Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Suppose velocity (in miles per hour) is given by \(v(t)=3 t\), where \(t\) is measured in hours. We are interested in the distance traveled from \(t=0\) to \(t=k\), where \(k\) is a constant. (a) By solving the differential equation \(d s / d t=3 t\) and using the initial condition \(s(0)=s_{0}\), nd the distance function \(s(t) .\) Using \(s(t)\), nd i. \(s(0)\). ii. \(s(k)\). iii. the distance traveled between \(t=0\) and \(t=k\). (b) Find the area under the graph of \(v(t)\) from \(t=0\) to \(t=k\). Verify that your answers to part (a) iii and (b) are the same.

Step-by-Step Solution

Answer

The function for the distance travelled s(t) is \(1.5t^2 + s_0\). For \(t = 0\) it is \(s_0\) and for \(t = k\) it is \(1.5k^2 + s_0\). The distance travelled from \(t = 0\) to \(t = k\) is \(1.5k^2\), which coincides with the area under the function \(v(t) = 3t\) from 0 to k.

Step 1: Solve differential equation

The differential equation given is \(ds / dt = 3t\), which can be rewritten as \(ds = 3tdt\). Integrating both sides gives the function \(s(t) = 1.5t^2 + s_0\), where \(s_0\) is the constant of integration.

Step 2: Evaluate for s(0)

With the initial condition \(s(0) = s_0\), evaluating \(s(t)\) when \(t = 0\) gives \(s(0) = 1.5*0^2 + s_0 = s_0\). Therefore, the initial distance \(s_0\) for \(t = 0\) is \(s_0\). It usually represents the starting distance which, in the context of this problem, is usually 0.

Step 3: Evaluate for s(k)

To determine \(s(k)\), substitute \(t = k\) into \(s(t) = 1.5t^2 + s_0\). This gives \(s(k) = 1.5k^2 + s_0\). The result represents the distance at time \(t = k\).

Step 4: Calculate the distance travelled

The distance traveled from \(t = 0\) to \(t = k\) is given by the difference \(s(k) - s(0)\). This evaluates to \(1.5k^2 + s_0 - s_0 = 1.5k^2\).

Step 5: Determine area under the velocity function

The area under the velocity function, which represents the distance traveled, is given by the integral of the velocity function \(v(t) = 3t\) from \(t = 0\) to \(t = k\). Calculating this integral gives the result \(\frac{1}{2}*3t^2\) evaluated from \(t = 0\) to \(t = k\), which equals \(1.5k^2\).

Step 6: Verify results

Comparing the results obtained in Step 4 and Step 5, it's observed that both are \(1.5k^2\), thus confirming the equivalence of the methods and satisfying the condition mentioned in part (b) of the problem.

Key Concepts

velocity

Velocity is a measure of how fast something is moving in a specific direction. In our exercise, the velocity is given by the function \(v(t) = 3t\). Here, \(v(t)\) represents velocity in miles per hour, and \(t\) is the time in hours. This tells us that as time progresses, the velocity increases linearly.
Understanding velocity as a function is crucial. It not only tells us the speed, but also how the speed changes over time. For example, if \(t = 1\), the velocity is \(3 \, \text{mph}\), and when \(t = 2\), the velocity doubles to \(6 \, \text{mph}\). This shows the object's speed is continuously increasing as time goes by.
Velocity is essential in solving differential equations because it links closely with distance through integration. By knowing how velocity changes over time, we can determine the distance traveled over a given period.

distance traveled

Distance traveled is the total length of the path an object covers during its motion. To find this in the exercise, we start by solving the differential equation \(\frac{ds}{dt} = 3t\). This equation states that the rate of change of distance \(s(t)\) with respect to time is equal to the velocity at any time \(t\).
To find the distance function \(s(t)\), we integrate both sides of this equation, resulting in \(s(t) = 1.5t^2 + s_0\), where \(s_0\) is the initial distance—a constant, often zero in such problems.
Let's break this down:

• Start at \(t = 0\) by calculating \(s(0)\). With \(s_0 = 0\), \(s(0) = 0\), showing that the distance is zero at the start.
• For any time \(t = k\), evaluate \(s(k) = 1.5k^2 + s_0\). If \(s_0\) is zero, \(s(k)\) gives us the total distance traveled from the start to time \(k\).
• The distance traveled is the difference between \(s(k)\) and \(s(0)\), simplifying to \(1.5k^2\).

This demonstrates how we use integration to move from velocity to distance, underscoring the relationship between velocity and cumulative distance.

integration

Integration is a fundamental concept in calculus that helps in calculating quantities like the distance traveled from a varying velocity. In the exercise, we use integration to transform the rate of change in position, given as \(\frac{ds}{dt} = 3t\), into the distance function \(s(t)\).
Integration involves finding the antiderivative, or the "inverse," of a derivative. For our velocity function \(3t\), integrating with respect to time \(t\) yields a new function: \(s(t) = 1.5t^2 + s_0\).
Key points about integration:

• Integrating velocity provides a distance function, which tells us the position of the object at any time \(t\).
• It requires solving for the constant of integration, \(s_0\), using given initial or boundary conditions, such as \(s(0) = 0\).
• In practical terms, integration allows us to calculate quantities like areas under curves, which is another method to find the distance traveled.

Overall, integration is a powerful tool that bridges the gap between changing rates (velocity) and accumulated totals (distance). It fundamentally underpins our ability to model and solve problems involving motion.

area under a curve

The area under a curve in a velocity-time graph represents the distance traveled. In our exercise, calculating this area provides the distance traversed from \(t=0\) to \(t=k\) by integrating the velocity function \(v(t) = 3t\).
Here's how it works:

• The integral of \(v(t) = 3t\) from \(t=0\) to \(t=k\) symbolizes the total distance, or the area under the curve, from the start to time \(k\).
• The calculation is carried out by solving \[ \int_{0}^{k} 3t \, dt = \left[ 1.5t^2 \right]_0^k = 1.5k^2 - 0 = 1.5k^2 \].
• This solution matches the distance function obtained through integration in earlier steps, suggesting a consistent method to determine distance.

Understanding this concept is important because it illustrates how graphical representation of velocity can elucidate physical movement.
This method not only ties back to the algebraic solution, confirming the distance traveled, but also enhances comprehension of how graphical methods relate to real world movement.