Problem 13
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. $$v=\frac{1}{2 t+1}(\mathrm{m} / \mathrm{s}), \text { for } 0 \leq t \leq 8 ; n=4$$.
Step-by-Step Solution
Verified Answer
Answer: The approximate displacement of the object over the specified interval is \(1.248\) meters.
1Step 1: Determine the width of each subinterval
Since there are 4 subintervals and the total range is \(8\), the width of each subinterval is \(\Delta t = \frac{8}{4} = 2\).
2Step 2: Calculate the left endpoints' times
The left endpoint of each subinterval represents the time at which we'll evaluate the velocity function. For our 4 subintervals with a width of 2, the left endpoints are: \(t_0=0\), \(t_1=2\), \(t_2=4\), \(t_3=6\).
3Step 3: Evaluate the velocity function at the left endpoints
Using the velocity function \(v(t) = \frac{1}{2t+1}\), we evaluate it at the left endpoints:
\(v(t_0) = v(0) = \frac{1}{2(0)+1} = 1\)
\(v(t_1) = v(2) = \frac{1}{2(2)+1} = \frac{1}{5}\)
\(v(t_2) = v(4) = \frac{1}{2(4)+1} = \frac{1}{9}\)
\(v(t_3) = v(6) = \frac{1}{2(6)+1} = \frac{1}{13}\)
4Step 4: Calculate the Riemann sum
Now, we multiply each velocity value by the width of the subinterval and then sum the results:
Displacement \(\approx \Delta t[v(t_0) + v(t_1) + v(t_2) + v(t_3)]\)
Displacement \(\approx 2[1 + \frac{1}{5} + \frac{1}{9} + \frac{1}{13}]\)
Displacement \(\approx 2[\frac{365}{585}]\)
5Step 5: Simplify the final result
Multiply and simplify the expression for the displacement:
Displacement \(\approx \frac{730}{585} \approx 1.248\) m
The approximate displacement of the object over the specified interval is \(1.248\) meters.
Key Concepts
Velocity FunctionDisplacement ApproximationSubinterval Partitioning
Velocity Function
A velocity function represents the rate of change of an object's position with respect to time. In mathematics and physics, this function is crucial for describing how the speed of an object varies over time. In the given exercise, the velocity function is defined as
\( v(t) = \frac{1}{2t+1} \),
where \( t \) is the time in seconds, and \( v(t) \) is the velocity in meters per second (m/s). It is essential to understand that, in this context, velocity denotes instantaneous speed, meaning the rate at which an object is traveling at a specific instance in time.
By inputting different values of time into this function, one can determine the velocity of the object at those exact moments. For example, if you substitute time \( t = 0 \) into the function, you'd find that the velocity is \( v(0) = 1 \) m/s, indicating that at the very start, the object is moving at a speed of 1 meter per second.
\( v(t) = \frac{1}{2t+1} \),
where \( t \) is the time in seconds, and \( v(t) \) is the velocity in meters per second (m/s). It is essential to understand that, in this context, velocity denotes instantaneous speed, meaning the rate at which an object is traveling at a specific instance in time.
By inputting different values of time into this function, one can determine the velocity of the object at those exact moments. For example, if you substitute time \( t = 0 \) into the function, you'd find that the velocity is \( v(0) = 1 \) m/s, indicating that at the very start, the object is moving at a speed of 1 meter per second.
Displacement Approximation
Displacement approximation is a method used to estimate the total movement of an object over a certain time interval when its velocity varies throughout that interval. Instead of exact calculations, which might require complex integrations, we can use various numerical methods. In particular, the problem at hand uses a Riemann sum to approximate the displacement.
The essence of displacement approximation is to calculate the area under the velocity-time graph, which represents the object's displacement. This is typically done by dividing the total time interval into smaller subintervals, evaluating the velocity function at specific points within these subintervals, and then summing the areas of rectangles constructed with these velocities as heights. As the number of subintervals increases, the approximation typically becomes more accurate because the rectangles better conform to the shape of the curve.
The essence of displacement approximation is to calculate the area under the velocity-time graph, which represents the object's displacement. This is typically done by dividing the total time interval into smaller subintervals, evaluating the velocity function at specific points within these subintervals, and then summing the areas of rectangles constructed with these velocities as heights. As the number of subintervals increases, the approximation typically becomes more accurate because the rectangles better conform to the shape of the curve.
Subinterval Partitioning
Subinterval partitioning is essential in the process of estimating the area under a curve, which, in this context, is applied to calculate the displacement of an object. The technique involves dividing the total interval into smaller 'subintervals.' Here, the partition of the interval \(0 \leq t \leq 8\) is done by creating four equal subintervals, each of length \(\Delta t = 2\).
By selecting a point within each subinterval, here the left endpoint, and evaluating the velocity function at these points, we can construct rectangles. The area of each rectangle is then a product of the interval width and the velocity at the left endpoint. Summing up the areas of these rectangles forms a Riemann sum, which provides the approximate displacement. The Riemann sum can be thought of as a collection of guesses of what the actual displacement might be, with each rectangle's area serving as one such guess.
By selecting a point within each subinterval, here the left endpoint, and evaluating the velocity function at these points, we can construct rectangles. The area of each rectangle is then a product of the interval width and the velocity at the left endpoint. Summing up the areas of these rectangles forms a Riemann sum, which provides the approximate displacement. The Riemann sum can be thought of as a collection of guesses of what the actual displacement might be, with each rectangle's area serving as one such guess.
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