Problem 18 - Applied Calculus

Question

Your velocity is given by \(v(t)=t^{2}+1\) in \(\mathrm{m} / \mathrm{sec},\) with \(t\) in Eeconds. Estimate the distance, \(s\), traveled between \(t=0\) and \(t=5 .\) Explain how you arrived at your estimate.

Step-by-Step Solution

Answer

The estimated distance traveled is approximately 46.67 meters.

Step 1: Understanding the Problem

To find the distance traveled, we need to calculate the integral of the velocity function over the given time interval. The velocity function is given as \( v(t) = t^2 + 1 \).

Step 2: Set Up the Integral

The distance traveled is the integral of the velocity function from \( t = 0 \) to \( t = 5 \). This is represented as: \[s = \int_{0}^{5}(t^2 + 1)\,dt\]

Step 3: Integrate the Function

We need to compute the integral \( \int (t^2 + 1)\, dt \). The antiderivative of \( t^2 \) is \( \frac{t^3}{3} \) and the antiderivative of \( 1 \) is \( t \). Therefore, the integral becomes \( \frac{t^3}{3} + t \).

Step 4: Evaluate the Definite Integral

Evaluate the antiderivative from \( 0 \) to \( 5 \). This results in: \[s = \left[ \frac{t^3}{3} + t \right]_{0}^{5} = \left( \frac{5^3}{3} + 5 \right) - \left( \frac{0^3}{3} + 0 \right)\]

Step 5: Calculate the Result

Calculate \( \frac{5^3}{3} + 5 \):\( \frac{5^3}{3} = \frac{125}{3} \approx 41.67 \) and \( 5 \) results in:\( s = 41.67 + 5 = 46.67 \).

Step 6: Conclude the Estimate

The estimated distance traveled between \( t = 0 \) and \( t = 5 \) is approximately 46.67 meters.

Key Concepts

Velocity Function

The velocity function describes how an object's speed changes over time. In this exercise, the velocity function is given by \( v(t) = t^2 + 1 \), representing the velocity in meters per second. Here, \( t \) stands for time in seconds. This quadratic function tells us how fast the object is moving at any given moment.

The term \( t^2 \) suggests that the object's velocity increases as time progresses, indicating acceleration. Meanwhile, the constant \( +1 \) ensures that the velocity is never below 1 meter per second, even at \( t = 0 \).

• At \( t = 0 \), the velocity is 1 m/s.
• At \( t = 1 \), velocity becomes \( 1^2 + 1 = 2 \) m/s.
• Similarly, at \( t = 5 \), it is \( 5^2 + 1 = 26 \) m/s.

Understanding this function helps calculate how much distance the object covers over a time interval by integrating it.

Definite Integral

A definite integral is used to calculate the total area under a curve, or in this case, the total distance an object travels over a time interval. In mathematical terms, the definite integral of a function \( f(t) \) from \( a \) to \( b \) is written as:
\[\int_{a}^{b} f(t)\]In this context, we are using the definite integral to find the total distance traveled by integrating the velocity function \( v(t) = t^2 + 1 \) from \( t = 0 \) to \( t = 5 \). This integral represents the sum of all the infinitesimal changes in distance over this time period.

The definite integral provides an exact measure for calculating total travel distance based on how velocity changes over time. In this case, it translates into finding the integral from \( t = 0 \) to \( t=5 \):

• \( \int_{0}^{5} (t^2 + 1) \, dt \)

Calculating this integral gives us the precise distance the object has traveled.

Distance Calculation

Distance calculation using integration involves determining how far an object has traveled over a specific time interval by evaluating the definite integral of its velocity function. Once we set up the definite integral to calculate the distance, we focus on solving it.

First, we find the antiderivative of our given velocity function \( v(t) = t^2 + 1 \). The antiderivative is calculated as:

• For \( t^2 \), the antiderivative is \( \frac{t^3}{3} \)
• For the constant \( 1 \), it's \( t \)

Thus, the integral becomes \( \frac{t^3}{3} + t \). We then proceed to evaluate this antiderivative from \( t = 0 \) to \( t = 5 \):
\[t = \left\{ \left( \frac{5^3}{3} + 5 \right) - \left( \frac{0^3}{3} + 0 \right)\right\}\]Calculating this gives an approximate distance of 46.67 meters traveled. This method ensures an accurate computation of the total distance, taking into account the changes in velocity over time.

Antiderivative

An antiderivative, in calculus, is a function whose derivative gives back another function. In simpler terms, if you take a derivative of an antiderivative, you get the original function back. In the exercise at hand, we use the antiderivative to solve the definite integral of the velocity function to determine distance.

Finding the antiderivative is a key step in solving integrals. For the function \( v(t) = t^2 + 1 \), the antiderivatives are:

• \( t^2 \) becomes \( \frac{t^3}{3} \)
• and \( 1 \) simply becomes \( t \)

This transforms the problem of finding the integral into evaluating \( \frac{t^3}{3} + t \) across the desired interval.Using the antiderivative effectively converts a function that describes a rate, such as velocity, into a value that can be used to calculate total change, such as distance. It's a pivotal concept ensuring that we can transition from understanding how fast something moves to how much ground it actually covers.