Math Problem Statement

Speedometer readings for a vehicle (in motion) at 7-second intervals are given in the table.

t (sec) 0 7 14 21 28 35 42 v (ft/s) 0 8 27 48 61 59 43

Estimate the distance traveled by the vehicle during this 42-second period using the Riemann sums L6, R6 and M3

Solution

To estimate the distance traveled by the vehicle over the 42-second interval, we'll use Riemann sums for approximating the area under the speed-time curve. This area represents the total distance traveled. We have three types of Riemann sums to calculate:

  1. Left Riemann Sum (L6) – Using left endpoints.
  2. Right Riemann Sum (R6) – Using right endpoints.
  3. Midpoint Riemann Sum (M3) – Using midpoints with fewer intervals.

Each time interval, Δt\Delta t, is 7 seconds.

Step 1: Left Riemann Sum (L6)

Using 6 intervals, we evaluate each interval's height based on the left endpoint.

L6=Δt(v(0)+v(7)+v(14)+v(21)+v(28)+v(35))L6 = \Delta t \cdot (v(0) + v(7) + v(14) + v(21) + v(28) + v(35))

Substitute the values: L6=7(0+8+27+48+61+59)=7203=1421 ftL6 = 7 \cdot (0 + 8 + 27 + 48 + 61 + 59) = 7 \cdot 203 = 1421 \text{ ft}

Step 2: Right Riemann Sum (R6)

Using 6 intervals, we evaluate each interval's height based on the right endpoint.

R6=Δt(v(7)+v(14)+v(21)+v(28)+v(35)+v(42))R6 = \Delta t \cdot (v(7) + v(14) + v(21) + v(28) + v(35) + v(42))

Substitute the values: R6=7(8+27+48+61+59+43)=7246=1722 ftR6 = 7 \cdot (8 + 27 + 48 + 61 + 59 + 43) = 7 \cdot 246 = 1722 \text{ ft}

Step 3: Midpoint Riemann Sum (M3)

Using 3 intervals, we evaluate each interval's height based on the midpoint of each interval.

The midpoints of each interval are:

  1. From t=0t = 0 to t=14t = 14: midpoint is t=7t = 7, so v(7)=8v(7) = 8
  2. From t=14t = 14 to t=28t = 28: midpoint is t=21t = 21, so v(21)=48v(21) = 48
  3. From t=28t = 28 to t=42t = 42: midpoint is t=35t = 35, so v(35)=59v(35) = 59

M3=2Δt(v(7)+v(21)+v(35))M3 = 2 \cdot \Delta t \cdot (v(7) + v(21) + v(35))

Substitute the values: M3=14(8+48+59)=14115=1610 ftM3 = 14 \cdot (8 + 48 + 59) = 14 \cdot 115 = 1610 \text{ ft}

Summary of Results:

  • Left Riemann Sum (L6): 1421ft1421 \, \text{ft}
  • Right Riemann Sum (R6): 1722ft1722 \, \text{ft}
  • Midpoint Riemann Sum (M3): 1610ft1610 \, \text{ft}

Would you like more details on any of these steps?

Related Questions

  1. How is the trapezoidal rule different from the Riemann sums for estimating distances?
  2. Why do we expect the Midpoint Riemann Sum to be more accurate than L6 and R6?
  3. How would the result change if we used a different interval width?
  4. Can Riemann sums be used to estimate the area under curves for non-uniform intervals?
  5. What practical applications use Riemann sums beyond distance estimation?

Tip

When estimating areas under curves, Midpoint Riemann sums generally offer better accuracy because they balance the overestimation and underestimation across each interval.

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Math Problem Analysis

Mathematical Concepts

Calculus
Riemann Sums
Distance Estimation

Formulas

Left Riemann Sum: L6 = Δt * (v(t0) + v(t1) + ... + v(t5))
Right Riemann Sum: R6 = Δt * (v(t1) + v(t2) + ... + v(t6))
Midpoint Riemann Sum: M3 = 2 * Δt * (v(t1) + v(t3) + v(t5))

Theorems

Riemann Sum

Suitable Grade Level

Undergraduate - Calculus I