Version 034 – Exam02 – Gilbert – (56380)
1
This
printout
should
have
12
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
Determine a Cartesian equation for the
curve given in parametric form by
x
(
t
) = 3 ln(16
t
)
,
y
(
t
) =
√
t .
1.
y
=
1
3
e
x/
4
2.
y
=
1
4
e
6
/x
3.
y
=
1
3
e
x/
8
4.
y
=
1
3
e
8
/x
5.
y
=
1
4
e
x/
3
6.
y
=
1
4
e
x/
6
correct
Explanation:
We have to eliminate the parameter
t
from
the equations for
x
and
y
.
Now from the
equation for
x
it follows that
t
=
1
16
e
x/
3
.
But then
y
=
parenleftBig
1
16
e
x/
3
parenrightBig
1
/
2
=
1
4
e
x/
6
.
002
10.0 points
Let
Q, R
be the points where the ray of
angle
θ
intersects circles centered at the origin
as shown in
2
4
6
8
−
2
−
4
−
6
−
8
2
4
6
8
−
2
−
4
−
6
−
8
θ
P
R
Q
and let
P
be the point of intersection of the
horizontal line through
Q
and the vertical line
through
R
.
As
θ
varies,
P
traces out a curve.
Write
this curve in parametric form
(
x
(
θ
)
, y
(
θ
))
,
0
≤
θ
≤
2
π .
(Hint:
find the coordinates of
Q
and
R
.)
1.
(7 sin
θ,
4 cos
θ
)
2.
(4 cos
θ,
7 sin
θ
)
3.
(4 tan
θ,
7 sec
θ
)
4.
(7 sec
θ,
4 tan
θ
)
5.
(7 cos
θ,
4 sin
θ
)
correct
6.
(4 sec
θ,
7 tan
θ
)
Explanation:
From the graph it follows that the inner
circle has radius 4, while the outer circle has
radius 7.
On the other hand, any circle of
radius
r
centered at the origin can be written
in parametric form as
(cos
θ,
sin
θ
)
,
0
≤
θ
≤
2
π ,
Thus
Q
and
R
are the points
Q
(4 cos
θ,
4 sin
θ
)
,
R
(7 cos
θ,
7 sin
θ
)
.
But
P, Q
have the same
y
coordinate, and
P, R
have the same
x
coordinate.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Version 034 – Exam02 – Gilbert – (56380)
2
Consequently, the curve traced out by
P
has the parametric form
(7 cos
θ,
4 sin
θ
)
for 0
≤
θ
≤
2
π
. Eliminating
θ
, we see that
P
traces out the ellipse
x
2
49
+
y
2
16
= 1
whose relation to the circles is shown in
2
4
6
8
−
2
−
4
−
6
−
8
2
4
6
8
−
2
−
4
−
6
−
8
P
003
10.0 points
Find an equation for the tangent line to the
curve given parametrically by
x
(
t
) =
e
2
t
,
y
(
t
) = 3
t
2
+
t
+ 2
at the point
P
(1
,
2).
1.
y
=
1
2
x
+
5
2
2.
y
=
−
1
2
x
+
3
2
3.
y
=
x
+
3
2
4.
y
=
1
2
x
+
3
2
correct
5.
y
=
x
+
5
2
6.
y
=
−
1
2
x
+
5
2
Explanation:
Notice first that
P
(1
,
2) is the point corre
sponding to the choice
t
= 0. We can thus use
the point slope formula with
t
= 0 to find an
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 TextbookAnswers
 Multivariable Calculus, parametric form

Click to edit the document details