t}{2} \left[f(a) + 2(f(x_1) + f(x_2) + \ldots + f(x_{n-1})) + f(b)\right] \]

In our exercise, we use the Trapezoidal Rule to estimate the integral of the velocity function \(v(t) = \ln(t^2 + 1)\) over the interval \[0, 3\].
This method becomes more accurate as \(n\), the number of subintervals, increases, capturing the function's behavior more precisely.

Simpson's Rule

Simpson's Rule is another powerful numerical integration technique used to estimate the area under a curve.
This rule is generally more accurate than the Trapezoidal Rule when the function is reasonably smooth over the interval.

• Simpson's Rule works by fitting a quadratic (parabola) to each pair of subintervals, effectively capturing the curve's changes.
• The formula for Simpson's Rule is \[ \int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{3} \left[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \ldots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)\right] \]

The choice between Simpson's Rule and other numerical methods, like the Trapezoidal Rule, typically depends on the required precision. When the velocity function \(v(t) = \ln(t^2 + 1)\) is interpolated with a parabola, Simpson's Rule can deliver a closer estimate to the true integral, especially when the function varies significantly in shape over the interval.

Velocity Function

The velocity function in a physical context describes the speed at which an object travels over time. It is a function of time \(t\).
In our exercise, the velocity function is given as \(v(t) = \ln(t^2 + 1)\) in feet per second.

• This describes how fast and in what direction the object is moving at any given point between \(t = 0\) and \(t = 3\).
• The natural logarithmic form indicates that the velocity might change rapidly or gradually as time progresses.

Understanding the velocity function is critical since the definite integral of this function over a given time interval gives the total distance traveled. Interpreting these kinds of functions involves considering both their magnitude (speed) and mathematical behavior (how the logarithm affects changes).

Definite Integral

A definite integral is a fundamental concept in calculus that allows us to compute the net area under a curve over a specific interval. For our purposes, it is the key to determining the total distance traveled when the velocity function is known.

• The notation for a definite integral from \(a\) to \(b\) of a function \(f(x)\) is \[ \int_{a}^{b} f(x) \, dx \]
• In our exercise, \(\int_{0}^{3} \ln(t^2 + 1) \, dt\) represents the total distance the particle travels from \(t = 0\) to \(t = 3\).

Definite integrals provide a precise way to accumulate or aggregate quantities, such as distance, over time. They are key to solving real-world problems in physics and engineering, where understanding changes over intervals is crucial. By estimating these integrals through numerical methods when an analytical solution is difficult or impossible, we can still gain meaningful insights from them.