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.
Whoops, just realized I can't edit it.
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
#1. the average value is just the area divided by the width. It's just the height of a rectangle with the same area as under the graph.
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
Hey, I know a lot of these concepts are pretty basic so I sound really dumb but I didn't really get a solid hold on them before, yknow, I missed the whole unit, but I'll try my best to answer them with your help so please check these!
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
1. River Currents:
A. To estimate the average river current speed from 6:00 AM until 6:40 AM using trapezoids, you can start by dividing the time interval into smaller intervals, such as every 10 minutes. For each interval, calculate the average of the upper and lower current speeds, then multiply that average by the duration (10 minutes in this case). Add up the products for all intervals and divide by the total duration (40 minutes) to get the average current speed.
Here's an example calculation for the first interval from 6:00 AM to 6:10 AM:
Average current speed = (0 mph + 1.5 mph) / 2 = 0.75 mph
Duration = 10 minutes
Product = 0.75 mph * 10 minutes = 7.5 mile
Repeat this calculation for all intervals and then sum up the products. Finally, divide the sum by the total duration (40 minutes) to find the average current speed.
B. To estimate the time at which the river current had the average velocity, you can choose a point within the time interval and determine its average speed. If it matches the average current speed calculated in part A, then that is the estimated time. Alternatively, you can calculate the average current speed for smaller intervals within the time interval, such as every minute, to get a more accurate estimate of the time when the river current had the average velocity.
C. To determine how far south the bottle will be at 6:40 AM, you can estimate the distance traveled by multiplying the average current speed (calculated in part A) by the duration (40 minutes). If there are different average speeds within the time interval, you will need to consider the different speeds during the time the bottle would have traveled.
D. To find the average acceleration of the bottle from 6:00 AM to 6:40 AM, you need to know the initial velocity of the bottle at 6:00 AM and the final velocity of the bottle at 6:40 AM. With these values, you can use the formula for average acceleration: average acceleration = (change in velocity) / (time interval). Note that there might be changes in the velocity of the current during the time interval, so the average acceleration could vary.
2. Region R:
A. To find the area of region R, you can calculate the integral of the function y = 1/x^3 from x = 1 to x = 3. Integration will yield the area bounded by the curve, the x-axis, and the vertical lines x = 1 and x = 3.
B. To find the value of h such that the vertical line x = h divides region R into two equal areas, you can set up an equation and solve for h. The equation should involve integrating the function y = 1/x^3 from x = 1 to x = h, setting it equal to half of the total area of R found in part A, and then solving for h.
C. To find the volume of the solid generated when region R is revolved about the x-axis, you can use the method of cylindrical shells or disks. Set up an integral that represents the volume of an infinitesimally thin shell or disk and integrate it from x = 1 to x = 3.
D. To find the value of k such that the vertical line x = k divides region R into two regions that generate solids with equal volumes when revolved about the x-axis, you can set up an equation similar to part B, but this time integrating from x = 1 to x = k and setting it equal to half of the total volume of the solid generated when R is revolved about the x-axis.