huang (th22624) – HW03 – Gilbert – (55015)
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001
10.0 points
Find the
y
intercept of the tangent line at
P
(2
,
1) to the graph of the curve defined para
metrically by
x
(
t
) = 4 cos 2
t ,
y
(
t
) = 2 sin
t .
1.
y
intercept = 3
2.
y
intercept =
−
3
3.
y
intercept =
3
2
correct
4.
y
intercept =
−
3
2
5.
y
intercept =
3
4
6.
y
intercept =
−
3
4
Explanation:
The point
P
(2
,
1) on the graph of
x
= 4 cos 2
t ,
y
= 2 sin
t
is the point at which
t
=
π/
6. On the other
hand,
x
′
(
t
) =
−
8 sin 2
t ,
y
′
(
t
) = 2 cos
t ,
and so the slope of the tangent line at
P
is
given by
y
′
parenleftBig
π
6
parenrightBigslashBig
x
′
parenleftBig
π
6
parenrightBig
=
−
√
3
4
√
3
=
−
1
4
.
But then by the pointslope formula,
y
−
1 =
−
1
4
(
x
−
2)
is an equation for the tangent line at
P
, which
after simplification becomes
y
+
1
4
x
=
3
2
.
Consequently, this tangent line has
y
intercept =
3
2
.
002
10.0 points
Find
dy
dx
for the curve given parametically
by
x
(
t
) = 2 + 3
t
2
,
y
(
t
) = 2
t
2
+
t
3
.
1.
dy
dx
=
6
1 + 6
t
2.
dy
dx
=
1
3
+
t
3.
dy
dx
=
1
3
+ 2
t
4.
dy
dx
=
2
3
+
t
5.
dy
dx
=
2
3
+
1
2
t
correct
6.
dy
dx
=
3
4 + 3
t
7.
dy
dx
=
6
4 + 3
t
8.
dy
dx
=
3
1 + 6
t
Explanation:
Differentiating with respect to
t
we see that
x
′
(
t
) = 6
t ,
y
′
(
t
) = 4
t
+ 3
t
2
.
Consequently,
dy
dx
=
y
′
(
t
)
x
′
(
t
)
=
4
t
+ 3
t
2
6
t
=
2
3
+
1
2
t
.
003
10.0 points
Find
d
2
y
dx
2
when
x
(
t
) = 5 +
t
2
,
y
(
t
) =
t
ln(
t
4
)
.
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huang (th22624) – HW03 – Gilbert – (55015)
2
1.
d
2
y
dx
2
=
−
2 ln
t
t
3
2.
d
2
y
dx
2
=
ln
t
t
3
3.
d
2
y
dx
2
=
ln
t
t
2
4.
d
2
y
dx
2
=
−
ln
t
t
3
correct
5.
d
2
y
dx
2
=
2 ln
t
t
2
6.
d
2
y
dx
2
=
−
2 ln
t
t
2
Explanation:
First notice that
y
(
t
)
=
4
t
ln
t
.
After
differentiation with respect to
t
, therefore, we
see that
x
′
(
t
) = 2
t ,
y
′
(
t
) = 4(1 + ln
t
)
.
Thus
dy
dx
= 2
parenleftBig
1 + ln
t
t
parenrightBig
.
On the other hand, by the Chain Rule,
d
dt
parenleftBig
dy
dx
parenrightBig
=
dx
dt
braceleftBig
d
dx
parenleftBig
dy
dx
parenrightBigbracerightBig
=
dx
dt
d
2
y
dx
2
,
in which case
d
2
y
dx
2
=
d
dt
parenleftBig
dy
dx
parenrightBigslashBig
dx
dt
.
But
d
dt
parenleftBig
dy
dx
parenrightBig
= 2
parenleftBig
t
( 1
/t
)
−
(1 + ln
t
)
t
2
parenrightBig
=
−
2 ln
t
t
2
.
Consequently,
d
2
y
dx
2
=
−
ln
t
t
3
.
004
10.0 points
Determine all values of
t
for which the curve
given parametrically by
x
= 3
t
3
−
2
t
2
+ 1
,
y
= 2
t
3
+ 3
t
2
−
1
has a vertical tangent?
1.
t
= 0
,
1
2.
t
= 0
,
−
1
3.
t
= 0
,
−
4
9
4.
t
= 1
5.
t
= 0
,
4
9
correct
6.
t
=
−
1
7.
t
=
−
4
9
8.
t
=
4
9
Explanation:
After differentiation with respect to
t
we
see that
y
′
(
t
) = 6
t
2
+ 6
t ,
x
′
(
t
) = 9
t
2
−
4
t .
Now
dy
dx
=
y
′
(
t
)
x
′
(
t
)
,
so the tangent line to the curve will be vertical
at the solutions of
x
′
(
t
) =
t
(9
t
−
4) = 0
,
hence at
t
= 0
,
4
9
.
005
10.0 points
Determine all points
P
at which the tangent
line to the curve given parametrically by
x
(
t
) =
t
3
−
6
t ,
y
=
−
3
t
2
is parallel to the line (
t,
2
t
).
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 Spring '07
 Gilbert
 Distance

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