Question

The velocity (in miles/hour) of a hiker walking along a straight trail is given by \(v(t)=3 \sin ^{2}(\pi t / 2),\) for \(0 \leq t \leq 4\) Assume that \(s(0)=0\). a. Determine and graph the position function, for \(0 \leq t \leq 4\). b. What is the distance traveled by the hiker in the first 15 min of the hike? (Hint: \(\sin ^{2} t=\frac{1}{2}(1-\cos 2 t)\) .) c. What is the hiker's position at \(t=3 ?\)

Short answer

Question: Write the position function, distance traveled in the first 15 minutes, and the hiker's position at \(t = 3\) for a hiker with a velocity function \(v(t) = 3\sin^2(\pi t/2)\) and initial position \(s(0) = 0\). Answer: The position function is \(s(t) =\frac{3}{2}(t - \frac{1}{\pi}\sin(\pi t))\). The distance traveled in the first 15 minutes is \(\frac{3}{8}\) miles. The hiker's position at \(t=3\) is \(\frac{9}{2}\) miles.

Step-by-step solution

Find the position function

To find the position function, integrate the velocity function with respect to time: $$s(t) = \int v(t) dt = \int 3\sin^2(\pi t / 2) dt.$$

Use the trigonometric identity

Replace \(\sin^2 t\) using the given identity: \(\sin^2 t = \frac{1}{2}(1 - \cos2t)\). So, $$s(t) = \int 3\cdot\frac{1}{2}(1 - \cos(\pi t)) dt. $$

Solve the integral

To solve the integral, first distribute the constants and then integrate term by term: $$s(t) = \frac{3}{2} \int (1-\cos(\pi t))dt = \frac{3}{2} (t - \frac{1}{\pi}\sin(\pi t)) + C.$$ As \(s(0) = 0\), we have \(\frac{3}{2} \cdot 0 - \frac{3}{2\pi} \cdot \sin(0) + C = 0\). So, the position function is: $$s(t) =\frac{3}{2}(t - \frac{1}{\pi}\sin(\pi t)).$$ Now that we have the position function, we can answer the other parts of the question.

Find the distance traveled in the first 15 min

To find the distance traveled in the first 15 minutes, first convert the time to hours (since velocity is in miles/hour): \(15 \,\text{min} = 15/60=1/4 \,\text{hour}.\) Now, integrate the absolute value of the velocity function from 0 to 1/4: $$\int_0^{1/4} |v(t)| dt = \int_0^{1/4} 3\sin^2(\pi t/2) dt$$ $$ = 3 \int_0^{1/4} \frac{1}{2}(1-\cos(\pi t)) dt.$$

Evaluate the integral

Evaluate the integral using the antiderivative obtained in Step 3: $$3\cdot\frac{1}{2}(\left[t - \frac{1}{\pi}\sin(\pi t)\right]_0^{1/4})= \frac{3}{8}$$ So, the hiker travels a distance of \(\frac{3}{8}\) miles in the first 15 minutes.

Find the hiker's position at \(t=3\)

To find the hiker's position at \(t=3\), simply plug in \(t=3\) into the position function: $$s(3) = \frac{3}{2}(3 - \frac{1}{\pi}\sin(3\pi))= \frac{9}{2}.$$ So, the hiker's position at \(t=3\) is \(\frac{9}{2}\) miles.

Key concepts

Position Function

Let's delve into the concept of a position function, which is a crucial topic in calculus for modeling and understanding movement. In simple terms, a position function, denoted as \(s(t)\), represents the location of an object along a path as a function of time \(t\). It is fundamental in tracking how far and in what direction an object has traveled.
To find the position function given a velocity function, one must integrate the velocity function. In our example, the velocity function is \(v(t)=3 \sin ^{2}(\pi t / 2)\). Upon integration, it becomes:
\[s(t) = \int 3 \sin^2(\pi t / 2) dt.\]To complete the integral, we'd replace \(\sin^2 t\) using a trigonometric identity. The resultant position function, \(s(t)\), lets us calculate specific positions at given times and discuss the motion over time accurately.

Velocity Function

A velocity function describes how fast an object's position changes over time. It tells us the rate at which the position is moving at any given point in time. In calculus, it is derived from the derivative of the position function, \(s(t)\).
For our exercise, the velocity function provided is \(v(t)=3\sin^2(\pi t / 2)\). By analyzing this velocity function, we can predict the hiker's speed as time progresses from \(0\) to \(4\) hours. This data is crucial for understanding periods of acceleration or rest.
In practice, integrating this function will yield the position function, allowing us to deduce where the hiker is situated along the trail at different times.

Trigonometric Identities

Trigonometric identities are mathematical tools simplifying complex trigonometric expressions. They are essential in calculus, particularly when working with functions involving sine, cosine, and other trigonometric functions.

• In this exercise, we employ the identity \(\sin^2 t = \frac{1}{2}(1 - \cos 2t)\) to simplify the velocity function \(v(t)=3\sin^2(\pi t/2)\).

By simplifying it using identities, integration and further calculations become more manageable.
Understanding and applying these identities help in transforming challenging integrals or derivatives into more comfortable forms, making them less convoluted and easier to solve.

Integration

Integration is a fundamental concept in calculus, used to find areas under curves, among other things. It is the reverse operation of differentiation.
In the given exercise, integration helps us transform the velocity function into the position function. We compute:
\[s(t) = \int v(t) dt = \int 3\sin^2(\pi t / 2) dt.\]Using integration, we first identify the antiderivative and then apply boundaries, confirming constants by initial conditions (e.g., \(s(0)=0\)).

• This process allows for precise calculations of motion, giving insights into the trajectory and behavior of moving objects.

Understanding integration deeply can enhance problem-solving skills in various mathematical and real-world scenarios.

Distance Traveled

Distance traveled is another critical output of calculus, indicating the total length covered by an object over time, regardless of its direction.
In this context, it is determined by integrating the absolute value of the velocity function over the given time interval. For the hiker's first 15 minutes (converted to 0.25 hours), the calculation is:
\[\int_0^{1/4} |v(t)| dt = 3 \int_0^{1/4} \frac{1}{2}(1-\cos(\pi t)) dt.\]
After evaluating the integral, we find that the distance traveled is \(\frac{3}{8}\) miles.

• This concept is key to understanding the actual path length, which considers all changes, stops, and reversals in direction, calculated as total movement without regard to direction.

It showcases the utility of calculus in providing a complete picture of motion dynamics.