ACT245H1S
1. For what values of % is O% b %O  ?
A) c  %  B) %  c or % € PROBLEM SET 1
C) %  c D) %  E) % € 2. If % b %& b & ~ and % b l%& b & ~ then % b & ~ ?
A)
l
b B) C)
lbl³
²
D) E) 3. Transform to rectangular coordinates: ² c ³ ~ l .
A) % b & ~ c l
B) % b & ~ C) % b & ~ l
D) % c & ~ c E) % c & ~ c l
%
c%
4. If & ~ c
, then find % in terms of &.
&
c&
A) % ~ c
B) % ~ &
C) % ~ ²& b l& b ³
D) % ~ &
E) % ~ ²& b & ³
5.
5
lim 5 [ ~ :
5¦B
A) B) C) D) b B
E) None of A,B,C,D
6. Let ²%³ be a function which has a continuous second derivative at % ~ .
Let ²³ ~ , Z ²³ ~ c , and ZZ ²³ ~ . Let ²%³ ~ % ²%³ .
Calculate ZZ ²³.
A) c B) C) D) E) 7. Let ²%³ ~ % % . Determine the th derivative of at % ~ .
A) B) C) ² c ³² c ³
D) ² c ³² c ³² c ³
E) [
8. The point ²Á ³ lies on the line tangent to % b %& b & b ~ at ²Á c ³.
Calculate the value of .
A) c B) c C) c D) c E) c 9. Let ²%³ ~ % b% b% for c B  %  B, and let be the inverse function for .
What is Z ² ³?
A) B) C) D) E) 10. Find lim ² b ³ (Á £ ).
A) ¦bB
B) C) 11. Let ' ~ ²%Á &³ where
D) c
C'
C%
E) c
~ & b , C'
C& ~ % b & , % ~ ! and & ~ c ! .
Determine C'
C! .
A) c ! b !
B) c ! b ! b D) c ! b ! c E) c ! b ! c C) c ! b ! c 12. Let * be the curve defined by % ~ ! b ! ~ and & ~ ! c ! c .
Determine the %-intecept for the line tangent to * at ²Á c ³.
A) c 1
B) c C) D) 1
E) &
C 13. Let (%Á &³ ~ %& b % for % € and & € . Find C% C& .
&
&
&
%
%
A) % c %
B)
b
C)
c
D)
&
&
%
&
%
% b &
&
E) c %
& c %
14. Let ²%³ ~ % c % c O%O for c  %  . Let ? be the value of % that
minimizes ²%³ and let ? be the value of % that maximizes (%³. Calculate ? c ? .
A) c B) c C) c D) c E) c 15. In a certain industrial process, the rate of change of a resource 9 is proportional to the
amount of 9 remaining. During the first hour, 3000 tons of 9 are reduced to 2000 tons.
Calculate the amount of 9 , in tons, that will remain after two more hours.
A) B) C) D) 500
E) Questions 16 and 17 are based on the following parametrically defined curve:
%²!³ ~ ! Á &²!³ ~ ! at the point ! ~ À
16. Find the equation of the tangent line to the curve at ! ~ .
17. Find
&
%
.
18. Two curves are represented parametrically as follows:
Curve 1 % ~ ! b Á & ~ ! for all real !
Curve 2 % ~ ! b Á & ~ ! b ! b for all real !
The two curves are tangent to one another at the ²%Á &³ point ( b Á ³ .
What is the value of ?
A. B. C. c D. b E. c b 19. Express the linear relationship % c & b ~ in polar coordinates in
the form ~ ²³ .
20.
lim ´ % c ²% b l% b ³µ ~
%¦bB
A) c B
B) c C) D) E) B
ACT245S PROBLEM SET 1 SOLUTIONS
1. c  % b %  S % b % b € and % b % c  .
% b % b € for all % , since the quadratic equation % b % b ~ has no real
cf c
roots (roots are % ~
~ c f l c ).
l
% b % c  S ²% b ³²% c ³  S c  %  (since one of the factors
must be negative and one must be positive, but it is impossible to have % b  and
% c € , thus we must have % b € and % c  ). Answer: A
2. ²% b &³ ~ % b %& b & ~ b %& , and
l%& ~ c ²% b &³ S %& ~ c ²% b &³ b ²% b &³
b
S %& ~ c ²% b &³ b b %& S % b & ~ ~ .
3. ² b ³ ~ ² ³² ³ c ² ³² ³
l
Answer: D
l
S ² c ³ ~ ² ³² c ³ c ² ³² c ³ ~ l
S c%b& ~.
Answer: D
4. Multiply equation by % to get
l
% c &% c ~ &f & b
S % ~
~ & f l& b . We ignore the negative root since % € .
Then % ~ ²& b l& b ³ .
Answer: C
5.
5
lim 5 [ ~ hhhhhhhhh
 [ h ² ³B ~ .
hhhhh
hhhh
5¦B
Answer: A
6. Z ²%³ ~ % ²%³ b % Z ²%³ Á ²³ ²%³ ~ ²%³ b % Z ²%³ b % ²³ ²%³
²³ ²³ ~ ²³ b Z ²³ b ²³ ²³ ~ À
Answer: C
7. 1st derivative - % ²% b % ³ ; 2nd derivative - % ²% b % b % ³
3rd derivative - % ²% b % b % b %³
4th derivative - % ²% b % b % b % b ³
It might be possible to determine a general expression for the th derivative in terms of then substitute % ~ . However, note that the 1st, 2nd and 3rd derivatives ² ~ Á Á )
must be if % ~ , but the 4th derivative is not at % ~ . This eliminates answers
A, B, C and E.
Answer: D
8. The slope of the tangent line at the point ²Á c ³ is &Z ²³. Using implicit
differentiation with respect to %, we have % b & b %&Z b & &Z ~ , so that
c%c&
%b&
cc²c³
b²c³
~ c .
Since ²Á ³ is on the tangent line, the slope of the tangent line must also be equal to
&Z ~
c²c³
c
, and at the point ²Á c ³ we have &Z ²³ ~
~ c , and therefore, ~ c .
Answer: C
9. If ²&³ is the inverse function of & ~ ²%³ then Z ²&³ ~ Z ²%³ .
Thus, Z ² ³ ~
, where % is the value such that ²%³ ~ .
% b% b%
²% b%b³
It is clear from the definition of ²%³ that this value of % is % ~ (since bb ~ ).
Thus, Z ² ³ ~ .
Answer: B
²b ³
10. Taking natural log we find lim ² b ³ ~ lim °
²c° ³°²b ³
~ lim
c°
¦bB
¦bB
¦bB
~ . Then the original limit is .
Answer: C
&
C' %
C'
11. C'
C! ~ C% h ! b C& h ! ~ ² c ! b ³²!³ b ²! b c !³² c ³ ~ c ! b ! c Alternatively, it is possible to find ' in terms of % and & (and an arbitrary constant ) C'
since C%
~ & b it follows that ' ~ %& b % b ²&³ , but then
C'
C&
~ % b & ~ % b Z ²&³ so that Z ²&³ ~ & and ²&³ ~ & b , and then
' ~ %& b % b & b ~ ²! ³² c !³ b ! b ² c !³ b .
From this we have C'
C! ~ c ! b ! c .
&
&
Answer: D
! c
12. The slope of the tangent line is % ~ ! , %
! ~ !b À The point
²%Á &³ ~ ²Á c ³ corresponds to a value of ! such that % ~ ! b ! b ~ and
& ~ ! c ! c ~ c À Solving the second equation results in ! ~ Á , but only ! ~ satisfies the first equation. Thus, ²%Á &³ ~ ²Á c ³ corresponds to ! ~ , and the slope of
the tangent line is
! c
!b
~ À The equation of the tangent line is
the %-intercept occurs where & ~ , so that % ~ À
C &
C C
C
%
´ C& µ ~ C%
´ c %& b &
13. C% C& ~ C%
% µ ~ c & c % .
&c²c³
%c
~ Á and
Answer: E
Answer: E
14. ²%³ ~ H
% c%b%~% b% for c%
% c%c%~% c% for %
The critical points occur where Z ²%³ ~ ¢
% b ~ S % ~ c Á % c ~ S % ~ À
The max and min occur at a critical point, interval end-point, or point at which the
derivative does not exist. These points are % ~ c Á c Á Á Á with
² c ³ ~ c Á ² c ³ ~ c Á ²³ ~ Á ²³ ~ c Á ²
³ ~ c À
? ~ Á ? ~ S ? c ? ~ c .
Answer: E
15. 9
! ~ 9 . This is a differential equation that can be written in the form
9²!³
9²!³
~ ! , and integrating both sides with respect to ! results in 9²!³ ~ ! b ,
or equivalently, 9²!³ ~ h ! , where ~ . Then
9²³ ~ ~ Á 9²³ ~ ~ ~ S ~ S 9²!³ ~ ² ³! À
Two hours later, at time ! ~ , we have 9²³ ~ ² ³ ~ .
Answer: E
16.
&
%
line is
so that
17.
&
%
&°!
!
%°! ~ !
&c
%c ~ . Or,
&
% ~ is the
~
&
%
! ² % ³. !
&
%° S %
~
Or, & ~
~
!
. At ! ~ , slope is 3, %²³ ~ Á &²³ ~ ,
&
& ~ %° S %
~ %° , and at ! ~ , % ~ ,
slope of the tangent line.
~
°
!
~
!
Á
&
%
~
&
%
! ² % ³. !
~
c°!
!
~
c
!3
.
~ c %c° .
18. The ²%Á &³ point ² b Á ³ corresponds to ! ~ for Curve 1, and since ! ~ % c for Curve 2, it follows that the intersection/tangent point corresponds to ! ~ for
Curve 2 as well. Then, on Curve 2 at the intersection/tangent point, we must have the
common value of & ~ , so that ~ b b . We must also have the common
&
&°!
first derivative (for the curves to be tangent), so that % ~ %°! are the same for both
curves at the intersection/tangent point. For Curve 1 the first derivative at ! ~ is
&°!
%°!
~ ! e
!~ &°!
~ , and for Curve 2, it is %°! ~ 2!b
e!~ ~ b which must be as well, and thus, ~ c . Then, since ~ b b , it follows
that ~ b .
Answer: D
19. c b ~ S ~
20. % c ²% b l% b ³ ~ ´
as % S b B .
Answer: B
c %
.
µ ~ ´
µ
%bl% b
bmb S ´ b
µ ~ c