Maple Lab 4 (Spring 2007 Math 251)
Section 05-07
Answer Key
Exercises
Note: Use appropriate colors and plot options to make your plots look nice. The appearance of
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Please delete unused execution groups.
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1. Define a vector field
and a parametric curve
where
.
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, `*`(`^`(y, 2), `*`(z)))), y)], [`+`(`*`(2, `*`(`^`(x, 3), `*`(y, `*`(z)))), x)], [`+`(`*`(`^`(x, 3), `*`(`^`(y, 2))), cos(z...](images/math251-spring2007-lab4_answer-key_6.gif) |
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], [sin(t)], [t]], [](images/math251-spring2007-lab4_answer-key_8.gif) |
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1(a). Plot the parametric curve.
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1(b). Compute the line integral
where
is the curve parametrized by
. Compute the integral by plugging in the parametrization and computing
directly.
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1(c). Compute the line integral
by computing
.
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![Typesetting:-mprintslash([`:=`(rPrimeLength, `*`(`^`(2, `/`(1, 2))))], [`*`(`^`(2, `/`(1, 2)))])](images/math251-spring2007-lab4_answer-key_19.gif) |
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, `*`(`^`(2, `/`(1, 2)), `*`(sin(t))))))], [`+`(`*`(`/`(1, 2), `*`(`^`(2, `/`(1, 2)), `*`(cos(t)))))], [`+`(`*`(`/`(1, 2), `*`(`...](images/math251-spring2007-lab4_answer-key_21.gif) |
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1(d). Check to see if the vector field
conservative.
If it is, compute
using the fundmental theorem
of line integrals.
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(7) |
Since
is the zero vector field,
must be conservative.
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![Typesetting:-mprintslash([`:=`(f, `+`(`*`(`^`(x, 3), `*`(`^`(y, 2), `*`(z))), `*`(y, `*`(x)), sin(z)))], [`+`(`*`(`^`(x, 3), `*`(`^`(y, 2), `*`(z))), `*`(y, `*`(x)), sin(z))])](images/math251-spring2007-lab4_answer-key_31.gif) |
(8) |
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, `*`(Pi)))]], [](images/math251-spring2007-lab4_answer-key_33.gif) |
(9) |
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(10) |
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2. [Section 16.4 #6]
Let
be the curve which consists of the parabola
from
to
and then the line segment from
to
.
Plot
and define the functions:
and
.
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![Typesetting:-mprintslash([`:=`(P, proc (x, y) options operator, arrow; VectorCalculus:-`*`(`*`(`^`(y, 2)), sin(x)) end proc)], [proc (x, y) options operator, arrow; VectorCalculus:-`*`(`*`(`^`(y, 2)),...](images/math251-spring2007-lab4_answer-key_48.gif) |
(12) |
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![Typesetting:-mprintslash([`:=`(Q, proc (x, y) options operator, arrow; VectorCalculus:-`*`(`*`(`^`(x, 2)), sin(y)) end proc)], [proc (x, y) options operator, arrow; VectorCalculus:-`*`(`*`(`^`(x, 2)),...](images/math251-spring2007-lab4_answer-key_50.gif) |
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2(a). Compute the line integral
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)]], [](images/math251-spring2007-lab4_answer-key_53.gif) |
(14) |
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)], [`+`(1, `-`(t))]], [](images/math251-spring2007-lab4_answer-key_55.gif) |
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![Typesetting:-mprintslash([`:=`(A, `+`(24, `-`(`*`(14, `*`(cos(1)))), `-`(`*`(19, `*`(sin(1))))))], [`+`(24, `-`(`*`(14, `*`(cos(1)))), `-`(`*`(19, `*`(sin(1)))))])](images/math251-spring2007-lab4_answer-key_57.gif) |
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![Typesetting:-mprintslash([`:=`(B, `+`(`-`(`*`(2, `*`(cos(1)))), `-`(`*`(4, `*`(sin(1)))), 4))], [`+`(`-`(`*`(2, `*`(cos(1)))), `-`(`*`(4, `*`(sin(1)))), 4)])](images/math251-spring2007-lab4_answer-key_59.gif) |
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![Typesetting:-mprintslash([`:=`(LHS, `+`(28, `-`(`*`(16, `*`(cos(1)))), `-`(`*`(23, `*`(sin(1))))))], [`+`(28, `-`(`*`(16, `*`(cos(1)))), `-`(`*`(23, `*`(sin(1)))))])](images/math251-spring2007-lab4_answer-key_61.gif) |
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2(b). Use Green's Theorem to turn the line integral from part 2(a)
into an equivalent double integral and verify the Green's
Theorem holds.
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![Typesetting:-mprintslash([`:=`(RHS, `+`(28, `-`(`*`(16, `*`(cos(1)))), `-`(`*`(23, `*`(sin(1))))))], [`+`(28, `-`(`*`(16, `*`(cos(1)))), `-`(`*`(23, `*`(sin(1)))))])](images/math251-spring2007-lab4_answer-key_63.gif) |
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3. Consider the surface
where
.
Plot this surface.
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3(a). Find formulas for both orientations of this surface using
the gradient operator. Decide which orientation points
"upward" and which points "downward" and label the
orientations accordingly.
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, `*`(`^`(`+`(`*`(`^`(x, 2)), `*`(`^`(y, 2))), `/`(1, 2))))], [`/`(`*`(y), `*`(`^`(`+`(`*`(`^`(x, 2)), `*`(`^`(y, 2))), `/`(1, 2))))], ...](images/math251-spring2007-lab4_answer-key_72.gif) |
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![Typesetting:-mprintslash([`:=`(gradLength, `*`(`^`(2, `/`(1, 2))))], [`*`(`^`(2, `/`(1, 2)))])](images/math251-spring2007-lab4_answer-key_74.gif) |
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, `*`(`^`(2, `/`(1, 2)), `*`(x))), `*`(`^`(`+`(`*`(`^`(x, 2)), `*`(`^`(y, 2))), `/`(1, 2)))))], [`+`(`/`(`*`(`/`(1, 2), `...](images/math251-spring2007-lab4_answer-key_76.gif) |
(23) |
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, `*`(`^`(2, `/`(1, 2)), `*`(x))), `*`(`^`(`+`(`*`(`^`(x, 2)), `*`(`^`(y, 2))), `/`(1, 2))))))], [`+`(`-`(`/`(`*`(`/`(1...](images/math251-spring2007-lab4_answer-key_78.gif) |
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3(b). This surface can be parametrized using the equations:
,
, and
where
and
.
Define a such a parametrization:
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))], [`*`(u, `*`(sin(v)))], [u]], [](images/math251-spring2007-lab4_answer-key_86.gif) |
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3(c). Compute both orientations for this surface again, but
this time use the parametrization. Again label them as
either "upward" or "downward".
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))))], [`+`(`-`(`*`(u, `*`(sin(v)))))], [`+`(`*`(`^`(cos(v), 2), `*`(u)), `*`(`^`(sin(v), 2), `*`(u)))]], [](images/math251-spring2007-lab4_answer-key_88.gif) |
(26) |
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![Typesetting:-mprintslash([`:=`(ruXrvLength, `*`(`^`(2, `/`(1, 2)), `*`(u)))], [`*`(`^`(2, `/`(1, 2)), `*`(u))])](images/math251-spring2007-lab4_answer-key_90.gif) |
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, `*`(`^`(2, `/`(1, 2)), `*`(cos(v))))))], [`+`(`-`(`*`(`/`(1, 2), `*`(`^`(2, `/`(1, 2)), `*`(sin(v))))))], [`+`(`*`(`/`(1,...](images/math251-spring2007-lab4_answer-key_92.gif) |
(28) |
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, `*`(`^`(2, `/`(1, 2)), `*`(cos(v)))))], [`+`(`*`(`/`(1, 2), `*`(`^`(2, `/`(1, 2)), `*`(sin(v)))))], [`+`(`-`(`*`(`/`(1, 2),...](images/math251-spring2007-lab4_answer-key_94.gif) |
(29) |
3(d). Plot this surface (using
's parametric plot) along
with a few sample normal vectors (using your downward orientation).
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