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Chapter 5: Problem 24

A particle starts out from the origin. Its velocity at time \(t\) is given by $$v(t)=4 t^{3}+2 t$$ Estimate how far it travels between the times \(t=0\) and \(t=3\) using a Riemann sum with \(n=6\).

Short Answer

Expert verified
The particle travels approximately 63.75 units.

Step by step solution

01

Understand the problem

We need to estimate the distance traveled by a particle moving along a line. We are given the velocity function \(v(t) = 4t^3 + 2t\) and need to estimate the distance using a Riemann sum with \(n=6\) from \(t=0\) to \(t=3\).
02

Determine the width of each subinterval

The total interval length is from \(t=0\) to \(t=3\), so the length \([a, b] = [0, 3]\). We divide this interval into 6 equal parts, each with a width of \[ \text{width} = \frac{b-a}{n} = \frac{3-0}{6} = 0.5 \]
03

Identify the sample points

For the left Riemann sum, we use the left endpoint of each subinterval. The sample points are: 0, 0.5, 1, 1.5, 2, and 2.5.
04

Calculate the velocity at each sample point

Evaluate the velocity function \(v(t)\) at each sample point:\[ \begin{align*} v(0) & = 4(0)^3 + 2(0) = 0 \ v(0.5) & = 4(0.5)^3 + 2(0.5) = 4(0.125) + 1 = 1.5 \ v(1) & = 4(1)^3 + 2(1) = 4 + 2 = 6 \ v(1.5) & = 4(1.5)^3 + 2(1.5) = 13.5 + 3 = 16.5 \ v(2) & = 4(2)^3 + 2(2) = 32 + 4 = 36 \ v(2.5) & = 4(2.5)^3 + 2(2.5) = 62.5 + 5 = 67.5\end{align*} \]
05

Multiply the velocities by the width of the subintervals and sum

Multiply each velocity by the width (0.5), then sum these products: \[ \begin{align*} \text{Distance} & = 0.5[v(0) + v(0.5) + v(1) + v(1.5) + v(2) + v(2.5)] \ & = 0.5[0 + 1.5 + 6 + 16.5 + 36 + 67.5] \ & = 0.5[127.5] \ & = 63.75 \text{units} ewline \end{align*} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance Traveled
To understand how far an object moves, we need to consider its velocity over time.
By summing small chunks of distance where the velocity is approximately constant, we can estimate the total distance.
In this exercise, we're using a Riemann sum to achieve this.
The total distance traveled is equivalent to the area under the velocity-time graph between the given intervals.
Velocity Function
The velocity function describes the speed and direction of the particle at any given time.
In our problem, the velocity function is given as: \[ v(t)=4 t^{3}+2 t \].
This function lets us calculate the velocity of the particle at specific points in time.
By evaluating this function at certain time intervals, we get the particle鈥檚 velocity at those moments.
Subinterval Width
Subinterval width is the length of each of the small intervals into which we divide the total time span.
Here, the total time interval is from \[ t=0 \] to \[ t=3 \], and we divide it into 6 parts, hence \[ n=6 \].
The width of each subinterval can be found using the formula: \[ \text{width} = \frac{b-a}{n} = \frac{3-0}{6} = 0.5 \].
This width will be used to multiply the velocities to approximate the distances in each subinterval.
Sample Points
Sample points are specific points at which we evaluate the velocity function.
For a left Riemann sum, we use the left-end points of each subinterval.
The sample points in this exercise are: 0, 0.5, 1, 1.5, 2, and 2.5.
These points help us to determine the approximate distances traveled in each subinterval.
Left Riemann Sum
The left Riemann sum is a method to approximate the integral of a function.
It sums up rectangular areas under the curve, using the left end of each subinterval to evaluate the function.
In our exercise: \[ \text{Distance} = \text{width} \times [v(0) + v(0.5) + v(1) + v(1.5) + v(2) + v(2.5)] \].
Finally, this sum gives us an estimate for the distance traveled, which in this case, is \[ 63.75 \text{units} \].

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