Math 180, Section 104
Midterm 2 — November 9th 2010
Page 1 of 7
This midterm has 4 questions on 7 pages, for a total of 33 points.
Duration: 50 minutes
• Read all the questions carefully before starting to work.
• Give complete arguments and explanations for all your calculations, answers without
justifications will not be marked.
• Continue on the back of the previous page if you run out of space.
• Attempt to answer all questions for partial credit.
• This is a closed-book examination. None of the following are allowed: documents,
cheat sheets or electronic devices of any kind (including calculators, cell phones, etc.)
Full Name (including all middle names):
Student Number/ID:
Signature:
Question:
1
2
3
4
Total
Points:
15
6
6
6
33
Score:
Page 1 of 7
Math 180, Section 104
Midterm 2 — November 9th 2010
Page 2 of 7
SHORT ANSWER QUESTIONS.
Please show your work and also underline your final answer.
Each question is worth 3 marks, but an incorrect answer will be given at most 1 mark.
Unless otherwise stated, it is not necessary to simplify your answers.
3 marks
)). Note: another notation for arctan is tan−1 .
1. (a) Find arctan(tan( 5π
4
Solution: Recall that the range of arctan is (− π2 , π2 ) so the answer cannot be
5π
since that’s outside the range! In fact, drawing a picture of tan x we realize
4
that all we have to do is shift x = 5π
back into the range of x-values for which
4
− π = π4 .
tan x is one-to-one. This is achieved by shifting by −π to get 5π
4
3 marks
(b) Find the derivative of y = arccos(e2x ). Note: another notation for arccos is cos−1 .
1
Solution: We remember that the derivative of arccos x is − √1−x
2 or if we forgot
this we derive this formula via implicit differentiation. Then we use chain rule:
2e2x
d
1
· e2x · 2 = − √
arccos(e2x ) = − p
dx
1 − e4x
1 − (e2x )2
2
Note: Exponents multiply so (e2x )2 6= e4x . This was a common error.
Page 2 of 7
Math 180, Section 104
3 marks
Midterm 2 — November 9th 2010
Page 3 of 7
(c) Find the equation of the tangent line at (1, 2) of the curve x2 + 2xy − y 2 + x = 2.
Solution: We need to find the slope via implicit differentiation.
2x + 2xy 0 + 2y − 2yy 0 + 1 = 0
Solving for y 0 we get
y0 =
1 + 2x + 2y
2y − 2x
Finally, we evaluate at (1, 2) and find
y 0 (1, 2) =
1+2+4
7
=
4−2
2
So the equation of the tangent line at (1, 2) is y − 2 = 27 (x − 1).
3 marks
(d) Find the derivative of y = (ln x)sin x . Make sure to express your final answer in terms
of x only.
Solution: We use logarithmic differentiation.
ln y = sin x · ln(ln x)
1 0
1 1
· y = cos x · ln(ln x) + sin x ·
·
y
ln x x
sin x
y 0 = (ln x)sin x · (cos x · ln(ln x) +
)
x ln x
Page 3 of 7
Math 180, Section 104
3 marks
Midterm 2 — November 9th 2010
Page 4 of 7
2
(e) Use an appropriate linear approximation to compute (8.06) 3 and express your final
n
where n is an integer.
answer in the form 300
Solution: The closest nicely “cube-rootable” number to 8.06 is 8. So we’ll find
2
1
the tangent line to f (x) = x 3 at x = 8 and plug in 8.06. Since f 0 (x) = 32 x− 3 we
find
1
2 1
L(x) = f (8) + · 1 (x − 8) = 4 + (x − 8)
3 83
3
Now plug in 8.06
1
L(8.06) = 4 + (8.06 − 8)
3
0.06
=4+
3
6
=4+
100 · 3
1200
6
=
+
300
300
1206
=
300
Page 4 of 7
Midterm 2 — November 9th 2010
Math 180, Section 104
Page 5 of 7
FULL-SOLUTION PROBLEMS
In questions 2–4, justify your answers and show all your work. If you need more space,
use the back of the previous page.
6 marks
2. Two sides of a triangle have fixed lengths 12 m and 15 m. The angle between them is
increasing at a rate of 2 degrees per minute. How fast is the length of the third side
increasing when the angle between the sides of fixed length is 60 degrees? A useful
formula for this problem will be the law of cosines: c2 = a2 + b2 − 2ab cos(θ) where c is
the side opposite the angle θ.
Solution: We draw a diagram of the triangle in which we call the side of length 15
a, the side of length 12 b, and call the angle between a and b θ. The triangle is not
necessarily a right triangle so we’ll have to use the Law of Cosines to find the length
= 2 which we
of the side opposite the angle θ which we’ll call c. We know that dθ
dt
π
dc
convert into π radians to get dθ
=
.
We
are
looking
for
which
we’ll find by
dt
90
dt
differentiating the Law of Cosines with respect to time and the plug in the unknown
values one by one. We start by differentiating which is quite easy since a and b are
fixed:
2cc0 = 2ab sin θ · θ0
ab · θ0 sin θ
c0 =
c
Now we would like to evaluate c0 at θ = π/3. The only thing in the above equation
we don’t know yet is c(π/3). This can be found again by using the Law of Cosines:
p
c(π/3) = 152 + 122 − 2 · 15 · 12 cos(π/3)
√
√
√
= 152 + 122 − 12 · 15 = 189 = 3 21
since cos(π/3) = 1/2. Now we’re ready for the final answer:
c0 (π/3) =
π
sin(π/3)
90
√
3 21
15 · 12 ·
√
π 3
180 · 90
√ 2
=
3 21
√
π 3
= √
3 21
√
since sin(π/3) =
3
.
2
Page 5 of 7
Math 180, Section 104
6 marks
Midterm 2 — November 9th 2010
Page 6 of 7
3. The power generator at one of the Cryonics Institute’s cryopreservation facilities has
failed and some frozen bodies are starting to warm up. They have been kept at 88
Kelvin in a room at 15◦ C = 288 Kelvin. It is known that if they heat up beyond 138
Kelvin they will become zombies. How long does the institute have to reactivate its
generator before the Undead will walk the Earth? It is known that human bodies heat
up at a rate of 1 Kelvin per minute at 88 Kelvin in a 15◦ C room. Recall Newton’s Law
of Cooling: T 0 (t) = k(T (t) − Ts ). Advice: Make sure not to mix up your units!
Solution: We want to find the time after which the bodies have reached a temperature of 138 Kelvin. We assume that they heat up according to Newton’s Law of
Cooling T 0 (t) = k(T (t) − Ts ). We are told that T 0 = 1 when T = 88 which gives
us a nice quick way of finding k: we just plug into the Law to get 1 = k(88 − 288)
1
so k = − 200
where we used that we are told the surrounding temperature is 288
Kelvin. Next we recall that the solution to Newton’s Law of Cooling is given by
T (t) = Ts + (T (0) − Ts )ekt or if we forgot it we derive it quickly using the substitution
method. Since we know k now we solve this equation for when T (t) = 138:
138 = 288 + (88 − 288)ekt
−150 = −200ekt
150
) = kt
ln(
200
−200 ln(3/4) = t
Note that this answer makes sense since ln(3/4) is negative so that we end up getting
a positive time. So the Cryonic’s Institute has −200 ln(3/4) minutes which is about
two hours to turn their generators back on.
Page 6 of 7
Math 180, Section 104
6 marks
Midterm 2 — November 9th 2010
Page 7 of 7
4. Find the third degree Taylor polynomial about a = 1 of f (x) = ln(2 − x). What is an
upper bound for the worst error that could result from using this Taylor polynomial to
(n+1)
approximate ln(0.98)? Lagrange’s Remainder Formula is Rn (x) = f (n+1)!(c) (x − a)(n+1)
where c lies between x and a.
Solution: Here are the first four derivatives which were all found using power rule
and chain rule in conjunction:
f (x) = ln(2 − x)
1
f 0 (x) = −
2−x
1
f 00 (x) = −
(2 − x)2
2
f 000 (x) = −
(2 − x)3
2·3
f 0000 (x) = −
(2 − x)4
Evaluating at x = 1 we get
f (0) = 0
f 0 (0) = −1
f 00 (0) = −1
f 000 (0) = −2
So we compute the ck by dividing by k!:
c0
c1
c2
c3
=0
= −1
= −1/2
= −1/3
So
1
1
T3 (x) = −(x − 1) − (x − 1)2 − (x − 1)3 .
2
3
So now we have that f (x) = ln(2 − x) ≈ T3 (x) when x near 1. So ln(0.98) =
ln(2 − 1.02) ≈ T3 (1.02). Note that this means we are plugging 1.02 into the Taylor
polynomial not 0.98. This was another common error. To find the error in our
approximation we use Lagrange’s Remainder Formula with x = 1.02 and a = 1:
2·3
− (2−c)
4
f 0000 (c)
4
|R3 (1.02)| = |
(1.02 − 1) | = |
· 0.024 |
4!
4!
where 1 ≤ c ≤ 1.02. To make the denominator of the derivative smallest, we need to
pick c = 1.02 so our final answer becomes
|R3 (1.02)| ≤
6
0.024
1
4
·
0.02
=
=
0.984 · 4!
4 · 0.984
4 · 494
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