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September 18th, 2010, 07:28 AM   #1
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Slope field from graph

These are all similar questions, so as not to clutter up the forum, they're all in one post. Q15 is the important one (Q14 and 16 I'm pretty confident about, but would like to confirm).

Q15: This graph (without the axis numbers, though) is given in the y x f(y) plane. Question: Suppose we know that the graph is the rhs, f(y), of the diffeq dy/dt=f(y). Give a rough sketch of the slope field.

I'm confused by the axis choices (y is horz axis, f(y) is vert axis). In all prev. questions, the horz axis was t. Since the rhs has no t terms, the slope depends solely on y. I was tempted to make all the vertically aligned slope marks parallel, but the book's answer is given in the ty-pane, and looks like this. I am as confused as can be on this one. Any clues at all?

Q18: This is a trick question or am I so smart that it seems simple (actually, I know the second part of that statement is false ):
y(t)=2 for all t is the solution to dy/dt=f(t,y). (a) What does that tell you about the slope field (how much of it can you sketch)? (b) What about the solutions?

Can sketch all of it. The slope of y(t)=2 is zero for all t. The whole slope field is horz lines. There is only one solution, a flat line at level y(t)=the initial value.

Q14: This graph (without the numbers on the axis) is given on the t x f(t)-plane (horz axis is t, vert axis f(t)). The question: Suppose we know that this is the graph of the right hand side of f(t) of the diff eq dy/dt=f(t). Give a rough sketch of the corresponding slope field.

Since the graph corresponds to f(t), which is the rhs of dy/dt, and the rhs does not contain any y terms, I think the slope marks at a given t are all the same slope. IOW, all vertically aligned slope marks are parallel. So if the peak of the graph is at t_1<0, then all slope marks at t_1 are parallel and have slope=0 (all the way up and down). All slope marks at 0<t_2=the valley are also slope=0. For all values of t, the slope marks will be parallel to the graph shown.

Q16: A graph in the ty-plane with the shape of a cosine function is given. Point y(0)=1 is marked; it is the peak of one of the cycles. Suppose we know the graph is of a solution to dy/dt=f(t): (a) How much of the slope field can you sketch (hint: the diffeq only depends on t); (b) What can you say about the solution with y(0)=2?

(a) This slope field is dependent only on t, so all vertically aligned slope marks are parallel and they should be parallel to the shown graph at the same value of t.
(b) y(0)=2 probably has the identical solution, but with double the amplitude. The slope marks will mostly be steeper, except near the peaks/valleys.
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September 18th, 2010, 08:35 AM   #2
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Re: Slope field from graph

For question 18, is it not possible that the solution given is a trivial solution, where y(t) = 2 is a horizontal asymptote of the actual solution space? For a simple example, consider:

y'(t) = y(t) - 2

where the solution is:

y = Ce^t + 2

when C = 0, then y(t) = 2 for all t. However, all other choices for C are not horizontal lines.
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