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APPM 1350 Exam 2 Solutions SUMMER 2006
- (25 points, 5 points each)
(a) g′(y) = (y)
cos
y
y^2
y (b) s′(r) = (2 tan r^3 )(sec^2 r^3 )(3r^2 )
(c) f ′(x) =
1 + tan
x + (^1) x
[
sec^2
x +
x
)] (
x^2
(d) p′(t) = 2
[(
1 +^2
t
] [
1 +^2
t
t^2
]
(e) h′(x) = [cos (sec (cos x))][sec (cos x) tan (cos x)](− sin x)
- (20 points, 5 points each)
(a) (^) tlim→∞
3 t^3 /^4 + t^5 /^6 + t^2 4 t^7 /^3 + (^25)
t
= lim t→∞
3 t−^19 /^12 + t−^3 /^2 + t−^1 /^3 4 + 25 t−^11 /^6
(b) (^) x→−∞lim (^ |xx^ + 3+ 3) |= (^) x→−∞lim
1 for x ≥ 0 − 1 for x < 0 =^ x→−∞lim −1 =^ −^1
(c) (^) rlim→∞
r^2 + 1 − r = lim r→∞(
r^2 + 1 − r)
√r^2 + 1 +^ r r^2 + 1 + r
= lim r→∞
√^1
r^2 + 1 + r
(d) (^) y→−∞lim^ y
3 y^2 + 1
= (^) y→−∞lim^ y 1 + (^) y^12
Dominant term: y y^2 + 1 |y^3 −y^3 − y −y
so y
3 y^2 + 1
= y − y y^2 + 1
and the dominant term is y.
- (15 points) Consider the curve called the conchoid of Nicomedes described by the equation x^2 y^2 = (y + 1)^2 (4 − y^2 ) (a) Find (^) dxdy.
(x^2 )(2y dy dx
) + (2x)(y^2 ) = 2(y + 1)( dy dx
)(4 − y^2 ) + (y + 1)^2 (− 2 y dy dx
2 x^2 y
dy dx + 2xy
(^2) = 2 dy dx (y^ + 1)(4^ −^ y
(^2) ) − 2 y dy dx (y^ + 1)
2
dy dx
· (2x^2 y − 2(y + 1)(4 − y^2 ) + 2y(y + 1)^2 ) = − 2 xy^2 dy dx
= −xy
2 x^2 y − (y + 1)(4 − y^2 ) + y(y + 1)^2
(b) Find the tangent to the conchoid at the point (0,-2). dy dx |(0,−2)^ = 0^ ⇒^ y^ =^ −^2
APPM 1350 Exam 2 Solutions Page 2 SUMMER 2006
- (20 points) f (x) = 2 cos x + sin 2x for 0 ≤ x ≤ 2 π
f ′(x) = −2 sin x+2 cos 2x = −2 sin x+2(1−2 sin^2 x) = −2(2 sin^2 x+sin x−1) = −2(2 sin x−1)(sin x+1)
Critical points: sin x =^12 or sin x = − 1 ⇒ x = π 6 , 56 π , 32 π
f ′′(x) = −2 cos x − 4 sin 2x = −2 cos x − 8 cos x sin x = −2 cos x(1 + 4 sin x) Possible inflection points: cos x = 0 or sin x = −
4 ⇒^ x^ =^
π 2 ,^
3 π 2 and at^ x^ =^ α^1 , α^2 where sin^ αn^ =^ −^
π^4 and 2 < α^1 <^
3 π 2 and^
3 π 2 < α^2 <^2 π Increasing/Decreasing:
(0, π 6 ) ( π 6 , 56 π ) ( 56 π , 32 π ) ( 32 π , 2 π) f ′(x) + − + +
Concave Up/Down:
(0, π 2 ) ( π 2 , α 1 ) (α 1 , 32 π ) ( 32 π , α 2 ) (α 2 , 2 π) f ′′(x) − + − + −
-2 2 4 6 8
1
2
3 H Π 6 ,^3
!!! 3 2 L
H Π 2 ,0L
H ^56 Π , -^3
!!! 3 2 L
H ^32 Π ,0L
HΑ 1 ,fHΑ 1 LL
HΑ 2 ,fHΑ 2 LL
H0,2L
H 2 Π,2L
