University of Guyana
Faculty of Technology
EMT 111 - Final Examination - Solutions
December 20, 2010
Directions: Answer all questions. Show all work. Calculators ARE allowed.
1. Evaluate.(9 points)
1
(a) log64 4 . Since 43 = 64 ⇒ 64 3 = 4 ⇒ log64 4 = 13
√
√
(b) log 10. Here log 10 = log 101/2 = 21 log 10 = 12
(c) (2x3 + x − 1) ÷ (x2 + 1). Using long division we get 2x with −x − 1
as a remainder.
2. Solve for x.(9 points)
(a) log(x + 2) + log(x − 1) = 1 ⇒ (x + 2)(x − 1) = 10
⇒ x2 + x − 12 = 0 ⇒ (x + 4)(x − 3) = 0 ⇒ x = 3
(b) x2 − x ≤ 6 ⇒ x2 − x − 6 ≤ 0 ⇒ (x − 3)(x + 2) ≤ 0
⇒ x ∈ [−2, 3]
(c) 10x+4 = 100 ⇒ 10x+4 = 102 ⇒ x + 4 = 2
⇒ x = −2
3. Compute the derivative of the following functions.(20 points)
(a) s(t) =
(b) h(x) =
1
t
−
1
t2
−
x5 −4x2
x3
1
√
t
⇒ s0 (t) = − t12 +
= x2 −
4
x
⇒
2
t3
−
h0 (x) = 2x +
1
√
2 t3
4
x2
(c) f (x) = esec x ⇒ f 0 (x) = esec x sec x tan x
sin x
(d) g(x) = ln(cos x) ⇒ g 0 (x) = − cos
x = − tan x
p
√
√
(e) s(t) = sin( 1 − t2 ) ⇒ s0 (t) = cos( 1 − t2 ) ( 1 − t2 )0
|
{z
√−2t
2
√
⇒ s0 (t) =
2
−t cos(
√ 1−t )
1−t2
1
1−t2
}
4. Evaluate the following integrals. (24 points)
1
(a)
x
+ 2 dx =
2
= 13 (x2 + 2)3/2 + C
Z
p
x2
Z 1
(b)
2
xex dx =
0
1
2
Z 1
0
Z p
1
2
x2 + 2 d(x2 + 2) = (x2 + 2)3/2 · + C
2
3
1
2
ex d(x2 ) = (e1 − e0 ) =
2
1
2 (e
− 1)
cos(1 − x)
dx = − sin(1−x)
+C
3
3
Z
x3
√
(d)
dx
4 − x2
LetZx = 2 sin θ ⇒ dx Z= 2 cos θdθ
x3
8 sin3 θ · 2 cos θdθ
⇒ √
dx =
2
2 cos θ
Z 4−x
Z
Z
(c)
=8
sin3 θdθ = 8
(1 − cos2 θ) sin θdθ
√
2 3/2
cos3 θ
2
+ (4−x24) ] + C
] + C = 8[− 4−x
2
3
Z
Z
Z
5x − 2
−3dx
8dx
(e)
dx =
+
= −3 ln |x − 1| + 8 ln |x − 2| + C
(x − 1)(x − 2)
x−1
x−2
= 8[− cos θ +
π
cos4 x 4
cos x sin x dx = −
cos x d(cos x) = −
4
0
0
0
Z
(f)
=
π
4
Z
3
−( √12 )4
4
+
π
4
1
−1 1
=
+ =
4
16
4
3
3
16
5. Use the fact that 12 fluid ounces is approximately 6.89π cubic inches
to find the dimensions of the 12-ounce can that can be constructed
using the least amount of metal. (5 points)
V = πr2 h = 6.89π ⇒ h = 6.89
r2
SA = 2πr2 + 2πrh = 2πr2 + 2πr 6.89
= 2π[r2 + 6.89
r ]
r2
Need r that would minimize the surface area, SA.
3
SA0 (r) = 2π[2r − 6.89
] = 2π[ 2r −6.89
]
r2
r2
√
Setting SA0 (r) = 0 we have 2r3 = 6.89 and r = 3 3.445 = 1.51
6.89
⇒ h = 1.51
2 =3.02
Therefore the can must have a radius of 1.51 inches and a height of
3.02 inches.
6. Find the absolute minimum and maximum values of the function
f (x) = 13 x3 − 9x + 2 on the interval [0, 2]. (5 points)
f 0 (x) = x2 − 9 = (x − 3)(x + 3) = 0 ⇒ x = 3,x = −3 are critical
numbers.
However, x = 3 and x = −3 are outside the interval [0, 2].
Therefore we need not evaluate the function at these points.
2
Now f (0) =2 ← absolute maximum
and f (2) =
8
3
− 18 + 2 = − 40
3 ← absolute minimum
1
1
1−sin x − 1+sin x (5 points)
1+sin x−(1−sin x)
1
1+sin x =
1−sin2 x
1
=
2
tan
x
sec x = LHS
cos x
7. (a) Prove: 2 tan x sec x =
1
RHS = 1−sin
x −
2 sin x
2 sin x
= cos2 x = cos x ·
(b) An ant starts at the point (1,0) on the unit circle and walks
counterclockwise 4 units around the circle. Find the coordinates
of the final location of the ant. (5 points)
Coordinates are (x, y) = (r cos θ, r sin θ)
but r = 1 (unit circle) and θ = l = 4
Therefore coordinates of the ant is (cos 4, sin 4) where the angles
are in radians.
(c) Show that the triangle with sides 3,4 and 5 is a right-angled
triangle. (3 points)
If 3,4,5 represent the sides of a right triangle then 52 must equal
32 + 42 (By the Pythagorean Theorem).
Let’s check:
?
52 = 32 + 42
?
25 = 9 + 16
?
25 = 25 yes!
This implies that a triangle with sides 3, 4, 5 is a right triangle.
8. When a resistor of R ohms is connected across a battery with electromotive force E volts and internal resistance r ohms, a current of I
amperes will flow, generating P watts of power,
E
where I = r+R
and P = I 2 R.
Assuming r is a constant, what choice of R results in maximum power?
(5 points)
2
E
E2R
R = (r+R)
2
r+R
(r+R)2 E 2 −2E 2 R(r+R)
0
=0
P (R) =
(r+R)4
2
⇒ (r + R)E [r + R − 2R] = 0 ⇒ r − R
P = I 2R =
= 0 ⇒ R = r gives maximum
power.
9. A fish pond is nearly circular, is 20 feet deep at its center ,and has a
radius of 200 feet. The bottom of the pond can be modeled by
y = 20[(0.005x)2 − 1]
How much water can this pond hold? (5 points)
UsingZ the method of shells, we have
200
V =
2πx(0 − 20[(0.005x)2 − 1]) dx
0
3
Z 200
= 2π
0
Z
= 40π
"0
20x[1 − (0.005x)2 ] dx
200
(x − 25 × 10−6 x3 ) dx
x2
x4
= 40π
− 25 × 10−6 ×
2
4
#200
= 40π
0
40000 25 × 10−6 × 16 × 108
−
2
4
= 40π[20000 − 10000] = 400000π = 1,256,637 cubic ft.
10. John is a pilot with TransGuyana Airways. Assume that during one
storm he flies with the wind (a tailwind) at 300 miles per hour. Leaving the storm, he flies against the wind(a headwind) at 225 miles per
hour. If he did not adjust his plane’s speed, find the speed of the plane
in still air and the speed of the wind. (5 points)
Let x = speed of plane in still air, and y = speed of the wind then
x + y = 300
x − y = 225
⇒ 2x = 525 ⇒ x = 262.5 mph
⇒ 2y = 75 ⇒ y = 37.5 mph
4
!