part one: given y=0.5e^-t -0.2e^-2t +0.1sint -0.3cos t
and another curve
y=Ae^-t + Be^-2t + 0.1e^t(sint -cost) A B constants
How do you know that for large values of t that the first one is bounded and the second one is unbounded? And what does bounded mean?
part two: Given x = Ae^-0.04t +Be^-0.02t +1875
and y = -Ae^-0.04t + Be^-0.02t +625
How do you know that they grow exponentially like a normal exponential. I would have thought that they would be reflected in the y or x axis(corrosponding to each) one. And the y curve has a dip in it, how are you supposed to know that.
Tanswers to these questions would be greatly aprecciated I have an exam soon and this is quite hard. Thanks!!!!
Not really. I don't think it means in relation to each function. The mark scheme says that the first one(x) is bounded oscillations and the second one (y) is unbounded oscillations. I can't figure out what it means by large t. I know this would make the first part of the functios zero but then for the sin and cos parts it doesn' really make sense. Is it referring to the fact that for th y function the line connecting its msximum points as t tends to large is a decsyng eponential in which case it is 'not bounded' and for the second one in theory it would be a straight line making it bounded? From an anayltical standpoint however putting a large value of t into the sin and cos values doesnt really get you anywhere,,,,,,is this correct what i'm thinking thnx
So for large values of t the first function approaches to simplify this you need to use an "R" formula. are you familiar with how to write as ? but it should be clear that is the a simple sine curve and cannot exceed a particular value.
for the second one you apply a similar argument so the function approaches then simplify into the form of so your function becomes .
as then by multiplying by you get . What can you deduce form this inequality ?
Oh ri thanks. So for the first one the max value is R. I still don't see how that makes it bounded erm and the second one if its limits are between |0.1Re^t| then how is it unbounded. I would have thought in theory this would decay and have limits(max and min) according to its exponential curve. Sill a bi confused sorry!
I attached a graph to illustrate this.
because the limits the seconds ones lies between increase without limits for large t. have you tired sketching the curve ? I have attach a image of that to help you get the general idea.the second one if its limits are between |0.1Re^t| then how is it unbounded.
oh right yeah!!! I guess this means it is unbounded by having an exponential boundary. Ok so ill use the form Rsin(t+/- alpha) since this is easier to analyse for larger t I suppose. K thanks alot I think I understand now if im correct in presuming this???