MATHEMATICS 184 (Section 921) - TERM EXAM 1
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SOLUTIONS
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INSTRUCTIONS: Notes, books and calculators are not permitted. Answers
may receive no credit if accompanying work is not provided. Simplify all
answers as much as possible, and be sure to clearly indicate the final answer
for each question. Please use the back of the pages if you need extra space.
1. Two populations of bacteria are raised in a laboratory for an experiment.
Colony A is provided with a limitless supply of food, and so is found to double
in size each day. Colony B is not provided with any food and so is found to
decrease each day to 1/3 of the previous day’s size.
(a) If the initial population size of bacteria colony B is PB , write an expression for the number of bacteria in colony B after t days.
[2]
P(t) = PB ( 13 )t
where
t = number of days
P(t) = population after t days
(b) At this rate of decrease, how long will it take for bacteria colony B to be
reduced to 20% of its original size?
This is asking for the value of t at which
P(t) = 15 PB
So,
1
P = PB ( 13 )t
5 B
1
= ( 13 )t
5
ln( 15 ) = (ln( 13 ))t
ln( 15 ) = tln( 13 )
t=
ln( 1
)
5
ln( 1
)
3
So colony B will be reduced to 20% of its original size after
days. (roughly 1.465 days)
ln( 1
)
5
ln( 1
)
3
[4]
(c) If there were initially ten times more bacteria in colony B than in colony
A, after how many days would there be the same number of bacteria in both
colonies?
1
for colony A: P2 (t) = 10
PB (2)t
for colony B: P1 (t) = PB ( 31 )t (from part a)
The question is asking for the value of t at which P2 (t) = P1 (t). So,
PB ( 13 )t =
( 13 )t =
10 =
1
P (2)t
10 B
1
(2)t
10
(2)t
(1
)t
3
10 = ( 21 )t
3
10 = 6t
log(10) = log(6t )
1 = tlog(6)
t=
1
log(6)
So the two colonies have the same number of bacteria after
days. (approximately 1.285 days)
1
log(6)
[6]
2. (a) John Dough has purchased a guaranteed investment certificate (GIC)
that pays 4.5% nominal interest compounded annually for the first year, followed by 5% nominal interest compounded monthly for all subsequent years.
If he initially invested 1 million dollars, how much would Mr. Dough’s investment be worth after 5 years? (You may express your answer in millions
of dollars)
[7]
Here we must do 2 compound interest calculations.
For the first year
P(t) = P0 (1.045)t
so after one year he’d have
P(1) = 1 · (1.045) = 1.045 million dollars
We then use this as the initial investment for the formula for the
next 4 years
P(t) = P0 (1 + nr )nt
Here t=4, n=12, r=0.05, P0 = 1.045 million, so after the next 4
years he’d have
P(4) = 1.045(1 +
0.05 48
)
12
and we can’t simplify any more so:
Mr. Dough’s investment would be worth 1.045(1 +
dollars after 5 years. (approximately 1.276 million)
0.05 48
)
12
million
(b) What would John Dough’s investment be worth after the same 5 years if
instead he had invested his million dollars at a competing bank that offered
4% compounded continuously?
For this investment we use the formula
P(t) = P0 ert
here r = 0.04, t = 5 and P0 is 1 million dollars so,
P(5) = 1 · e0.04·5
P(5) = e0.2
So John’s investment would be worth e0.2 million dollars after 5
years if he chose this option. (approximately 1.221 million dollars)
[4]
3. Suppose that a ball is dropped from the upper observation deck of the CN
tower, 490m above the ground. The height, h (in meters), of the ball above
the ground after t seconds is given by the equation h = 490 − 4.9t2 .
(a) What is the average velocity of the ball during its descent to the ground?
[6]
First, we need to know how long it takes the ball to reach the
ground. So we want to know the value of t for which h = 0.
0 = 490 − 4.9t2
490 = 4.9t2
t2 = 100
t = 10
So the ball hits the ground after 10 seconds. So we need the average
velocity over the time interval [0,10].
average velocity during descent =
s(10)−s(0)
10−0
where s(t) = h = 490 − 4.9t2
avg vel =
490−4.9(100)−490
10
= − 490
= −49 meters/sec.
10
So the average velocity during the ball’s descent is -49 m/sec.
(b) How fast is the ball travelling at the moment it hits the ground?
Here we want the instantaneous velocity at t=10.
inst. vel. at t=a = limh→0
s(a+h)−s(a)
h
Here a = 10, so
= limh→0
490−4.9(10+h)2 −(490−4.9(10)2 )
h
= limh→0
490−4.9(100+20h+h2 )−0
h
= limh→0
−98h−4.9h2
h
= limh→0 − 98 − 4.9h
= −98
So the velocity of the ball when it hits the ground is -98 m/sec.
However, since speed = |velocity|, the speed of the ball when it
hits the ground is 98 m/sec.
[6]
4. Let
√
f (x) =
2x − 1
(a) What is the value of f 0 (3)?
[6]
f 0 (a) =
(a)
limh→0 f (a+h)−f
h
Here a=3, so
f (3+h)−f (3)
h
f 0 (3) = limh→0
√
√
2(3+h)−1− 2(3)−1
= limh→0
h
√
√
√
√
5+2h− 5 √5+2h+√5
= limh→0
·
h
5+2h+ 5
= limh→0
(5+2h)−(5)
√
√
h( 5+2h+ 5)
= limh→0
√ 2h √
h( 5+2h+ 5)
= limh→0
2 √
√
( 5+2h+ 5)
Now plug in h=0, to get f 0 (3) =
2
√
2 5
=
√1
5
(b) Find an equation for the straight line that passes through the point (3, f (3))
and that has slope f 0 (3) (from part (a)).
The equation of a line is:
y = mx + b
Here we know that m = f 0 (3), so
y=
√1 x
5
+b
So we only need to find b. We are also√ given that the line passes
through (3, f (3)) which is the point (3, 5), so we’ll use these as x
and y and solve for b:
√
5 = √15 (3) + b
√
b = 5 − √35
b=
√5
5
−
√3
5
=
√2
5
So the equation of the line is
y=
√1 x
5
+
√2
5
[4]