#$&* course MTH 279 2/25 11:48 am q_a_08110216
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Given Solution: Self-critique: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Solve the equation y ' = 3 y^2 * sqrt(t) for the initial condition y(1) = .5. Evaluate your solution for t = 1.1 and t = 1.2, and plot the resulting two points on your graph of the direction field. How close was the first new point obtained in the preceding problem to the actual solution? How close was the second new point obtained in the preceding problem to the actual solution? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1/y^2= 3 sqrt t dt -1/y= 2t^1.5+C y= C-1/2t^-1.5
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Given Solution: Self-critique: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. We could get the t = 1.1 approximation by using the same process, with two intervals, each of .05. What is your approximation to the y value for t = 1.05, based on the slope at (1, .5)? What is your approximation to the y value for t = 1.10, based on the slope at the point you just found? How far from the known solution is your new t = 1.10 approximation? Originally we got to t = 1.1 by a single step. We have now used two steps to get a new approximation. By what factor did our approximation error change? Did doubling the number of steps reduce the error in the approximation by a factor of two, less than two or more than two? Why did the approximation error change as it did? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 0.5375 y= 0.5666 0.567-0.5666= 0.0004 So much closer to and actually over the actual value. Our error decreased by 1/20 or 0.05 It was reduced significantly greater than 2. It changed in relation to the value of the interval, that is with a 0.1 interval we had a 0.1 error and with at 0.05 interval we had a 0.05 error. So we divided our by 1/2 but our error changed by 1/20.
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Given Solution: Self-critique: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. BB data: Shot at the same trajectory the BB follows a path similar to that of a good sand wedge shot in golf, but I believe it starts out with greater velocity and loses speed more quickly. In the sunlight against a blue sky you can track the thing for just about its entire flight. Pretty neat. Shot at the Moon, without correcting for gravity, and didn't really miss it by all that much. Don't want to steal NASA's thunder to I'll leave it at that. Shot into water at a low enough trajectory it will skip without losing too much speed. Shot at a nearly vertical trajectory the water stops it within a few centimeters. All this appears to be consistent with the gun's rating of 0.3 Joules and the BB's .12 gram mass (which is in turn consistent with the 250 ft/sec muzzle velocity). At that speed the BB should climb for about 1000 feet, if shot in a vacuum, which is what leads me to believe it's losing velocity. How could differential equations be used to provide models for this discussion? What tests are suggested by the solutions of those equations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have the Fnet Eqs that involve drag changing to the square of the velocity. Of course a drag coefficient would be involved to distinguish from the water and the air. Seems that coefficient would be related to density, maybe. We could use energy considerations since we have an energy rating and that would involve KE, PE and Work by NonCons Forces, namely the fluid resistance. Actually a Bernoulli's eq comes to mind for some reason, I guess just thinking about an object moving through a fluid but that's not the same as the motion of the fluid. Maybe something like: 1/2mv^2 - mgy - W_NC where v = dy/dt = y' and W_NC = F_frict*dy Or m dv/dt - kv Sub in dv/dy for the vertical shots and k is drag coefficient.