UNIVERSITY OF TORONTO
DEPARTMENT OF MATHEMATICS
MAT 235 Y – CALCULUS II
FALL-WINTER 2008-09
ASSIGNMENT #1. DUE ON OCTOBER 3.
PROBLEMS AND SOLUTIONS
1.
A brief review of single variable calculus.
a) Evaluate
(
)
2
0
lim
9
17ln
9ln
3
1
ln
ln
x
2
x
x
x
+
→
+
+
−
+
+
x
, if the limit exists.
Solution:
Let
2
9
17ln
9ln
A
x
x
=
+
+
and
2
3
1
ln
ln
B
x
x
=
+
+
.
Now,
2
2
A
B
A
B
A
B
−
−
=
+
, where
.
2
2
2
2
9
17ln
9ln
9(1
ln
ln
)
8ln
A
B
x
x
x
x
−
=
+
+
−
+
+
=
x
Notice that
2
2
2
2
9
17
9
17
9
17ln
9ln
ln
(
9)
ln
9
ln
ln
ln
ln
A
x
x
x
x
x
x
x
x
=
+
+
=
+
+
=
+
+
and similarly,
2
1
1
3 ln
1
ln
ln
B
x
x
x
=
+
+
. So,
2
2
9
17
1
1
ln
(
9
3
1 )
ln
ln
ln
ln
A
B
x
x
x
x
x
+
=
+
+
+
+
+
.
Notice also that
and
0
lim ln
x
x
+
→
= −∞
ln
ln
x
x
= −
when
0 <
x
< 1 .
Therefore,
(
)
0
0
2
2
8ln
8
4
lim
lim
3
3
3
9
17
1
1
ln
(
9
3
1 )
ln
ln
ln
ln
x
x
x
A
B
x
x
x
x
x
+
+
→
→
−
=
= −
= −
+
+
+
+
+
+
.
b) Let
P
(
t
,
f
(
t
) )
be any point on the curve
2
2
( )
(1
)
x
f x
x
=
+
and let
Q
be the point at which the
y
-axis
intersects the line that is tangent to the given curve at the point
P
. Suppose that
A
(
t
)
denotes the area of the
triangle
OPQ
, where
O
= ( 0 , 0 ) . Is there an absolute maximum value for the area
A
(
t
) ? If so, find this
absolute maximum value and determine the coordinates of all the points
P
for which that maximum is reached.
Solution:
An equation of the line that is tangent to the given curve at the given point is
y
–
f
(
t
) =
f
′
(
t
) (
x
–
t
) .
If
x
= 0
then the value of
y
in this equation is
y
=
f
(
t
) –
t
f
′
(
t
) . So,
Q
is the point with coordinates
( 0 ,
f
(
t
) –
t
f
′
(
t
) )
and
OPQ
is a triangle whose base has a length of
( )
( )
f
t
t f
t
′
−
and whose height has
a length of
t
. Therefore, the area of the triangle
OPQ
is
1
( )
(
( )
( ))
2
A t
t
f
t
t f
t
′
=
−
.
Now,
2
2
3
1
3
( )
(1
)
x
f
x
x
−
′
=
+
, then
2
4
2
2
2
3
2
3
1
(1
3
)
( )
(
)
2
(1
)
(1
)
(1
)
t
t
t
t
A t
t
t
t
−
=
−
=
+
+
+
2
t
.
Finally,
3
2
2
4
4
(2
)
( )
(1
)
t
t
A
t
t
−
′
=
+
which shows that
A
(
t
)
reaches its absolute maximum at
2
t
= ±
.
The absolute maximum value for the area of the triangle
OPQ
is
8
(
2 )
27
A
±
=
and this maximum area is
reached only when the tangency point
P
has coordinates
2
(
2 ,
9
±
±
)
.

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