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(math013)[2008](f)final~2816^_10006.pdf
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Math013 Calculus I Final Exam Solution, Fall 08
1. (12=3+3+3+3points) Hereisatableof somevaluesof threedi.erentiablefunctions a, b, and
c. The derivatives a , b and c are assumed to be continuous.
x
0 2 5
a(x) a (x) b(x) b(x) c(x) c (x)
1
.2
51 .11
2
00920 .1 .1 .1 19 3 23
2
Answer thefollowing: An answer of true means truein all situationsAn answer offalse means there is a case where the assertion is false.
(i) (3 points)The function c has an inverse.
True False Brief reason.
Solution The function c might not be one-to-one, e.g., we could have c(1)=0 = c(2).
(In fact, the function can not have an inverse. c(2)=0 and c (2)= .1 imply that c would decrease
into negative values before taking c(5) = 3, hence by continuity, c(x0)=0 = c(2) for some point
2 <x0 < 5; i.e., c is not one-to-one.)
(ii) (3 points)The derivative a takes on the value 1 in the interval 0 x 2.
True False Brief reason.
a(2).a(0)
Solution Notethat =1. By themeanvaluetheorem, a (x0)=1forsome0 <x0 <
2.0
2.
Compute the following:
(iii) (3 points)g(2)and g (2), whereg(x)= b(a(x))
Solution g(2)= b(a(2))= b(0)=5,
g (2)= b (a(2))a (2)= b (0)a (2)=1 0 =0
(iv) (3 points) Suppose b is an increasing function, and h is the inverse function to b. Compute h(9) and h (9).
Solution h(9)=2, since b(2)=9;
11
h (9)= = by the chain rule: b(h(x))= x =. b (h(x))h (x)=1.
b(2) 2 2. (10=5+5points) A6-ft-tall maniswalking awayfroma24-ft-talllamppostatarateof3ft/s.
(i) (5 points) At what rate is the length of his shadow changing when he is 16 ft away from the lamppost?
s 61 sx x
Solution Noting that = = ,or = ,wehave s = , and
s + x 24 4624.63
ds . 1dx . 1
== 3 =1 (ft/s)
dt 3 dt 3
x=16 x=16
24 ft
6 ft xs shadow
(ii) (5 points)At what rate is the distance between the lamp and the tip of his shadow changing when he is 16 ft away from the lamppost?
Solution Noting that L2 =(s + x)2 +242, we have
dL2 d(s + x)2 d242
=+
dt dt dt
dL ds dx
2L =2(s + x)+
dt dt dt
16 . 8
Putting in x =16, s = , and L = (16/3)+16)2 +242 = 145, we have
3
3
.
8 dL 64
145 = (1+3)
3 dt 3
x=16
dL . 32 32
= = 145 (ft/s)
dt x=16 145 145
( 2.6575 ft/s.)
3. (10=5+5points) Considertheparabola y = x2 .
(i) (5 points)Write down the equations of the tangent and normal lines to the parabola at the point (b,b2 ).
Solution y =2x, hence:
slope of thetangentline at(b,b2 ) =2b,
slope of the normalline at(b,b2)= . 1 ,if b .=0.
2b
Tangent line equation is y =2bx .b2
Normal line equation is y = . x +1+ b2 if b .0;otherwise x =
=0
2b 2
(ii) (5 points)The .gure shows a circle of radius 2 inscribed in the parabola. Determine the center of the circle.
Solution
By symmetry,the centerislocated at(0,a).
If(b,b2 )is one of the intersection points of the circle and the
parabola, we have
1
(b2 .a)2 +b2 =4 + b2 =4
4
b2..
.a 1 b