First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has been left blank for your workings.
Quiz 4.1: Page 1 of 4
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
(a) Consider a function, h(x), whose third-degree Maclaurin polynomial is 5 − 31 x2 + 2x3 .
What is h0 (0)?
Answer: 0
Solution: Since T3 (x) = h(0) + h0 (0) · x + 12 h00 (0) · x2 + 16 h000 (0) · x3 , we must have that
h0 (0) = 0.
(b) You are driving at 100kph. The light ahead changes to red and you brake smoothly to a
stop in 8 seconds with constant deceleration. How fast were you traveling 1 second before
you stop?
Answer: 12.5kph
Solution: Since v(t) = 100 − ct and v(8) = 0, c = 100/8 = 25/2. Thus at v(7) =
100 − 7c = 25/2 = 12.5kph.
Quiz 4.1: Page 2 of 4
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) A sample of that very rare metal plotdeviceium was measured at 9AM yesterday as
weighing 10 grams. It was measured again today at 10AM as weighing 8 grams. What is
the half-life of plotdeviceium? Leave your answer in calculator ready form.
Solution: We know that
W (t) = 10e−kt
W (25) = 8 = 10e−25k
So
e−25k =
4
5
k = − log(4/5)/25 =
log 5 − log 4
25
Then to find half-life we work out when W (t) = 5:
5 = 10e−kt
1
= e−kt
2
− log 2 = −kt
t = log(2)/k =
(b) Estimate
√
25 log 2
log 5 − log 4
35 using a linear approximation. Leave your answer as a fraction.
Answer: 71/12
Solution: We use the function f (x) =
approximation since we know that
√
x and point a = 36 as the centre of our
f (a) = f (36) =
We compute f 0 (x) =
1
√
;
2 x
√
36 = 6.
so
1
1
f 0 (36) = √ = .
12
2 36
√
So, a linear approximation of 35 = f (35) is
T1 (35) = f (36) + f 0 (36) · (35 − 36) = 6 −
Quiz 4.1: Page 3 of 4
1
71
= .
12
12
Long answer question — you must show your work
3. 4 marks Two ships are sailing in the ocean. At 2PM, Ship A is 10km east of the origin, and
travels west at a constant rate of 2km/h. At the same time Ship B is 12km north of the origin
and travels south at a constant rate of 3km/h. What is the rate of change of the distance
between the two ships when Ship A is 4km east of the origin?
Solution:
• We compute the distance z(t) between the ships as
z 2 (t) = x(t)2 + y(t)2 ,
where x(t) is the position of ship A east of the origin at time t (in hours) y(t) is the
position of ship B north of the origin at the same time t.
• We differentiate the above equation with respect to t and get
2z · z 0 = 2x · x0 + 2y · y 0 ,
• We are told that x0 = −2 and y 0 = −3. Further it will take 3 hours for particle A to
reach x = 4, and in this time particle B will reach y = 3.
• Alternatively (equivalently??) write x = 10 − 2t, y = 12 − 3t to get t = 3, x0 =
−2, y 0 = −3, y = 3.
p
√
• At this point z = x2 + y 2 = 32 + 42 = 5.
• Hence
10z 0 = 8 · (−2) + 6 · (−3) = −34
34
17
z 0 = − = − km/h.
10
5
Quiz 4.1: Page 4 of 4
First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has been left blank for your workings.
Quiz 4.2: Page 1 of 4
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
(a) Consider a function, h(x), whose third-degree Maclaurin polynomial is 5 − 21 x2 + 2x3 .
What is h00 (0)?
Answer: −1
Solution: Since T3 (x) = h(0) + h0 (0) · x + 12 h00 (0) · x2 + 16 h000 (0) · x3 , we must have that
h00 (0) = −1.
(b) You are driving at 60kph. The light ahead changes to red and you brake smoothly to a
stop in 5 seconds with constant deceleration. How fast were you traveling 1 second before
you stop?
Answer: 12 kph
Solution: Since v(t) = 60−ct and v(5) = 0, c = 60/5 = 12. Thus at v(4) = 60−4c =
12kph.
Quiz 4.2: Page 2 of 4
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) A sample of that very rare metal plotdeviceium was measured at 9PM yesterday as
weighing 8 grams. It was measured again today at 10AM as weighing 6 grams. What is
the half-life of plotdeviceium? Leave your answer in calculator ready form.
Solution: We know that
W (t) = 8e−kt
W (13) = 6 = 8e−13k
So
3
4
e−13k =
k = − log(3/4)/13 =
log 4 − log 3
13
Then to find half-life we work out when W (t) = 4:
4 = 8e−kt
1
= e−kt
2
− log 2 = −kt
t = log(2)/k =
(b) Estimate
√
13 log 2
log 4 − log 3
15 using a linear approximation. Leave your answer as a fraction.
Answer: 31/8
Solution: We use the function f (x) =
approximation since we know that
√
x and point a = 16 as the centre of our
f (a) = f (16) =
We compute f 0 (x) =
1
√
;
2 x
√
16 = 4.
so
So, a linear approximation of
√
1
1
f 0 (16) = √ = .
8
2 16
15 = f (15) is
T1 (15) = f (16) + f 0 (16) · (15 − 16) = 4 −
Quiz 4.2: Page 3 of 4
1
31
= .
8
8
Long answer question — you must show your work
3. 4 marks Two ships are sailing in the ocean. At 1PM, Ship A is 11km east of the origin, and
travels west at a constant rate of 4km/h. At the same time Ship B is 14km north of the origin
and travels south at a constant rate of 5km/h. What is the rate of change of the distance
between the two ships when Ship A is 3km east of the origin?
Solution:
• We compute the distance z(t) between the ships as
z 2 (t) = x(t)2 + y(t)2 ,
where x(t) is the position of ship A east of the origin at time t (in hours) y(t) is the
position of ship B north of the origin at the same time t.
• We differentiate the above equation with respect to t and get
2z · z 0 = 2x · x0 + 2y · y 0 ,
• We are told that x0 = −4 and y 0 = −5. Further it will take 2 hours for ship A to
reach x = 3, and in this time particle B will reach y = 4.
• Alternatively (equivalently??) write x = 11 − 4t, y = 14 − 5t to get t = 2, x0 =
−4, y 0 = −5, y = 4.
p
√
• At this point z = x2 + y 2 = 32 + 42 = 5.
• Hence
10z 0 = 6 · (−4) + 8 · (−5) = −64
64
32
z 0 = − = − km/h.
10
5
Quiz 4.2: Page 4 of 4
First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has birdn left blank for your workings.
Quiz 4.3: Page 1 of 4
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
(a) Consider a function, h(x), whose third-degree Maclaurin polynomial is 5 − 31 x2 + 2x3 .
What is h(0)?
Answer: 5
Solution: Since T3 (x) = h(0) + h0 (0) · x + 12 h00 (0) · x2 + 16 h000 (0) · x3 , we must have that
h(0) = 5.
(b) You are driving at 120kph. The light ahead changes to red and you brake smoothly to
a stop in 10 seconds with constant deceleration. How fast were you traveling 4 seconds
after you applied the brake?
Answer: 72 kph
Solution: Since v(t) = 120 − ct and v(10) = 0, so c = 120/10 = 12. Thus v(4) =
120 − 12(4) = 120 − 48 = 72kph.
Quiz 4.3: Page 2 of 4
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) A sample of rainon, a fictional radioactive isotope prevalent in Vancouver, was measured
at 8AM yesterday as weighing 10 grams. It was measured again today at 12PM as
weighing 7 grams. What is the half-life of rainon? Leave your answer in calculator ready
form.
Solution: We know that
W (t) = 10e−kt
W (28) = 7 = 10e−28k
So
7
10
−28k = log 7 − log 10
log 10 − log 7
k=
28
e−28k =
Then to find half-life we work out when W (t) = 5:
5 = 10e−kt
1
= e−kt
2
− log 2 = −kt
t = log(2)/k =
(b) Estimate
√
3
28 log 2
log 10 − log 7
9 using a linear approximation. Leave your answer as a fraction.
Answer:
Solution: We use the function f (x) =
approximation since we know that
√
3
x and point a = 8 as the centre of our
f (a) = f (8) =
We compute f 0 (x) = 13 x−2/3 =
1
√ 2;
33x
25
12
√
3
8 = 2.
so
1
1
f 0 (8) = √
=
.
2
12
338
Quiz 4.3: Page 3 of 4
So, a linear approximation of
√
3
9 = f (9) is
T1 (9) = f (8) + f 0 (8) · (9 − 8) = 2 +
Quiz 4.3: Page 4 of 4
1
25
= .
12
12
Long answer question — you must show your work
3. 4 marks Two birds are flying directly towards the same tree. At 1PM, bird A is 11 metres
east of the tree, flying west at a constant rate of 2 metres per second. At the same time,
bird B is due north of the tree, flying south. The distance between the two birds is 17 metres
at 1PM, and is decreasing at a constant rate of 3 metres per second. How fast is bird B flying
when bird A is 3 metres from the tree?
Solution:
• We compute the distance D(t) between the birds as
D2 (t) = A(t)2 + B(t)2 ,
where A(t) is the distance (in metres) from bird A east to the tree at time t (in
seconds), B(t) is the distance from bird B to the tree at the same time t.
• We differentiate the above equation with respect to t and get
2D · D0 = 2A · A0 + 2B · B 0
D · D 0 = A · A0 + B · B 0
• We are told that A0 = −2 and D0 = −3. Further, it will take 4 seconds for bird A to
reach A = 3, and in this time the distance between the birds will be D = 17 − 12 = 5
metres.
• Alternatively (equivalently??) write A = 11 − 2t, D = 17 − 3t to get t = 4, A0 =
−2, D0 = −3, D = 5.
√
√
• At this point B = D2 − A2 = 52 − 32 = 4.
• Hence
(5)(−3) = (3)(−2) + (4)B 0
−15 = −6 + 4B 0
−15 + 6
9
B0 =
= − m/s.
4
4
So Bird A is travelling south at 9/4 metres per second.
Quiz 4.3: Page 5 of 4