Chapter 9 Quiz Sec. 1
1. Plot the polar curve
r
(
θ
) = 1 + cos(
θ
), 0
≤
θ
≤
2
π
.
Solution:
M FFFFFFFF
11
Π
12
M FFFFFF
5
Π
6
M FFFFFF
3
Π
4
M FFFFFF
2
Π
3
M FFFFFF
7
Π
12
M FFF
Π
2
M FFFFFF
5
Π
12
M FFF
Π
3
M FFF
Π
4
M FFF
Π
6
M FFFFF
Π
12
0
FFFFF
Π
12
FFF
Π
6
FFF
Π
4
FFF
Π
3
FFFFFF
5
Π
12
FFF
Π
2
7
Π
12
2
Π
3
3
Π
4
5
Π
6
FFFFFFFF
11
Π
12
Π
M
2
M
1
1
2
M
2
M
1
1
2
2. Let (
x
(
t
)
, y
(
t
)) parametrize a curve in the plane. Explain why
dy
dx
=
y
′
(
t
)
x
′
(
t
)
.
(That is, why is this true?)
Solution:
There are several reasons why this is true. Here are two.
Reason #1. The velocity vector (
x
′
(
t
)
, y
′
(
t
)) is tangent so it has the same slope as the tangent line.
Its slope is rise/run = (ycomponent)/(xcomponent) =
y
′
(
t
)
/x
′
(
t
).
Reason #2. Let
f
be a function whose graph coincides with the parametrized cirve. Then
y
(
t
) =
f
(
x
(
t
)). By the chain rule it follows that
y
′
(
t
) =
f
′
(
x
(
t
))
*
x
′
(
t
)
.
Therefore
f
′
(
x
) =
y
′
(
t
)
x
′
(
t
)
where
x
=
x
(
t
).
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 Fall '11
 GeorgeJennings
 Calculus, Derivative, Cos, parametrized cirve

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