Qatar University
Department of Mathematics, Statistics, and Physics
Question Score From
I
II
III
6
7
6
IV
V
VI
VII
4
22
7
4
VIII 4
Total 60

Question 1. ( 6 Point s = 3+3 ) Find the derivative dy dx
(a) y
 sec( x 2 
1) dy dx
 sec( x
2 
1) tan( x
2 x 2 x

1
1.5 + 1.5
(b) y


6
 cot
2 x

7 dy


2 x

6  
2cot x

.

 csc
2 x

1 + 1 + 1 dx xy
3 dy
Question 2. (7 points) Find if x and y are related by the equation dx
 sin y x 1
Differentiating with respect to x gives y 3

3 x y 2 dy d x
 dy dx

3 xy
2  cos
 co s y dy dx y


3 x 2

0
 
3 x
2  y
3
 dy


3 dx 3 xy 2 x
2

 y cos
3 y
2 + 2 + ½ + ½
1+1
Page 2 of 8

Page 3 of 8

Question 3. (6 points) Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always two times its height. Suppose the height of the pile increases at a rate of 3 cm/s. At what rate is the sand is accumulating at the instant when the pile is 10 cm high??

Let v be The volume of the pile and h its height at time t ( as functions of t). Given that the radius r = 2h. We have: v

1
3
 r
2 h

1
3

( 2 h )
2 h

4
3
 h
3

Given: dh dt

3 cm / s
1 + 1
½ + ½

Required to find dv dt h

10
 dv dt


4
3

4
 h
2
( 3 )
( 3 h
2
) dh
 dt
12
 h
2
 dv dt h

10

12

( 10 )
2 
1200
 cm
3
/ s
1 + 1
1

Page 4 of 8
Question 4. (4 points) Find the values of x at which
( )
 x
3 
3 x

1 has absolute extrema on
 
2,0

.
 f x

3 x
2   x
2 
1) . So f x
  x
2
( ) 0 ( 1) 0 x 1
We ignore x

1 because it does not belong to interval

.
The only critical point in ( 2,0) is x
 
1
½ + ¾ + ½

Checking f ( 1) 3 , f ( 2) 1 , f (0) 1 gives that ( ) has absolute minimum of -1 at x
 
2 and absolute maximum of 3 at x
 
1 on


2,0

.
3(¼) + ¾ + ¾
Question 5. (22 points = 6+6+4+6) Consider the function

3 x

2
whose first and x 3 second derivatives are
 

6( x

1) and x
4 f

( )

6(3 x

4)
. x
5
(a) Determine the intervals on which f(x) is increasing and decreasing. Find the local extrema (if any).
 
( ) 0 x 1 is the only critical point.
Though

( ) undefined gives x

0 but it is not taken as critical point as x

0 is not in the domain.

Next we analyze

( ) to know increasing/decreasing intervals
(

,0) (0,1) (1, ) Interval
Sign of

( ) Positive Positive Negative
0.5
1.5 + 1.5 + 1.5
Nature of ( ) Increasing Increasing Decreasing

Local maximum at (1,1) . No local minimum.
1

Page 5 of 8
(b) Determine the intervals on which f(x) is concave up or concave down. Find the points of inflection (if any).
 f

( ) 0 x
4
3 as only candidate point of inflection.

Next we analyze f

( ) to know concavity
0.5
Interval
Sign of f

Concavity
( )
(

,0)
4
(0, )
3
(4/3, ∞)
Positive Negative Positive
Up Down Up
1.5 + 1.5 + 1.5

Point of inflection x

4
.
3
1
(c) Find the vertical and horizontal asymptotes of f(x) .

Vertical asymptote is x

0 with lim x

0

3 x

2 x 3
 
and lim x

0

3 x

2 x 3
 
1 + ½ + ½

Horizontal asymptote is y

0 with x lim

3 x

2 x
3

0
1 + 1

(d) Sketch the graph of f(x).
1
1
1
1
1
Page 6 of 8
1

Page 7 of 8
Question 6. (7 points) An open-top rectangular tank with a square base and a volume of
500 ft 3
is to be constructed using steel. Find the dimensions of the tank that has minimum surface area.

Given
2 x y

500 (*)
1
 To minimize A
 x
2 
4 xy (**) 1

From (*), we have y

500
2
. x
Substituting in (**) implies that we need to minimize
( )
 x
2 
2000
for x

0 x
1
 
( )

2 x

2000 x
2
1


( ) 0 2 x

2000 x 2

2 x
3 
2000 0 x
3 
1000
2 x
3 
2000 x 2
10

0
1
 
( )
 
4000
 x
3
A

(10) 0 x 10 is a local minimum.
½
Since x=10 is the only point of local extremum on the domain (0,∞) of A, then it must be an absolute minimum.
½

So the dimensions that minimize surface area are x

10 and y

500

500

5
10 2 100
1

Page 8 of 8
Question 7.
(4 points) Show that
( )
 
2 x

1 satisfies the hypothesis of Mean Value Theorem on
 
. Find the number c that satisfies the conclusion of Mean Value Theorem.

( ) is a polynomial so it is continuous on
and differentiable on
 
.
 

By MVT, there is a point c in (0,1) such that: f

( c )
 f ( 1 )

1

0 f ( 0 )
1

This gives 2 c
   
1
2
1
Question 8. (4 points ) Find the function f(x) , where
  5 
1 and
7
) f (1)

3
2
1
1
   5
7
)

Integrating gives
( ) x 7 x
2
  
2
C

Using
3
2
1 f (1)

3
implies
2
1
 
1
2
C C

So ( )
 x 7
 x
2
2

1

( ) 7 x 6  x
1
1
1
1