MAT 137Y, 2006–2007 Winter Session, Solutions to Term Test 2
1.
(9%)
(i)
Find the equation of the tangent line to the graph of
y
=
sec
x

2cos
x
at the point
(
π
3
,
1
)
.
y
=
sec
x
tan
x
+
2sin
x
, so the slope of the tangent line is
y
(
π
3
) =
sec
π
3
tan
π
3
+
2sin
π
3
=
2
√
3
+
2
·
√
3
2
=
3
√
3.
Thus the equation of the tangent line is
y

1
=
3
√
3
(
x

π
3
)
or
y
=
3
√
3
x

π
√
3
+
1.
(9%)
(ii)
For the equation
x
2
+
4
xy
+
y
3
+
5
=
0, find
d
2
y
dx
2
at the point
(
2
,

1
)
.
Differentiating implicitly, we have 2
x
+
4
x
dy
dx
+
4
y
+
3
y
2
dy
dx
=
0
.
Plugging in
x
=
2
,
y
=

1 gives
us
4
+
4
dy
dx

4
+
3
dy
dx
=
0
=
⇒
dy
dx
(
2
,

1
)
=
0
.
Differentiating implicitly again, we have
2
+
4
x
d
2
y
dx
2
+
4
dy
dx
+
4
dy
dx
+
6
y
dy
dx
2
+
3
y
2
d
2
y
dx
2
=
0
.
Plugging in
x
=
2
,
y
=

1
,
dy
dx
=
0, we have
2
+
8
d
2
y
dx
2
+
3
d
2
y
dx
2
=
0
=
⇒
d
2
y
dx
2
=

2
11
.
2.
(10%)
(i)
A balloon is rising at a constant speed of
5
3
meters per second. A boy is cycling along a straight
road at a speed of 5 meters per second. When he passes under the balloon, it is 15 meters above
him. How fast is the distance between the boy and the balloon changing three seconds later?
Let
y
be the height of the balloon and
x
be the horizontal distance between the balloon and the
bicycle. If
z
is the distance between the boy and balloon, then
x
2
+
y
2
=
z
2
=
⇒
2
x
dx
dt
+
2
y
dy
dt
=
2
z
dz
dt
=
⇒
x
dx
dt
+
y
dy
dt
=
z
dz
dt
.
Three seconds after the bicycle passes the balloon, we have
x
=
15 and
y
=
20, so by the
Pythagorean Theorem
z
=
25. Since
dx
dt
=
5 and
dy
dt
=
5
3
, we get
15
(
5
)+
20
(
5
3
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 Spring '08
 UPPAL
 Derivative, Slope, Sin, Mathematical analysis, Limit of a function, Convex function

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