Download MA107 Precalculus Algebra Exam 4 Review Solutions

MA107 Precalculus Algebra
Exam 4 Review Solutions
12 April, 2008
1. Find the amount that results from each of the following investments:
a. 1000 dollars invested at 5% compounded semi-annually, after a period of 10 years.
(*) The compound interest formula is A = P(1 +
r nt
n) .
With P = 1000, n = 2, r = .05, t = 10, I get: A = 1000(1 +
.05 (2)(10)
2 )
= 1638.62.
b. 1000 dollars invested at 4.5% compounded monthy, after a period of 10 years.
(*) With P = 1000, n = 12, r = .045, t = 10, I get: A = 1000(1 +
.045 (12)(10)
12 )
= 1566.99.
c. 1000 dollars invested at 4% compounded continously, after a period of 10 years.
(*) The continuous interest formula is A = Pert .
With P = 1000, r = .04, t = 10, I get: A = 1000e(.04)(10) = 1491.82.
2. Find the present value of 5000 dollars invested at 10% interest, compounded daily.
(*) The present value formula is P = A(1 +
r nt
n) .
For n = 365, A = 5000, r = .10, I get:
P = 5000(1 +
.1 365t
.
365 )
1
3. At what interest rate would an initial investment of 3000 dollars double in value over 5 years? (Compounded continuously)
(*) We have A = Pert , with P = 3000, t = 5.
We want to solve for r when A = 6000.
6000 = 3000e5r
2 = e5r
ln(2)
= ln(e5r )
ln(2)
= 5r
r
=
ln(2)
5
4. The population of a colony of bacteria obeys the law of uninhibited growth. Initially there is a population of 2 million. After a period of 40 days, there is a population of 3 million.
a. What is the groth constant?
(*) I use the exponential growth model, P(t) = P0 ekt , where k is the growth constant, P0 is
the initial population, and t is time (days).
To find the growth constant, I need to plug known values into the model and solve
for k. I know P0 = 2000000, and when t = 40, P(t) = 3000000.
3000000 = 2000000e40k
3
2
= e40k
ln( 32 )
= ln(e40k )
ln( 32 )
= 40k
k =
ln( 3
2)
40
b. What is the size of the colony after 9 days?
(*) We now have the model P(t) = 2000000e
Plugging in 9 for t, see:
P(t) = 2000000e
ln( 3 )
2
40 9
≈ 2191041
2
ln( 3 )
2
40 t
.
c. When will the population double?
(*) We now have the model P(t) = 2000000e
ln( 3 )
2
40 t
.
I want to know t when P(t) = 4000000 (twice the initial population).
4000000 = 2000000e
2 =e
ln( 3 )
2
40 t
ln(2)
= ln(e
ln(2)
=
ln(2)
=t
ln( 3 )
2
40
ln( 3 )
2
40 t
3
)
ln( 2
40 t
t ≈ 60.38
3
ln( 3 )
2
40 t
)
5. A copy of this exam review has been sealed in a gold temple, awaiting to enlighten the mankind of
the future. After the temple has been opened and the mummy’s curse broken, archaeologists set to
work determining the age of the paper. They find it contains 0.3% of its original amount of Carbon-14.
Assuming a half-life of 5, 600 years, determine the age of the paper.
(*) The model for exponential decay is A(t) = A0 e−kt .
First we need to find k. The half-life tells me that when t = 5600, then A(t) =
So I plug these values into the model and solve for k.
1
2 A0
1
2 A0 .
= A0 e−5600k
1
2
= e−5600k
ln( 12 )
= ln(e−5600k )
ln( 21 )
= −5600k
ln( 1 )
2
− 5600
=k
ln( 1 )
2
Now that I have the model A(t) = A0 e 5600 t , I can work on determining the age of the
paper. I want to know t when A(t) = .003A0 . So I plug this information into the model
and solve for t.
.003A0
ln( 1 )
2
= A0 e 5600 t
ln( 1 )
2
.003 = e 5600 t
ln( 1 )
2
ln(.003)
= ln(e 5600 t )
ln(.003)
=
ln(.003)
ln( 1 )
2
5600
ln( 1
2)
5600 t
=t
t ≈ 46932.60
6. Convert 90 degrees to radians.
π
(*) 90◦ × 180
= π2 .
7. Convert
(*)
7
12 π
7
12 π radians to
◦
× 180
π = 105
degrees.
4
8. Suppose sin(θ) = 35 . The identities to find the exact value of each of the five remaining trigonometric
functions (cos(θ), tan(θ), csc(θ), sec(θ), cot(θ)).
(*) Consider a right triangle. The sine of an angle is the ratio of the lengths of the
“opposite” leg to the hypotenuse. So we can say the “opposite” leg has length 3, and the
hypotenuse has length 5.
To find the length of the “adjacent” leg, use the Pythagorean Theorem.
32 + x2
= 52
9 + x2
= 25
x2
= 16
x
=4
So the length of the “adjacent” leg is 4.
sin(θ)
=
opposite
hypotenuse
=
3
5
cos(θ)
=
adjacent
hypotenuse
=
4
5
tan(θ)
=
adjacent
hypotenuse
=
sin(θ)
cos(θ)
csc(θ)
=
1
sin(θ)
=
5
3
sec(θ)
=
1
cos(θ)
=
5
4
cot(θ)
=
1
tan(θ)
=
4
3
5
=
3
5
4
5
=
3
4
9. Suppose tan(θ) = 53 . The identities to find the exact value of each of the five remaining trigonometric
functions (sin(θ), cos(θ), csc(θ), sec(θ), cot(θ)).
(*) Again, consider a right triangle. Tangent is the ratio of the “opposite” leg to the
“adjacent” leg. We consider the “opposite” leg to be 3, and “adjacent” leg to be 5.
To get the hypotenuse, use the Pythagorean Theorem.
32 + 52
= x2
9 + 25 = x2
34 = x2
√
34 = x2
sin(θ)
=
opposite
hypotenuse
=
√3
34
cos(θ)
=
adjacent
hypotenuse
=
√5
34
tan(θ)
=
adjacent
hypotenuse
=
sin(θ)
cos(θ)
csc(θ)
=
1
sin(θ)
=
√
34
3
sec(θ)
=
1
cos(θ)
=
√
34
5
cot(θ)
=
1
tan(θ)
=
5
3
=
√3
34
√5
34
=
3
5
10. Without a calculator, determine the value of sec2 (83◦ ) − tan2 (83◦ ).
(*) Recall that 1 + tan2 (θ) = sec2 (θ). So:
sec2 (83◦ ) − tan2 (83◦ )
= (1 + tan2 (83◦ )) − tan2 (83◦ )
= 1 + tan2 (83◦ ) − tan2 (83◦ )
=1
6
11. Without a calculator, determine the value of (csc2 (13◦ ) − cot2 (13◦ )) × cos(13◦ ) × sec(13◦ ).
(*) Recall that 1 + cot2 (θ) = csc2 (θ). Also recall that sec(θ) =
1
cos(θ)
So:
(csc2 (13◦ ) − cot2 (13◦ )) × cos(13◦ ) × sec(13◦ )
=
((1 + cot2 (13◦ )) − cot2 (13◦ )) × cos(13◦ ) × sec(13◦ )
=
1 × cos(13◦ ) × sec(13◦ )
=
1 × cos(13◦ ) ×
=
1×1
=
1
1
cos(13◦ )
12. Without a calculator, determine the value of
cos(33◦ )
sin(57◦ ) .
(*) Because 33◦ + 57◦ = 90◦ , and cos and sin are cofunctions, then cos(33◦ ) = sin(57◦ ).
So I write
cos(33◦ )
sin(57◦ )
=
sin(57◦ )
sin(57◦ )
= 1.
13. Without a calculator, determine the value of cot(10◦ ) −
(*)
cos(10◦ )
sin(10◦ )
= cot(10◦ ), so cot(10◦ ) −
cos(10◦ )
sin(10◦ )
= 0.
7
cos(10◦ )
sin(10◦ ) .
14. Given cos(30◦ ) =
1
2
√
3, determine the value of
a. cos(60◦ ).
(*) In class we derived 12 .
b. sin2 (30◦ ).
(*) We know sin2 (30◦ ) + cos2 (30◦ ) = 1.
cos(30◦ ) =
√
3
2 ,
so cos2 (30◦ ) = (
√
3 2
2 )
= 34 .
Then
sin2 (30◦ ) +
3
4
=1
sin2 (30◦ )
=
1
4
sin(30◦ )
=
1
2
c. sec(30◦ ).
(*) sec(30◦ ) =
1
cos(30◦ )
=
√2 .
3
d. csc(30◦ ).
(*) By Complementary Angle Theorem, sin(30◦ ) = cos(60◦ ) = 12 . Then
csc(30◦ ) =
1
sin(30◦ )
= 2.
8
15. A ship, offshore from a vertical cliff known to be 900 feet in height, takes a sighting of the top of the
cliff. The angle of elevation is found to be 30◦ . How far offshore is the ship?
(*) We want to find x, the distance from the cliff.
I know tan(30◦ ) =
.577 =
opposite
, so
adjacent
900
x .
Solve for x to get x =
900
.577
≈ 1559.79.
9
16. To measure the height of the Watson Towers, two sightings are taken at a distance of 120 feet apart. If
the first angle of elevation is 66◦ , and the second is 60◦ , what is the height of the tower?
(*) The trick here is to look for right triangles. There are two. Both share the leg formed by
the tower (vertical red line). But the base of one is the horizontal red line, with a length of
x. The base of the other is the blue line, with a length of 120 + x.
The key is the tangent function. tan(θ) =
tan(66◦ ) = 2.246 =
h
x
tan(60◦ ) = 1.732 =
h
120+x
opposite
. I can get two relations.
adjacent
I have two equations and two variables. This gives me a system of two equations:
2.246 =
h
x
eqn 1
1.732 =
h
120+x
eqn 2
Solve equation 1 for x.
2.246 =
2.246x
x
h
x
=h
=
h
2.246
Now plug this solution into equation 2.
1.732 =
1.732(120 +
h
2.246 )
h
h
120+ 2.246
=h
207.84 + .771h = h
207.84 = .229h
h = 907.6
10
17. A point in the terminal side of an angle θ is (2, 3). (Assume the vertex of the angle rests on the origin).
Find each of the following:
a. sin(θ).
(*) The origin, the point (2, 0), and the point (2, 3) form a right triangle.
The line along the x-axis has a length of 2. Call this line a.
The line from (2, 0) to (2, 3) has a length of 3. Call this line b.
Then c2 = 22 + 32 = 4 + 9 = 13, so c =
sin(θ) =
b
c
=
√
13.
√3 .
13
b. cos(θ).
(*) cos(θ) =
a
c
=
√2 .
13
b
a
= 32 .
c. tan(θ).
(*) tan(θ) =
d. csc(θ).
(*) csc(θ) =
=
1
cos(θ)
=
1
tan(θ)
= 23.
e. sec(θ).
(*) sec(θ) =
√
1
sin(θ)
133.
√
132.
f. cot(θ).
(*) cot(θ) =
g. θ (in degrees and radians).
(*) tan(θ) =
b
a
= 32 , so θ = tan−1 ( 32 ) ≈ 56.31◦ .
11
18. The hypotenuse of a right triangle is 10 inches. One leg is 4 inches. Find the degree measure of each
angle.
(*) We just need one of the angles. Call it α. Because the sum of the angles of a triangle is
180◦ , and the triangle is a right triangle, the other angle will be 90◦ − α.
Let’s find the angle, α, that is formed by the hypotenuse and the known leg.
known leg is adjacent to the angle, so we’ll use the cosine relationship.
cos(α) =
adjacent
. So
hypotenuse
cos(α) =
4
10 .
The
4
Then α = cos−1 ( 10
) ≈ 66.42◦ .
Then the other angle is 90◦ − 66.42◦ = 23.58◦ .
19. Milton is sitting on a wall 4 feet above the ground. He is watching a squirrel in a tree 15 feet away. If
the angle of elevation is 55◦ , how tall is the tree?
(*)
Look at the diagram. We can use trigonometry to find “h”.
I know tan(55◦ ) =
1.428 =
opposite
, so
adjacent
h
15 .
Solve for h to get h = 15 × 1.428 = 21.42.
But the tree is an additional 4 feet tall. So the height is 25.42.
12
20. Solve the right triangle shown for each of the given parameters.
a. a = 4, b = 5.
(*) To get c, use the Pythagorean Theorem.
42 + 52
= c2
16 + 25 = c2
41 = c2
√
41 = c
opposite
= √541 .
hypotenuse
Then β = sin−1 ( √541 ) ≈ .896055385radians = 51.34◦ .
To get β, note that sin(β) =
To get α, note that the sum of the angles of a triangle is 180◦ .
180 − 90 − 51.34◦ = 38.66◦ .
b. α = 15◦ , a = 5.
(*) To get β, note that the sum of the angles of a triangle is 180◦ .
180 − 90 − 15◦ = 75◦ .
Now note that cos(β) =
adjacent
.
hypotenuse
So cos(75◦ ) = 5c .
5
Then c = cos(75
◦ ) ≈ 19.32.
Now note that sin(β) =
opposite
.
hypotenuse
b
So sin(75◦ ) = 19.32
.
Then b = 19.32 sin(75◦ ) ≈ 18.66.
13
So α must be
So β must be
c. α = 40◦ , β = 50◦ .
(*) There are multiple solutions to this problem. Here is how to derive one.
First note that sin(50◦ =
.766 =
opposite
, so
hypotenuse
b
a.
There are many choices for a and b. (Hence, why there are multiple solutions to the
problem). The easiest is to let b = .766 and a = 1.
Use the Pythagorean Theorem to get c:
(.7662 ) + 12 = c2
.586756 + 1 = c2
1.586756 = c2
1.260 ≈ c
14
21. Sketch a graph of the region bounded by x = 1, x = 3, y = x2 , y = 2x + 3 Where do the curves
intersect?
(*) To find the intersection, solve for x: x2 = 2x + 3.
x2
= 2x + 3
x2 − 2x − 3 = 0
(x − 3)(x + 1)
x=3
So I have the coordinates (3, 9) and (−1, 1).
15
=0
x = −1
22. Sketch a graph of the region bounded by y = x, y = x2 Where do the curves intersect?
(*) (note the scale on the graph is 1 line for every
1
4
unit.
To find the intersection, solve for x: x = x2 .
x
= x2
x − x2
=0
x(1 − x)
=0
x=0
x=1
So I have the coordinates (0, 0) and (1, 1).
16