Problem 24 - Calculus Early Transcendentals

Question

The velocity (in \(\mathrm{mi} / \mathrm{hr}\) ) 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 \(s(0)=0\) and \(t\) is measured in hours. a. Determine and graph the position function, for \(0 \leq t \leq 4\) (Hint: \(\sin ^{2} t=\frac{1}{2}(1-\cos 2 t)\) b. What is the distance traveled by the hiker in the first 15 min of the hike? c. What is the hiker's position at \(t=3 ?\)

Step-by-Step Solution

Answer

In this problem, the hiker's motion along the trail is represented by the velocity function, \(v(t) = 3 \sin^2(\frac{\pi t}{2})\). To find the position function, we integrated the velocity function and obtained \(s(t) = \frac{3}{2}(t - \frac{1}{\pi}\sin(\pi t))\). a) To graph the position function, plot the function over the given interval [0, 4]. b) To find the distance traveled in the first 15 minutes (0.25 hours) of the hike, we calculated the definite integral of the absolute value of the velocity function from 0 to 0.25: The hiker travels about 0.3572 miles in the first 15 minutes of the hike. c) To find the hiker's position at t=3 hours, we plugged t=3 into the position function: The hiker's position at t=3 is 4.5 miles along the trail.

Step 1: a. Determine and graph the position function for 0 ≤ t ≤ 4

To determine the position function, we need to integrate the velocity function. Given the hint, we can rewrite the velocity function as: \(v(t) = 3 \sin^2(\frac{\pi t}{2}) = \frac{3}{2} (1 - \cos(\pi t))\) Now, we find the integral with respect to t: \(s(t) = \int v(t) dt = \int \frac{3}{2}(1 - \cos(\pi t)) dt\) Now, integrate each term separately: \(s(t) = \frac{3}{2}(t - \int \cos(\pi t) dt)\) To find the integral of \(cos(\pi t)\), we can use a substitution: \(u = \pi t\) \(\frac{du}{dt} = \pi\) \(dt = \frac{du}{\pi}\) Now, replace t with u and dt with du: \(\int \cos(\pi t) dt = \int \cos(u) \cdot \frac{du}{\pi}\) Now, integrate: \(= \frac{1}{\pi}\sin(u) + C\) Now, resubstitute back to t: \( = \frac{1}{\pi}\sin(\pi t) + C\) Now, plug in the initial condition \(s(0) = 0\): \(0 = \frac{3}{2}(0 - \frac{1}{\pi}\sin(0) + C)\) \(C = 0\) So, the position function is: \(s(t) = \frac{3}{2}(t - \frac{1}{\pi}\sin(\pi t))\) To graph the function, plot the function over the given interval [0, 4].

Step 2: b. What is the distance traveled by the hiker in the first 15 minutes of the hike?

To find the distance traveled in 15 minutes (0.25 hours), we need to find the definite integral of the velocity function from 0 to 0.25: \(distance = \int_{0}^{0.25} v(t) dt = \int_{0}^{0.25} \frac{3}{2}(1 - \cos(\pi t)) dt\) \( = \frac{3}{2} (t - \frac{1}{\pi} \sin(\pi t)) |_0^{0.25}\) \( = \frac{3}{2}(0.25 - \frac{1}{\pi} \sin(\frac{\pi}{4})) - \frac{3}{2}(0 - \frac{1}{\pi} \sin(0))\) \(distance = \frac{3}{8} - \frac{3}{2 \pi} \approx 0.3572\) miles The hiker travels about 0.3572 miles in the first 15 minutes of the hike.

Step 3: c. What is the hiker's position at t=3?

To find the hiker's position at t=3, simply plug t=3 into the position function: \(s(3) = \frac{3}{2}(3 - \frac{1}{\pi}\sin(3\pi))\) Since \(\sin(3\pi) = 0\): \(s(3) = \frac{3}{2}(3 - 0)\) \(s(3) = 4.5\) miles The hiker's position at t=3 is 4.5 miles along the trail.

Key Concepts

Velocity

In calculus, **velocity** is a measure of how fast something is moving along a straight path. It's the rate of change of position with respect to time. When given a velocity function like \( v(t) = 3 \sin^2(\frac{\pi t}{2}) \), it describes how the hiker's speed varies over time. The function tells us how fast the hiker is walking at any given moment along their trek.
It's essential to understand that velocity is a vector quantity. This means it includes both a magnitude (how fast) and a direction (which way). In this problem, since the hiker walks along a straight trail, the direction is implicit and constant. But note, the velocity function itself gives only the speed, its direction here is constant.
When interpreting velocity functions, it's important to know:

• It can be positive, which means moving forward.
• It can also be zero or even negative, indicating rest or movement in the opposite direction, respectively.

Position Function

The **position function** gives the exact position of an object at any given time. It's derived by integrating the velocity function over time. If you know the initial position of the object (\(s(0)\)), you can determine its position at any time \(t\).
In our exercise, the position function \( s(t) \) was found by integrating the velocity function \( \frac{3}{2}(1 - \cos(\pi t)) \). This integration tells us how far the hiker has traveled at any point in time compared to where they started. The process begins with:
\[ s(t) = \int v(t) \, dt = \int \frac{3}{2}(1 - \cos(\pi t)) \, dt \]
Having integrated and applied the condition \(s(0) = 0\), we determined that:
\[ s(t) = \frac{3}{2} \left(t - \frac{1}{\pi}\sin(\pi t)\right) \]
This equation graphically represents the hiker's progress along the trail.

Integration

**Integration** is a fundamental concept in calculus, used to find the area under a curve, among other things. It essentially accumulates quantities to find totals, like distance from velocity. In our problem, integration allows us to convert the velocity function \( v(t) \) into the position function \( s(t) \).
Remember the basic steps of integration:

• Rewrite the given function in a suitable form (like using trigonometric identities).
• Integrate each term independently.
• Apply any known initial conditions to solve for constants.

Integration helps us transition from how fast something is going (velocity) to how far it has gone (position). The mathematical beauty of integration is in its ability to handle complex problems, providing clear, cumulative results.

Definite Integral

A **definite integral** provides the accumulated value of a function over a specific interval. In the context of motion, it can tell us the total distance traveled, as explored in our exercise for the hiker's first 15 minutes. To find the distance traveled in that window, we computed \[ \int_{0}^{0.25} v(t) \, dt \].
The definite integral calculates the total change in position over the interval between 0 and 0.25 hours. It answers how far the hiker journeyed in a specific time.
Key things to note about definite integrals:

• The upper and lower limits define the time period we're analyzing.
• It considers the function's exact area under its curve within those limits.
• Doesn't depend on any arbitrary constant because boundaries replace them.

This technique is crucial for analyzing real-world motion scenarios where pinpointing exact travel distances is necessary.