Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
To approach this problem, let’s break it down step-by-step.
### Part (a): Using 4 Subintervals
Given the velocity function: [tex]\( v(t) = 2t^2 + 1 \)[/tex] for [tex]\( 0 \leq t \leq 4 \)[/tex].
1. Divide the interval [tex]\([0,4]\)[/tex] into 4 subintervals:
- [tex]\([0, 1]\)[/tex]
- [tex]\([1, 2]\)[/tex]
- [tex]\([2, 3]\)[/tex]
- [tex]\([3, 4]\)[/tex]
2. Determine the midpoint of each subinterval:
- Midpoint of [tex]\([0, 1]\)[/tex] is [tex]\( t = 0.5 \)[/tex]
- Midpoint of [tex]\([1, 2]\)[/tex] is [tex]\( t = 1.5 \)[/tex]
- Midpoint of [tex]\([2, 3]\)[/tex] is [tex]\( t = 2.5 \)[/tex]
- Midpoint of [tex]\([3, 4]\)[/tex] is [tex]\( t = 3.5 \)[/tex]
3. Evaluate the velocity function at each midpoint to find the constant velocity over each subinterval:
- [tex]\( v(0.5) = 2(0.5)^2 + 1 = 2(0.25) + 1 = 0.5 + 1 = 1.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.5) = 2(1.5)^2 + 1 = 2(2.25) + 1 = 4.5 + 1 = 5.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.5) = 2(2.5)^2 + 1 = 2(6.25) + 1 = 12.5 + 1 = 13.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.5) = 2(3.5)^2 + 1 = 2(12.25) + 1 = 24.5 + 1 = 25.5 \, \text{ft/s} \)[/tex]
4. Calculate the displacement over each subinterval:
Displacement is the velocity times the time interval length. Each subinterval length is 1 second.
- [tex]\([0, 1]\)[/tex]: [tex]\( 1.5 \times 1 = 1.5 \, \text{ft} \)[/tex]
- [tex]\([1, 2]\)[/tex]: [tex]\( 5.5 \times 1 = 5.5 \, \text{ft} \)[/tex]
- [tex]\([2, 3]\)[/tex]: [tex]\( 13.5 \times 1 = 13.5 \, \text{ft} \)[/tex]
- [tex]\([3, 4]\)[/tex]: [tex]\( 25.5 \times 1 = 25.5 \, \text{ft} \)[/tex]
5. Sum the displacements to get the total displacement:
[tex]\[ 1.5 + 5.5 + 13.5 + 25.5 = 46 \, \text{ft} \][/tex]
Therefore, the approximate displacement using [tex]\( n = 4 \)[/tex] subintervals is [tex]\( 46 \, \text{ft} \)[/tex].
### Part (b): Using 8 Subintervals
1. Divide the interval [tex]\([0, 4]\)[/tex] into 8 subintervals:
- Each subinterval is [tex]\([0, 0.5]\)[/tex], [tex]\([0.5, 1]\)[/tex], [tex]\([1, 1.5]\)[/tex], [tex]\([1.5, 2]\)[/tex], [tex]\([2, 2.5]\)[/tex], [tex]\([2.5, 3]\)[/tex], [tex]\([3, 3.5]\)[/tex], [tex]\([3.5, 4]\)[/tex].
2. Determine the midpoint of each subinterval:
- Midpoint of [tex]\([0, 0.5]\)[/tex] is [tex]\( t = 0.25 \)[/tex]
- Midpoint of [tex]\([0.5, 1]\)[/tex] is [tex]\( t = 0.75 \)[/tex]
- Midpoint of [tex]\([1, 1.5]\)[/tex] is [tex]\( t = 1.25 \)[/tex]
- Midpoint of [tex]\([1.5, 2]\)[/tex] is [tex]\( t = 1.75 \)[/tex]
- Midpoint of [tex]\([2, 2.5]\)[/tex] is [tex]\( t = 2.25 \)[/tex]
- Midpoint of [tex]\([2.5, 3]\)[/tex] is [tex]\( t = 2.75 \)[/tex]
- Midpoint of [tex]\([3, 3.5]\)[/tex] is [tex]\( t = 3.25 \)[/tex]
- Midpoint of [tex]\([3.5, 4]\)[/tex] is [tex]\( t = 3.75 \)[/tex]
3. Evaluate the velocity function at each midpoint to find the constant velocity:
- [tex]\( v(0.25) = 2(0.25)^2 + 1 = 2(0.0625) + 1 = 0.125 + 1 = 1.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(0.75) = 2(0.75)^2 + 1 = 2(0.5625) + 1 = 1.125 + 1 = 2.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.25) = 2(1.25)^2 + 1 = 2(1.5625) + 1 = 3.125 + 1 = 4.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.75) = 2(1.75)^2 + 1 = 2(3.0625) + 1 = 6.125 + 1 = 7.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.25) = 2(2.25)^2 + 1 = 2(5.0625) + 1 = 10.125 + 1 = 11.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.75) = 2(2.75)^2 + 1 = 2(7.5625) + 1 = 15.125 + 1 = 16.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.25) = 2(3.25)^2 + 1 = 2(10.5625) + 1 = 21.125 + 1 = 22.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.75) = 2(3.75)^2 + 1 = 2(14.0625) + 1 = 28.125 + 1 = 29.125 \, \text{ft/s} \)[/tex]
4. Calculate the displacement over each subinterval:
Each subinterval length is 0.5 seconds.
- [tex]\([0, 0.5]\)[/tex]: [tex]\( 1.125 \times 0.5 = 0.5625 \, \text{ft} \)[/tex]
- [tex]\([0.5, 1]\)[/tex]: [tex]\( 2.125 \times 0.5 = 1.0625 \, \text{ft} \)[/tex]
- [tex]\([1, 1.5]\)[/tex]: [tex]\( 4.125 \times 0.5 = 2.0625 \, \text{ft} \)[/tex]
- [tex]\([1.5, 2]\)[/tex]: [tex]\( 7.125 \times 0.5 = 3.5625 \, \text{ft} \)[/tex]
- [tex]\([2, 2.5]\)[/tex]: [tex]\( 11.125 \times 0.5 = 5.5625 \, \text{ft} \)[/tex]
- [tex]\([2.5, 3]\)[/tex]: [tex]\( 16.125 \times 0.5 = 8.0625 \, \text{ft} \)[/tex]
- [tex]\([3, 3.5]\)[/tex]: [tex]\( 22.125 \times 0.5 = 11.0625 \, \text{ft} \)[/tex]
- [tex]\([3.5, 4]\)[/tex]: [tex]\( 29.125 \times 0.5 = 14.5625 \, \text{ft} \)[/tex]
5. Sum the displacements to get the total displacement:
[tex]\[ 0.5625 + 1.0625 + 2.0625 + 3.5625 + 5.5625 + 8.0625 + 11.0625 + 14.5625 = 46.5 \, \text{ft} \][/tex]
Therefore, the approximate displacement using [tex]\( n = 8 \)[/tex] subintervals is [tex]\( 46.5 \, \text{ft} \)[/tex].
### Part (a): Using 4 Subintervals
Given the velocity function: [tex]\( v(t) = 2t^2 + 1 \)[/tex] for [tex]\( 0 \leq t \leq 4 \)[/tex].
1. Divide the interval [tex]\([0,4]\)[/tex] into 4 subintervals:
- [tex]\([0, 1]\)[/tex]
- [tex]\([1, 2]\)[/tex]
- [tex]\([2, 3]\)[/tex]
- [tex]\([3, 4]\)[/tex]
2. Determine the midpoint of each subinterval:
- Midpoint of [tex]\([0, 1]\)[/tex] is [tex]\( t = 0.5 \)[/tex]
- Midpoint of [tex]\([1, 2]\)[/tex] is [tex]\( t = 1.5 \)[/tex]
- Midpoint of [tex]\([2, 3]\)[/tex] is [tex]\( t = 2.5 \)[/tex]
- Midpoint of [tex]\([3, 4]\)[/tex] is [tex]\( t = 3.5 \)[/tex]
3. Evaluate the velocity function at each midpoint to find the constant velocity over each subinterval:
- [tex]\( v(0.5) = 2(0.5)^2 + 1 = 2(0.25) + 1 = 0.5 + 1 = 1.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.5) = 2(1.5)^2 + 1 = 2(2.25) + 1 = 4.5 + 1 = 5.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.5) = 2(2.5)^2 + 1 = 2(6.25) + 1 = 12.5 + 1 = 13.5 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.5) = 2(3.5)^2 + 1 = 2(12.25) + 1 = 24.5 + 1 = 25.5 \, \text{ft/s} \)[/tex]
4. Calculate the displacement over each subinterval:
Displacement is the velocity times the time interval length. Each subinterval length is 1 second.
- [tex]\([0, 1]\)[/tex]: [tex]\( 1.5 \times 1 = 1.5 \, \text{ft} \)[/tex]
- [tex]\([1, 2]\)[/tex]: [tex]\( 5.5 \times 1 = 5.5 \, \text{ft} \)[/tex]
- [tex]\([2, 3]\)[/tex]: [tex]\( 13.5 \times 1 = 13.5 \, \text{ft} \)[/tex]
- [tex]\([3, 4]\)[/tex]: [tex]\( 25.5 \times 1 = 25.5 \, \text{ft} \)[/tex]
5. Sum the displacements to get the total displacement:
[tex]\[ 1.5 + 5.5 + 13.5 + 25.5 = 46 \, \text{ft} \][/tex]
Therefore, the approximate displacement using [tex]\( n = 4 \)[/tex] subintervals is [tex]\( 46 \, \text{ft} \)[/tex].
### Part (b): Using 8 Subintervals
1. Divide the interval [tex]\([0, 4]\)[/tex] into 8 subintervals:
- Each subinterval is [tex]\([0, 0.5]\)[/tex], [tex]\([0.5, 1]\)[/tex], [tex]\([1, 1.5]\)[/tex], [tex]\([1.5, 2]\)[/tex], [tex]\([2, 2.5]\)[/tex], [tex]\([2.5, 3]\)[/tex], [tex]\([3, 3.5]\)[/tex], [tex]\([3.5, 4]\)[/tex].
2. Determine the midpoint of each subinterval:
- Midpoint of [tex]\([0, 0.5]\)[/tex] is [tex]\( t = 0.25 \)[/tex]
- Midpoint of [tex]\([0.5, 1]\)[/tex] is [tex]\( t = 0.75 \)[/tex]
- Midpoint of [tex]\([1, 1.5]\)[/tex] is [tex]\( t = 1.25 \)[/tex]
- Midpoint of [tex]\([1.5, 2]\)[/tex] is [tex]\( t = 1.75 \)[/tex]
- Midpoint of [tex]\([2, 2.5]\)[/tex] is [tex]\( t = 2.25 \)[/tex]
- Midpoint of [tex]\([2.5, 3]\)[/tex] is [tex]\( t = 2.75 \)[/tex]
- Midpoint of [tex]\([3, 3.5]\)[/tex] is [tex]\( t = 3.25 \)[/tex]
- Midpoint of [tex]\([3.5, 4]\)[/tex] is [tex]\( t = 3.75 \)[/tex]
3. Evaluate the velocity function at each midpoint to find the constant velocity:
- [tex]\( v(0.25) = 2(0.25)^2 + 1 = 2(0.0625) + 1 = 0.125 + 1 = 1.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(0.75) = 2(0.75)^2 + 1 = 2(0.5625) + 1 = 1.125 + 1 = 2.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.25) = 2(1.25)^2 + 1 = 2(1.5625) + 1 = 3.125 + 1 = 4.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(1.75) = 2(1.75)^2 + 1 = 2(3.0625) + 1 = 6.125 + 1 = 7.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.25) = 2(2.25)^2 + 1 = 2(5.0625) + 1 = 10.125 + 1 = 11.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(2.75) = 2(2.75)^2 + 1 = 2(7.5625) + 1 = 15.125 + 1 = 16.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.25) = 2(3.25)^2 + 1 = 2(10.5625) + 1 = 21.125 + 1 = 22.125 \, \text{ft/s} \)[/tex]
- [tex]\( v(3.75) = 2(3.75)^2 + 1 = 2(14.0625) + 1 = 28.125 + 1 = 29.125 \, \text{ft/s} \)[/tex]
4. Calculate the displacement over each subinterval:
Each subinterval length is 0.5 seconds.
- [tex]\([0, 0.5]\)[/tex]: [tex]\( 1.125 \times 0.5 = 0.5625 \, \text{ft} \)[/tex]
- [tex]\([0.5, 1]\)[/tex]: [tex]\( 2.125 \times 0.5 = 1.0625 \, \text{ft} \)[/tex]
- [tex]\([1, 1.5]\)[/tex]: [tex]\( 4.125 \times 0.5 = 2.0625 \, \text{ft} \)[/tex]
- [tex]\([1.5, 2]\)[/tex]: [tex]\( 7.125 \times 0.5 = 3.5625 \, \text{ft} \)[/tex]
- [tex]\([2, 2.5]\)[/tex]: [tex]\( 11.125 \times 0.5 = 5.5625 \, \text{ft} \)[/tex]
- [tex]\([2.5, 3]\)[/tex]: [tex]\( 16.125 \times 0.5 = 8.0625 \, \text{ft} \)[/tex]
- [tex]\([3, 3.5]\)[/tex]: [tex]\( 22.125 \times 0.5 = 11.0625 \, \text{ft} \)[/tex]
- [tex]\([3.5, 4]\)[/tex]: [tex]\( 29.125 \times 0.5 = 14.5625 \, \text{ft} \)[/tex]
5. Sum the displacements to get the total displacement:
[tex]\[ 0.5625 + 1.0625 + 2.0625 + 3.5625 + 5.5625 + 8.0625 + 11.0625 + 14.5625 = 46.5 \, \text{ft} \][/tex]
Therefore, the approximate displacement using [tex]\( n = 8 \)[/tex] subintervals is [tex]\( 46.5 \, \text{ft} \)[/tex].
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
