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 ,  is not given
[At] t = 9, particle reaches final
position at D [on x-axis.]
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(a)
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Note: it is important to
consider  before
 .
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At point C, is not positive. This
is reached at by inspection of the graph of . This graph shows that for all values of t  , is not greater than zero.
 is not positive at
point C.
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At point C, is also not positive
at point C. Because is not greater than
zero, we know that the particle cannot “trace” back upon itself.
, because the particle is decreasing in y-magnitude from C  D in general, we
know that cannot be positive.
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(b)
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For  to be undefined, must = 0. Consider = 0.
From the graph of , = 0 only at t = 3 on  . At point B, t = 3.
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(c)
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t =
8  We know  (from slope of
tangent line).
@ t = 8 is  
=
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speed =  5.148 units/time unit
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(d)
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To find the distance
between A & D, we are only concerned with x values.
 = accumulated rate
of change of x = distance traveled
between t1 and t9
Distance between A and D =  39.255 units
Time it takes for the particle to get from A to D is from t = 0 to t =9.
At t = 0, the particle is at A. At t = 9, the particle is at D.
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Commentary by Mr. Hansen
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(a)
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Depending on how the rubric
is written, Isaac’s explanation may not be permitted. You cannot give as a
justification for the sign of an expression the observation that the graph
does not appear to have any positive values. Remember, you are not allowed to
“reason from a graph.” Safer explanation is to look at the two relevant
intervals of t values.
First, for  ,  goes from cos 0 to cos  (hence always  , and 0 only at t
= 3), while  goes from sin to sin  (hence always , and 0 only at t
= 3). [Reasoning like this is fine, by basic properties of trig functions.]
Second, for  , goes from cos to cos  (hence always  , and 0 only at t
= 3, 9), while goes from sin to  (hence always , and 0 only at t
= 3).
Therefore, for all  , the given expression for  is the product of –9
and two factors that never have a negative product. We conclude that is never positive.
Moreover, the product is 0 only at t
= 3 and t = 9. Thus is not positive
at C.
As for  , we know has the same sign as
at point C, since  is positive there by
inspection (slope of tangent line). Therefore is also not
positive at C. [Later on, after we know that point B corresponds to t = 3, we can say that and are both negative at
C. However, we have answered the questions posed, and we need to move on.]
[Note: This is a lot of work for what cannot possibly be more than 2 points
out of 9. Even though Isaac may not have earned both points, his method is a better
strategy, since it is time-efficient.]
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(b)
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Since both x(t) and y(t) are differentiable
[given], we know that the slope, , is undefined only if  . Isaac omitted only the first observation, and a picky rubric
would find fault with that, since one cannot assume in all cases that
undefined slope means denominator is zero.
Use of root-finder is permitted, and the only root on [0, 9] is at t = 3. Assuming that
part (b) is worth 2 points, Isaac’s solution would probably be 1 point out of
2.
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(c)
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Full points, probably 3 out
of 3.
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(d)
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Full points, probably 2 out
of 2. Note that a mere setup and answer would suffice:
39.255 units
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Final score for Isaac: 7 or
8 points out of 9, depending on how picky the rubric is. Either way, he is safely
in “AP 5” territory. Bravo!
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