Discover a wealth of information and get your questions answered on IDNLearn.com. Ask your questions and get detailed, reliable answers from our community of experienced experts.
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 value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
