Valencia (drv252) – assignment 1 – luecke – (58600)
1
This printout should have 13 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
HINTS. CalC11b16b:
sin/cos = tan;
CalC11b29s: The slope oF a line is the slope
oF the tangent at any point on the line;
CalC11b40a: What is the convention about
the square root sign?
001
10.0 points
IF the constant
C
is chosen so that the
parabolic arc
y
=
x
2
8
From (0
,
0) to (4
,
2) is given parametrically
by
(
Ct , y
(
t
) )
,
0
≤
t
≤
3
,
fnd the coordinates oF the point
P
on this arc
corresponding to
t
= 2.
1.
P
=
p
2
9
,
8
3
P
2.
P
=
p
8
3
,
8
9
P
correct
3.
P
=
p
8
9
,
8
3
P
4.
P
=
p
4
3
,
2
9
P
5.
P
=
p
2
9
,
4
3
P
6.
P
=
p
4
3
,
8
9
P
Explanation:
We have to determine
y
(
t
) and
C
so that
8
y
(
t
) =
C
2
t
2
,
while
x
(0) = 0
,
y
(3) = 2
,
3
C
= 4
.
Thus
C
=
4
3
,
8
y
(
t
) =
p
4
3
P
2
t
2
.
Consequently, when
t
= 2,
P
= (2
C, y
(2)) =
p
8
3
,
8
9
P
.
keywords: parametric curve, parabola
002
10.0 points
Determine
A
so that the curve
y
= 9
x
+ 42
can be written in parametric Form as
x
(
t
) =
t

5
,
y
(
t
) =
At

3
.
1.
A
=

11
2.
A
= 10
3.
A
=

9
4.
A
= 9
correct
5.
A
=

10
6.
A
= 11
Explanation:
We have to eliminate
t
From the parametric
equations For
x
and
y
. Now From the equation
For
x
it Follows that
t
=
x
+ 5. Thus
y
= 9
x
+ 42 =
A
(
x
+ 5)

3
.
Consequently
A
= 9
.
003
10.0 points
±ind a Cartesian equation For the curve
given in parametric Form by
x
(
t
) = 4
t
2
,
y
(
t
) = 8
t
3
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentValencia (drv252) – assignment 1 – luecke – (58600)
2
1.
x
= 2
y
4
/
3
2.
x
=
y
2
/
3
correct
3.
x
=
y
3
/
2
4.
x
= 2
y
2
/
3
5.
x
= 2
y
3
/
2
6.
x
=
y
4
/
3
Explanation:
We have to eliminate the parameter
t
from
the equations for
x
and
y
.
But from the
equation for
y
, it follows that
t
=
1
2
y
1
/
3
,
in which case
x
= 4
p
1
2
y
1
/
3
P
2
=
y
2
/
3
.
004
10.0 points
Find a Cartesian equation for the curve
given in parametric form by
x
(
t
) = 5 cos
2
2
t ,
y
(
t
) = 3 sin
2
2
t .
1.
3
x

5
y
= 15
2.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Holcombe
 Chemistry, 2 sec, Parametric equation, 2 feet, 15 feet, 13 Sec

Click to edit the document details