Solutions - Mu Alpha Theta

Solutions to Calculus Individual – FAMAT State Convention 2004
4 3 2
D 1. Use L’Hopital’s Rule twice to get D 8. V = πr dV = 4πr dr
3
2
dV = 4 π(2) (.01) dV = .16π
≈0.503
cos(2 x) + xcos x+ sin x−e
= lim
x→0
96xcos x+
96sin x
x
A 9. dw 1 1 dz 2
− 2sin(2 x) + 2cos x−xsin x−e
1
= + = 3( x + 2) 2
⋅2x
2
= lim
=
dz 27 z dx
x→0
192cos x−
96xsin x 192
dz dx 1 1 28
| = 54 | = + =
2 2
dx x= 1 dz z=
27 27 27 27
D 2. I. True by Intermediate Value Theorem
28 56
II. Not necessarily true because f is not
⋅ 54 =
2
27 27
necessarily differentiable on (2,5).
2
III. True by Intermediate Value Theorem C 10. d y 4y
2
= ⋅ 2y
−1=
2y
2
dx
2
2 2y
−1
2
A. 3. Solve 600 − 16t
= 0 f′ () t =−32t
2
d y
5 6 ⎛5 6 ⎞ | = 2(3) = 6
2
t =
f ' 2
⎜
8
2 ⎟ =− 0 6
dx y=
3
⎝ ⎠
x − x
B 11. f′ ( x) = e −e −2sinx
B 4.
f′ (0) = 0, f′ ( − .1) < 0, f′
(.1) > 0
7 x −10−5 7 − 10+ 5 7 −10−25
⋅ x =
x
f has a local minimum at x = 0.
10( x − 5) 7x− 10+ 5 10( x−5) ⋅ 7x− 10+
5
C 12. y′ = − 2sin(2 x) + 2cos(2 x) = 0
7x−35 7( x−5)
= =
1=
tan(2 x)
10( x−5) 7x− 10 + 5 10( x−5) 7x− 10 + 5
π π k
7 7
2 x= + kπ;
x= + π
lim =
+
x 5
10( 7 10 5)
→ x − +
4 8 2
100
The function has a period of π .
B 5.
y′ = x + x− =
2
3 4 4 1
2
3 4 15 0
x
+ x− =
(3x− 5)( x+ 3) = 0
1 f ⎛π
⎞ ⎛5π
⎞
⎜ ⎟= 2 and ⎜ ⎟=
2
⎝ 8 ⎠ f ⎝ 8 ⎠
−
The maximum value is 2,
5
x= or x=−
3
3
C 6.
h′ ( x) = f′ ( g( x)) ⋅g′
( x)
h′′ ( x) = f′ ( g( x)) ⋅ g′′ ( x) + g′ ( x) ⋅ f′′ ( g( x)) ⋅g′
( x)
h′′ (2) = f′ (2) ⋅ g′′ (2) + 4 f′′
(2)
=− g′′ (2) + 4 f′′ (2) =− f′′ (2) + 4 f′′ (2) = 3 f′′
(2)
3
1 2
A 7. y = 2 1 ( 2 1)
2
∫ x+ dx= ⋅ x+ + C
2 3
1
( 27 − 1
y(4) − y(0) )
3 26 13
= = =
4 4 12 6
B 13. I. True because f is strictly decreasing.
II. True because f ′ is decreasing.
III. Not necessarily true.
IV. True by the mean value theorem
applied to f ′ .
A 14. This series is equal to
π
∫
o
π
sin xdx = − cos x ] =−cosπ
−− cos0 = 2
E 15.
0
x
x
x e + 1−xe
g'( x) = ; g′′
( x)
=
x
e + 1 + 1
1
g′′ (0) =
2
x
( e ) 2

Solutions to Calculus Individual – FAMAT State Convention 2004
1
4
2
1
E 23. 6x+ 2 = 4 6x= 2 x=
D 16. u = x u(1) = 1 u(2) = 4
2
∫ e u du
3
π
1
⎛1⎞ 5
4
π
4⎜
⎟ − y = 3 y =−
C 17. ∫( cos x − sin x) dx+ ∫( sin x−cos
x)
dx=
⎝3⎠
3
2
0
π
5 ⎛1⎞
2 8
4
− = 3 ⎜ ⎟ + + k k = −
π
3 ⎝3⎠
3 3
4
π
( sin x+ cos x) ] + ( −cos x− sin x)
] = B 24.
0
π
I. . h(2) = f( g(2)) ≈ f(0) ≈5 NO
4
2 2 ⎛ 2 2 ⎞ II. h′ (3) = f′ ( g(3)) ⋅g′ (3) ≈ f′
(.25) ⋅+=−⋅+=− NO
+ − 1+ 1−0− − − = 2 2
2 2 ⎜ 2 2 ⎟ III. h′ (1) = f′ ( g(1)) ⋅ g′ (1) = f′
(1) ⋅−= 0 YES
⎝ ⎠
2
IV. h′′ ( x) = f′ ( g( x)) ⋅ g′′ ( x) + ( g′ ( x) ) ⋅ f′′
( g( x))
3
A 18. dy 2xy −6x
−12
2
| = | = =−3 ′′
2 2
h (4) = f′ ( g(4)) ⋅ g′′ (4) + ( g′ (4))
⋅ f′′
( g(4))
dx (2,0) 4−
3x y (2,0) 4
≈ 0+ 1⋅
f ′′(1) = a positive number
dy =− 3 dx dy =−3( − .03) = .09
2x cosx−
sinx
since f is concave up at 1. YES
A 19. v'( t) = a( t)
=
3
2
2x
A 25.
3 3π
3 dV 9π
2 dh
r = h V = h = h
a(3.5) < 0 and v(3.5) < 0
4 16 dt 16 dt
means speed is increasing.
9π
dh dh 2
2
6π
= ⋅16 ⋅ = A=
πr
16 dt dt 3
2
A 20. When x < 0 slopes are negative and
dA dr dA 1 dA ft
= 2 πr
= 2π
⋅3 ⋅ = 3π
when x > 0 slopes are positive. When
dt dt dt 2 dt min
x = 0 , slopes are 0.
D 26.
2 4
2
( )
B 21. y x
u 2 2 1
= x ∫ xf ′( x ) + 1 dx = ( ) 1
2
∫ f ′ u du + dx
3 4 3
′ = 20 + 15x = 5 x (4 + 3 x) = 0
∫
0 0
0
x= 0 or x=−4/3
1
4
1 1
= f( u)] + 2 = ( f(4) − f(0) ) + 2= ⋅ 12+ 2=
8
2 3 2
y′′ = 60x + 60x = 60 x (1 + x) = 0
2 0 2 2
x= 0 or x=−1
2 −4
B 27. f '( x) = f′′
( x) = f(0) = 0;
2x
+ 1 ( 2 + 1) 2
y′
x
+ − +
f′ (0) = 2; f′′
(0) =− 4 2 +− 4 =−2
-4/3 0
′′ y
D 22. ∫
−
+ +
−1
0
The graph has a local max, a local min,
and an inflection point.
3 4
∫
0 3
( )
x + 1dx + 2x −4
dx =
3
2
3 4
2
( x+ 1 ) 2 ] + ( x −4 x)
]
3
0 3
16 2 23
= − + 0+ 3=
3 3 3
−kx
−kx
C 28. f '( x) = e ( − kx + 1) f ′′( x) =−ke (2 −kx)
2 ⎛3⎞
f′′ ( x) = 0 when x= f′′ (0) < 0 f′′
⎜ ⎟>
0
k
⎝k
⎠
2
f is concave down for x<
.
k
B 29. The zeros of f are 1 and –7/2.
⎛ 7 ⎞
f′′ ( x) = 12x+ 6 f′′ (1) = 18 f′′
⎜− ⎟=−36
⎝ 2 ⎠
D 30.
g( − 1) =−2 f( − 2) =− 9 m= f′ ( −2) ⋅g′
( − 1) = 19
y− ( − 9) = 19( x+ 1) or 19x− y =−10