Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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**Problem:**

The velocity (in feet/second) of a projectile \( t \) seconds after it is launched from a height of 10 feet is given by \( v(t) = -15.6t + 147 \). Approximate its height after 3 seconds using 6 rectangles.

**Solution:**

First, we need to use the given velocity function to approximate the height of the projectile after 3 seconds. The velocity function is \( v(t) = -15.6t + 147 \).

To approximate the height after 3 seconds using 6 rectangles, we need to use the concept of Riemann sums where the number of rectangles (n) is 6.

Below are the steps to proceed:

1. **Divide the interval \([0, 3]\) into 6 subintervals:**
   - The width of each subinterval, \(\Delta t\) = \(\frac{3 - 0}{6} = 0.5\) seconds.

2. **Calculate the midpoints of each subinterval:**
   - For each subinterval, the midpoint, \( t_i \), can be calculated as follows:
     - \( t_1 = \frac{0 + 0.5}{2} = 0.25 \)
     - \( t_2 = \frac{0.5 + 1.0}{2} = 0.75 \)
     - \( t_3 = \frac{1.0 + 1.5}{2} = 1.25 \)
     - \( t_4 = \frac{1.5 + 2.0}{2} = 1.75 \)
     - \( t_5 = \frac{2.0 + 2.5}{2} = 2.25 \)
     - \( t_6 = \frac{2.5 + 3.0}{2} = 2.75 \)

3. **Evaluate the velocity function at each midpoint:**
   - \( v(t_1) = v(0.25) = -15.6(0.25) + 147 \)
   - \( v(t_2) = v(0.75) = -15.6(0.75) + 147 \)
   - \( v(t_3) = v(1.25) = -15.6
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Transcribed Image Text:**Problem:** The velocity (in feet/second) of a projectile \( t \) seconds after it is launched from a height of 10 feet is given by \( v(t) = -15.6t + 147 \). Approximate its height after 3 seconds using 6 rectangles. **Solution:** First, we need to use the given velocity function to approximate the height of the projectile after 3 seconds. The velocity function is \( v(t) = -15.6t + 147 \). To approximate the height after 3 seconds using 6 rectangles, we need to use the concept of Riemann sums where the number of rectangles (n) is 6. Below are the steps to proceed: 1. **Divide the interval \([0, 3]\) into 6 subintervals:** - The width of each subinterval, \(\Delta t\) = \(\frac{3 - 0}{6} = 0.5\) seconds. 2. **Calculate the midpoints of each subinterval:** - For each subinterval, the midpoint, \( t_i \), can be calculated as follows: - \( t_1 = \frac{0 + 0.5}{2} = 0.25 \) - \( t_2 = \frac{0.5 + 1.0}{2} = 0.75 \) - \( t_3 = \frac{1.0 + 1.5}{2} = 1.25 \) - \( t_4 = \frac{1.5 + 2.0}{2} = 1.75 \) - \( t_5 = \frac{2.0 + 2.5}{2} = 2.25 \) - \( t_6 = \frac{2.5 + 3.0}{2} = 2.75 \) 3. **Evaluate the velocity function at each midpoint:** - \( v(t_1) = v(0.25) = -15.6(0.25) + 147 \) - \( v(t_2) = v(0.75) = -15.6(0.75) + 147 \) - \( v(t_3) = v(1.25) = -15.6
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