Problem 16 - Calculus Early Transcendentals

Question

The velocity of an object is given by the following functions on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into \(n\) subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles. $$v=\frac{t+3}{6}(\mathrm{m} / \mathrm{s}), \text { for } 0 \leq t \leq 4 ; n=4$$.

Step-by-Step Solution

Answer

Answer: The approximate displacement of the object is \(\frac{9}{2}\) meters.

Step 1: Subdivide the time interval

First, subdivide the time interval \(0 \leq t \leq 4\) into 4 subintervals. To do this, calculate the width of each subinterval as follows: $$\Delta t=\frac{4-0}{4}=1$$ This means that each subinterval will be of width 1.

Step 2: Identify the left endpoints of each subinterval

Next, identify the left endpoints for each subinterval. These endpoints will be used to evaluate the velocity function and determine the height of the rectangles. For a width of 1, the left endpoints are \(t_0 = 0, t_1 = 1, t_2 = 2,\) and \(t_3 = 3\).

Step 3: Calculate the height of the rectangles

Now, evaluate the velocity function at each left endpoint to calculate the height of the rectangles: $$v(t_0)=\frac{0+3}{6}=\frac{1}{2}$$ $$v(t_1)=\frac{1+3}{6}= \frac{2}{3}$$ $$v(t_2)=\frac{2+3}{6}= \frac{5}{6}$$ $$v(t_3)=\frac{3+3}{6}=1$$

Step 4: Calculate the area of the rectangles and sum them up

Now, multiply the height of each rectangle by its width (\(\Delta t=1\)) to calculate the area of each rectangle, and sum up the areas to approximate the displacement: $$\text{Displacement} \approx \sum_{i=0}^3 v(t_i) \cdot \Delta t$$ $$\text{Displacement} \approx (\frac{1}{2} + \frac{2}{3} + \frac{5}{6} + 1) \cdot 1 = \frac{9}{2} \text{ meters}$$ The approximate displacement of the object within the time interval 0 to 4 seconds is \(\frac{9}{2}\) meters using 4 left Riemann sums.

Key Concepts

Displacement

Displacement refers to the change in position of an object over a certain period of time. In the context of velocity and Riemann sums, displacement can be approximated by breaking the time interval into smaller segments, calculating the velocity at certain points, and multiplying each velocity value by the time segment's width. Displacement is essentially an integral of the velocity function over time, representing the total distance the object has moved from its starting point.

• To approximate displacement, one common method is using Riemann sums, which divide the time period into small intervals and sum up the areas.
• Each area gives an estimate of the distance covered by the object within that time span.
• By using the left endpoint of each segment to calculate the velocity, we arrive at a rough estimate of the total displacement.

So in summary, displacement involves calculating the aggregate movement over time, providing invaluable insight into how far an object travels in a given timeframe.

Velocity Function

A velocity function, such as the one given in the problem, describes how the velocity of an object changes with respect to time. It's expressed as a mathematical equation that allows us to calculate the velocity at any instant within a certain interval.

• In this example, the velocity function is given by \(v = \frac{t+3}{6}\; \text{m/s}\).
• This equation implies that the velocity depends linearly on the time \(t\).
• As time progresses, the velocity of the object increases, reflecting a linear relationship between time and speed.

The velocity function helps us understand not just the speed at a given moment but also how the object's speed changes over the course of the interval. By evaluating this function at various points, we can make informed predictions about the object's motion.

Subdivision Method

The subdivision method is fundamental to estimating integrals like displacement via Riemann sums. It involves breaking the period—such as the interval \([0,4]\)—into equal smaller subintervals. Using a consistent strategy, such as the left endpoint method, enhances the approximation's accuracy.

• In the problem, the interval is subdivided into four equal subintervals, each of width \(\Delta t = 1\).
• The left endpoints \(t_0 = 0, t_1 = 1, t_2 = 2, t_3 = 3\) are used to evaluate the velocity function.
• This approach leads to rectangles, where each rectangle’s height is determined by the velocity function at the left endpoint, and width is determined by the time span of each subdivision.

By summing the areas of these rectangles, the subdivision method simplifies the process of finding approximate values for integral-related calculations, such as displacement, without requiring complex calculus techniques.