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Material Type: Exam; Professor: Ellermeyer; Class: Calculus III; University: Kennesaw State University; Term: Fall 2005;
Typology: Exams
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MATH 2203 ñExam 2 (Version 2) Solutions September 28, 2005
S. F. Ellermeyer Name
Instructions. Your work on this exam will be graded according to two criteria: mathe-
matical correctness and clarity of presentation. In other words, you must know what
you are doing (mathematically) and you must also express yourself clearly. In particular,
write answers to questions using correct notation and using complete sentences where
appropriate. Also, you must supply su¢ cient detail in your solutions (relevant calculations,
written explanations of why you are doing these calculations, etc.). It is not su¢ cient to just
write down an ìanswerî with no explanation of how you arrived at that answer. As a rule
of thumb, the harder that I have to work to interpret what you are trying to say, the less
credit you will get. You may use your calculator but you may not use any books or notes.
r 1 (t) =
p 3 t; 6 t 2
p 3 12
t + 14 8
p 3
and r 2 (t) =
3 t 2 ;
p 3 t; 8
Do the two particles collide? If so, at what time do they collide?
Solution: In order for the particles to be in the same place at the same time, there must be some time t such that all of the equations
3 t 2 = 3 p 3 t =
p 3 t
8 = 6t 2
p 3 12
t + 14 8
p 3
are satisÖed.
The Örst of these equations is satisÖed for t = 1 and so is the second. Let us see if t = 1 satisÖes the third equation:
2
p 3 12
p 3 = 6 + 8
p 3 12 + 14 8
p 3 = 8.
Thus the particles do collide at time t = 1.
r (t) =
3 t 2 ;
p 3 t; 8
at the point on the curve corresponding to t = 1.
Solution: Since r 0 (t) =
6 t;
p 3 ; 0
and jr 0 (t)j =
p 36 t^2 + 3,
the unit tangent vector (at any point on the curve) is
T (t) =
jr^0 (t)j
r 0 (t) =
6 t p 36 t^2 + 3
p 3 p 36 t^2 + 3
(It is easy to observe that jT (t)j = 1 for all t.) The unit tangent vector at the point on the curve corresponding to t = 1 is
p 39
p 3 p 39
r (t) = e t cos (t) ; e t sin (t) ; t
at the point (1; 0 ; 0). (You must, of course, include all calculations in detail.)
Solution: First we compute
r 0 (t) = e t sin (t) + e t cos (t) ; e t cos (t) + e t sin (t) ; 1
= e t (cos (t) sin (t)) ; e t (cos (t) + sin (t)) ; 1
r^00 (t)
= e t ( sin (t) cos (t)) + e t (cos (t) sin (t)) ; e t ( sin (t) + cos (t)) + e t (cos (t) + sin (t)) ; 0
= 2 et^ sin (t) ; 2 et^ cos (t) ; 0
and
jr 0 (t)j =
q (et^ (cos (t) sin (t)))
2
2
2
p e^2 t^ (1 2 sin (t) cos (t)) + e^2 t^ (1 + 2 sin (t) cos (t)) + 1
=
p 2 e^2 t^ + 1.
At t = 0 (which corresponds to the point (1; 0 ; 0)), we have
r 0 (0) = h 1 ; 1 ; 1 i
r 00 (0) = h 0 ; 2 ; 0 i
jr 0 (0)j =
p
Match the equations given in añe with the pictures. (Write each parametric formula
next to the graph that is described by the formula.)
(a) r(u; v) = hu 2
(b) r(u; v) = hu + v; u^2 ; v^2 i, 1 u 1 , 1 v 1
(c) r(u; v) = h2 sin (u) ; 3 cos (u) ; vi, 0 u 2 , 1 v 1
(d) r(u; v) = hu^3 ; u sin (v) ; u cos (v)i, 1 u 1 , 0 v 2
(e) r(u; v) = h(1 cos (u)) sin (v) ; u; (u sin (u)) cos (v)i, 0 u 2 , 0 v 2
r(u; v) = hu^3 ; u sin (v) ; u cos (v)i
r(u; v) = hu^2 + 1; v^3 + 1; u + vi
r(u; v) = h2 sin (u) ; 3 cos (u) ; vi