Hi, I missed school due to illness (and my school isn't very forgiving) so I would really appreciate some help answering these questions, tbh I have no idea how to do these (it was a short unit and I missed all of it) but If I just see how they are solved it would be a huge help and give me some time to study further. (also I'm sorry for the overflow but here is a part 2 if anyone would be willing to check it out [will update link as soon as I post it])
1. River currents (miles per hour) at a certain location are given below. The current direction at this location was “from the north” during the time interval shown, and the current did not exhibit any severe fluctuations other than those shown in the chart.
(chart) gyazo.com/59426eeda2730d4514647ac00a2db734
A. Using trapezoids, estimate the average river current speed from the north from 6:00 AM until 6:40 AM.
B. At approximately what time between 6:00 AM and 6:40 AM would you estimate that the river current had the average velocity?
C. A message in a bottle near this location is released at 6:00 AM. Assuming that the bottle travels along with the river’s current, approximately how far south will the bottle be at 6:40 AM?
D. What is the average acceleration of this bottle from 6:00 AM to 6:40 AM?
2. The region R is bounded by the x-axis, x = 1, x = 3, and y = 1/x^3.
A. Find the area of R.
B. Find the value of h, such that the vertical line x = h divides the region R into two Regions of equal area.
C. Find the volume of the solid generated when R is revolved about the x-axis.
D. The vertical line x = k divides the region R into two regions such that when these two regions are revolved about the x-axis, they generate solids with equal volumes. Find the value of k.
3 answers
Here is part 2, if anyone is willing to take a look at it
jiskha.com/questions/1788953/I-know-it-looks-like-a-lot-but-its-just-10-questions-part-1-It-also-explains
A. So, assuming you know how to find the area of a trapezoid, the first one has area 10(17+20)/2 = 175
figure the others, add them up, and divide by the width: 40
B. just draw the graph and estimate where its height is the average speed from A.
C. distance is the area under the speed graph
D. since the final speed is the same as the initial speed, no net acceleration.
#2. as always, sketch the graph. The area is just
A. a = ∫[1,3] 1/x^3 dx = -1/(2x^2) [1,3] = -1/18 + 1/2 = 4/9
B. You just want to find h such that
∫[1,h] 1/x^3 dx = ∫[h,3] 1/x^3 dx
-1/(2h^2) + 1/2 = -1/18 + 1/(2h^2)
1/h^2 = 5/9
h = 3/√5 ≈ 1.34
C. using discs of thickness dx,
v = ∫[1,3] πr^2 dx
where r=y=1/x^3
v = ∫[1,3] π(1/x^3)^2 dx = π∫[1,3]1/x^6 dx = -π/(5x^5) [1,3] = 242π/1215
Or, using shells of thickness dy,
v = ∫[1/27,1] 2πrh dy + π(1/27)^2 * 2
where r=y and h = x-1 = 1/∛y - 1
v = ∫[1/27,1] 2πy(1/∛y - 1) dy = 242π/1215
You have to add the cylinder of radius 1/27 to the hollow shape roated under the curve from y=1/27 to y=1
D. huh. we'll stick with discs for this part. As with #1, we want h such that
π∫[1,k]1/x^6 dx = π∫[k,3]1/x^6 dx
k = 3/121^(1/5) ≈ 1.15
1.
A.
area = (10/2) (17 + 2(20) + 2(22) + 2(21) + 17)
area = 5 (160)
area = 800
800/40 = 20
The average river current speed from the north from 6:00 AM until 6:40 AM, was 20mph.
B. (graph: gyazo.com/f7846a75d364fa0a2e0c71cf0173b991)
At around 6:10 AM and 6:32-6:33 (6:325), the river current had the average velocity.
C. Would that be the same area from earlier? 800 miles?
D. Are net acceleration and average acceleration not different?
2.
A. a = ∫[1,3] 1/x^3 dx = -1/(2x^2) [1,3] = -1/18 + 1/2 = 4/9
the area of r is 4/9
B. h = 3/√5 ≈ 1.34
C. I'm a bit confused on C, would the answer not just be 242π/1215?
"You have to add the cylinder of radius 1/27 to the hollow shape rotated under the curve from y=1/27 to y=1" how do I do this?
D. k = 3/121^(1/5) ≈ 1.15