1. No category

SAMPLE MIDTERM
1.
For the function  (), given in the
following figure, find (make
reasonable estimates from the grid, if neccessary)
(a)  (5) = 3
(b)  (−7) = −2
(c)  (−2) = −2
(d)  (7) = 3
(e)  (−5) = −3
(f)  (3) = 0
(g) lim  () = 2
→7
(h)
(i)
(j)
(k)
(l)
lim  () = 2
→−2+
lim  () = −2
→−2−
lim  () =  
→−2
lim  () = −5
→5−
lim  () = −5
→5+
(m) lim  () = −5
→5
(n)
(o)
lim
→−3
 () −  (−3)
=
 − (−3)
1
2
 () −  (−5)
=0
→−5
 − (−5)
lim
(p) the largest value in the following list:  0 (−7)  0  (3) = 0  (−5) = −3  (−3)  0
 0 (−3) = 12 =   
(q) the intervals where the function is negative ( ()  0).
(−8 −2), (3 5) and (5 57)
(r) the intervals where the function is positive ( ()  0)..
(−9 −8), (−2 3)  (57 9)
(s) the values of  where the function is not a continuous function.
 = −2  = 5  = 7
2. Limits such as (and similar problems!):
1
(a) lim +
=∞
→−5  + 5
(b)
(c)
lim −
→−5
lim
→5−
1
= −∞
+5
1
=
+5
1
10
5 + 32 − 1
=
→∞ 2 − 4 + 72
(d) lim
(e)
53 + 3 − 1
= − 57
→−∞
2 − 73
lim
(f) lim
→2
(g)
3
7
lim
−2
=
 + 2 − 6
→−3
1
5
+3
= − 15
 + 2 − 6
√
2−
√ = −2 2
(h) lim √
→2
− 2
√
√
42 +  − 3 − 
√
(i) lim
= − 12
→−∞
 + 4 + 1
(j) lim
sin 3
=
sin 5
(k) lim
 sin 
=1
1 − cos2 
→0
→0
3
5
 − 1
=1
→0 sin 
(l) lim
(m)
lim − = 0
→+∞
2
=0
→0 sin 
(n) lim
−3
= −6
(o) lim (1 + 2)
→0
µ
¶
1
1
(p) lim
− 
= 12
→0 
 −1
³
³  ´´
(q) lim  sin
=
→∞

(r) lim (1) = 1
→0
(s)
(t)
lim (tan  ln ) = 0
→0+
5
lim − (sec 3 cos 5) = −

3
→
2
(u) lim (( − ) cot ) = 1
→
3. Suppose the function  () is defined as  () =
½
2
2 + √
cos  2 + 1
if
if
0
. For what value of  is this
≥0
function continuous on the interval (−∞ +∞)?
¡
¢
lim−  () = lim− 2 + 5 + 2 = 2
→0
→0
p
lim+  () = lim+ cos  2 + 1 = 1
→0
→0
so 
2
= 1, or  = ±1

(and similar problems!) for


= 2 sin 3 + 32 cos 3
(a)  = 2 sin (3) :

¡
¢3 
¢2 ¡ 2
¢
¡
(b)  = 2 sin (3) :
3 cos (3) + 2 sin (3)
= 3 2 sin (3)

4. DERIVATIVES: Find
(c)  =
2

2 sin 3 − 32 cos 3
:
=
sin (3) 
sin2 (3)
√

1
3
ln  + 1 :
=
2

3 (ln  + 1) 3
µ√
¶
 cos 
2

1
(e)  = ln
− tan  −
:
= 2
1 + 2

1 + 2
!
µ 2
¶− 23 Ã
q
 −1

4
1
3 2 −1
(f)  = 2 +1 :
=3
2

2 + 1
(2 + 1)
!
µ
¶3
¶2 Ã 2
µ
3 + 1

3 − 2 (3 + 1)
3 + 1
(g)  =
:
=3
2

2
(2 )2
(d)  =
−1

2sin 
 ³ sin−1  ´
√
:
= (ln 2)
(h)  = 2
=
2


1 − 2
´
³
√
√
√
√ √


(i)  = 2  :
=
2  = 2  +  


¡ −1 ¢  ¡ ¡ −1 ¢¢
1
√
(j)  = ln cos  :
ln cos  = −
−1

(cos ) 1 − 2
sin−1 
¢
¡
¢
¡ ¢
¡

(k)  = 2 sin3 3 + ln2  : 
2 sin3 3 + ln2  = 2 3 sin2 3 (cos (3)) (3) + 2 (ln ) 1 =

18 sin2 3 cos 3 + 2 ln

¡
¢
¡
¢


(l)  = ln 2  :  0 = 
(2 ln  + ) = 2 + 1
ln 2 + ln  = 
5. Find the given derivative for the given function:
21
(a)  () = sin (3) : Find 
21 ( ())
 = sin (3)
 0 = 3 cos (3)
 00 = −32 sin (3)
 000 = −33 cos (3)
 (4) = 34 sin (3)
 (5) = 35 cos 3
 (6) = −36 sin 3
 (7) = −37 cos 3
 ( ) = ±3 cos 3 :
so
(b)  () = ln (1 + ) : Find
3
3
 00
 000
4
4
0
 () =
00 () =
000 () =
(4) () =
positive for 1 5 9 13 17 21 
negative for 3 7, 11 15 19 ...
21
(sin 3) = 321 cos 3
21
¸
( ())
0
(c)  () = 3 ln  : Find
∙
1
= (1 + )−1
1+
−2
= − (1 + )
2
−3
= 2 (1 + ) =
(1 + )3
=
( ())
µ ¶
1
3 ln  + 
= 32 ln  + 2

µ ¶
1
2
6 ln  + 3
+ 2 = 6 ln  + 3 + 2 = 6 ln  + 5

µ ¶
1
6 ln  + 6
+ 5 = 6 ln  + 6 + 5 = 6 ln  + 11

6

2
3
6. Implicit Differentiation:
For the curve  +  2 + 2 =  + 1 :

(a) What is the derivative
?



 +  + 2
+ 2 = 1



( + 2) = 1 − 2 − 


1 − 2 − 
=

 + 2
(b) What is the equation of the tangent line on the curve  +  2 + 2 =  + 1 at the point (0 1)?
1 − 2 (0) − (1)
1 − 2 − 
=
|()=(01) =
=0
 + 2
(0) + 2 (1)
 − 1 = 0 ( − 0)
Tangent line is :  = 1
7. Find the equation of the tangent line for the curve  2 −  + 2 = 4 at the point (2 0)
2 0 − ( 0 + ) + 2 = 0
 − 2
0 =
2 − 
−4
 =
=2
−2
 = 2 ( − 2) = 2 − 4
8. Calculate  0 for the following curves (implicitly):
(a)  3 + 3 2 − sin  = 3 − 
 3 + 3 2 ( 0 ) + 6 ( 0 ) − cos  ( 0 ) = 32 − ( 0 )
3 2 ( 0 ) + 6 ( 0 ) − cos  ( 0 ) + ( 0 ) = 32 −  3
32 −  3
0 =
2
3 + 6 − cos  + 1
(b) sin () +  = 
cos () [ +  ( 0 )] +  ( 0 ) = 1
 cos () +  ( 0 ) cos () + ( 0 )  = 1
1 −  cos ()
0 =
 cos () + 
(c)  sin 2 =  cos 2
sin 2 +  (cos 2) (2 0 ) = ( 0 ) cos 2 +  (− sin 2) (2)
2 (cos 2) ( 0 ) − ( 0 ) cos 2 = −2 sin 2 − sin 2
−2 sin 2 − sin 2
0 =
2 cos 2 − cos 2
9. For the curve ( − )2 = 1 +  :

(a) What is the derivative
?

2 ( − )
µ
¶


1
−1
= 1 or
−1=


2 ( − )

1
= 1+

2 ( − )
2
(b) What is the equation of the tangent line on the curve ( − ) = 1 +  at the point (0 1)?
1
3
|()=(01) = 1 +
=
2 (1 − 0)
2
3
−1 =
( − 0)
2
3
Tangent line is  =
+1
2
10. The height (in meters) of a projectile shot vertically upward from a point 1.5 m above ground level with an initial
velocity 25.48 m/s is  = 15 + 2548 − 492 after  seconds.
(a) When does the projectile reach its maximum height?
maximum height - when velocity changes from positive to negative
2548
 () = 2548 − 98 = 0 for  =
= 26 s
98
(b) What is the maximum height?
2
maximum height is  (26) = 15 + 2548 (26) − 49 (26) = 34 624 m
:
11. A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cm/s. Find the rate
at which the area within the circle is increasing after (a) 1 s, (b) 3 s, and (c) 5 s.
 = 2


= 60 cm/s (constant speed = rate of change)

= 60 cm after  = 3  µ
= 180
¶ cm, and  = 5 means  = 300 cm

 ¡ 2¢

=
 = 2




= 2 (60 cm) (60 cm/s) = 7200 cm2 /s
1:


3:
= 2 (180 cm) (60 cm/s) = 21 600 cm2 /s


5:
= 2 (300 cm) (60 cm/s) = 36 000 cm2 /s

=  ·  = 60 and
after  = 1 s, the radius is 
At  =
At  =
At  =
12. A bacteria culture grows with constant relative growth rate. The bacteria count was 400 after 2 hours and 25,600
after 6 hours.
(a) What is the relative growth rate?

=  so  () =  

 (2) = 400 =  2
 (6) = 25600 =  6
6
 
25600
1
Divide:
=
or 4 = 64 so 4 = ln 64 and  = ln 64
2
 
400
4
(b) What is the initial size of the culture?
400
400
400
= 1 ln 64 =
= 50
2

8
2
(c) What is the number of bacteria after 5 hours? 10 hours?  hours?
1
 = 5 :  (5) = 50( 4 ln 64)5 ≈ 9 051
plug into first:  =
1
 = 10 :  (10) = 50( 4 ln 64)10 = 1 638 400
 hours
:

 () = 50 4 ln 64
(d) When will the population reach 50,000?

 () = 50000 = 50 4 ln 64


4 ln (1000)
50000
=  4 ln 64 or ln 64 = ln (1000) so  =
= 6 643 856 190
50
4
ln (64)
so it will take  ≈ 66 hours, or 6 hours 36 minutes ( 06 () = 06 (60 minutes) = 36 minutes).
13. A sample of tritium-2 decayed to about 94.5% of its original amount after a year.
(a) What is the half-life of tritium-3?
 () =  
 (1) =   = 0945 so  = 0945 or  = ln 0945
half-life :  needed to get 50% of 
1
1
 =   so =  ln 0945 so  ln 0945 = ln 05
2
2
ln 05
 =
≈ 12 25 = 12 years and 3 months
ln 0945
(b) How long would it take for the sample to decay to 20% of its original amount?
=   ln 0945 so ln 02 =  ln 0945
ln 02
 =
≈ 28 45 years ≈ 28 years 5 months and 12 days
ln 0945
020