**Unformatted text preview: **1 MAT272
Calculus II
Chapter 6: Inverse Functions: Exponential, Logarithmic, and Inverse Trigonometric Functions ………………3
6.6: Inverse Trigonometric Functions …………………………………………………………………………………………3
6.7: Hyperbolic Functions …………………………………………………………………………………………………………10
6.8: Indeterminate Forms and l’Hospital’s Rule ………………………………………………………………………..16
Chapter 7: Techniques of Integration ……………………………………………………………………………………………………21
7.1: Integration by Parts …………………………………………………………………………………………………………..21
7.2: Trigonometric Integrals ……………………………………………………………………………………………………..26
7.3: Trigonometric Substitution ……………………………………………………………………………………………….32
7.4: Integration of Rational Functions by Partial Fractions ……………………………………………………….39
7.5: Strategy for Integration …………………………………………………………………………………………………….48
7.6: Integration Using Tables and Computer Algebra Systems ………………………………………………….54
7.7: Approximate Integration ……………………………………………………………………………………………………56
7.8: Improper Integrals …………………………………………………………………………………………………………….61
Chapter 8: Further Applications of Integration ……………………………………………………………………………………..72
8.1: Arc Length …………………………………………………………………………………………………………………………72
8.2: Area of a Surface of Revolution …………………………………………………………………………………………79
Chapter 9: Differential Equations ………………………………………………………………………………………………………….83
9.1: Modeling with Differential Equations ………………………………………………………………………………..83
9.2: Direction Fields and Euler’s Method ………………………………………………………………………………….86
9.3: Separable Equations ………………………………………………………………………………………………………….89
9.4: Models for Population Growth ………………………………………………………………………………………….97
9.5: Linear Equations ………………………………………………………………………………………………………………105
9.6: Predator-Prey Systems …………………………………………………………………………………………………….110
Chapter 10: Parametric Equations and Polar Coordinates …………………………………………………………………..113
10.1: Curves Defined by Parametric Equations ……………………………………………………………………….113
10.2: Calculus with Parametric Curves ……………………………………………………………………………………120
10.3: Polar Coordinates ………………………………………………………………………………………………………….135
10.4: Areas and Lengths in Polar Coordinates …………………………………………………………………………144
10.5: Conic Sections ……………………………………………………………………………………………………………….155
10.6: Conic Sections in Polar Coordinates ………………………………………………………………………………164
Chapter 11: Infinite Sequences and Series ………………………………………………………………………………………….169
11.1: Sequences ……………………………………………………………………………………………………………………..169
11.2: Series …………………………………………………………………………………………………………………………….181
11.3: The Integral Test and Estimates of Sums ……………………………………………………………………….192
11.4: The Comparison Tests ……………………………………………………………………………………………………198
11.5: Alternating Series ………………………………………………………………………………………………………….202
11.6: Absolute Convergence and the Ratio and Root Tests …………………………………………………….207
11.7: Strategy for Testing Series …………………………………………………………………………………………….215 2
11.8: Power Series ………………………………………………………………………………………………………………….217
11.9: Representations of Functions as Power Series ………………………………………………………………222
11.10: Taylor and Maclaurin Series …………………………………………………………………………………………234
11.11: Applications of Taylor Polynomials ………………………………………………………………………………248 3 Chapter 6: Inverse Functions
Section 6.6: Inverse Trigonometric Functions
(Pages 481-482)
1. Find the exact value for the expression 5
13 5
cos �2 sin−1 � ��
13 Let = sin−1 � �. Then, from the double-angle formula for Cosine, we have that
cos �2 sin−1 � 5
�� = cos(2) = 1 − 2 sin2 ()
13 5
13 From right triangle trigonometry, sin−1 � � implies that, relative to , the opposite side has length 5, and the hypotenuse has length 13. Therefore, we have that 5
25
50
119
5 2
�=1−
=
cos �2 sin−1 � �� = 1 − 2 sin2() = 1 − 2 � � = 1 − 2 �
13
169
169 169
13 2. Simplify the expression
sin(2 arccos()) Let = arccos(). Then, from the double-angle formula for Sine, we have that
sin(2 arccos()) = sin(2) = 2 sin() cos() From right triangle trigonometry, arccos() implies that, relative to , the adjacent side has length ,
and the hypotenuse has length 1. Therefore, the opposite side has length √1 − 2 . So, we have that
sin(2 arccos()) = 2 sin() cos() = 2 � √1 − 2 � � � = 2�1 − 2
1
1 4
3. Prove that 1 [cot −1 ()] = −
1 + 2 Let = cot −1(). Then, = cot() and therefore, through implicit differentiation with respect to , we
have that
1 = − csc 2 () ⇒
= − sin2 () From right triangle trigonometry, cot −1() implies that, relative to , the adjacent side has length ,
and opposite side has length 1. Therefore, the hypotenuse has length √1 + 2 . So, we have that
2 1
1
= − sin2() = − �
� =− 1 + 2
√1 + 2 4. Find the derivative. From the Chain Rule, we have that 1 − �� =
�arctan �� 1 + =
= 1 − arctan ��
�
1 + 1 1 1 − −2 (−1)(1 + ) − (1)(1 − )
∙
∙�
�
2 2 �1 + �
(1 + )2
1
− 1 + ��
1 + �
1 1 1 + −2
1
1 + 1
1
∙ �
∙
=
∙�
∙−
2
2
1 − (1 + )
(1 + )2
1 − 1 + 1 + 2 1 − 1 + 1 + 1 + 1
1
∙�
∙−
=−
2
(1 + )
2
1 − 2√1 − 2 5
5. Find the derivative of the function and state the domains of the function and its derivative.
() = cos−1(3 − 2) From the definition of the inverse Cosine function, we have that the domain is
−1 ≤ 3 − 2 ≤ 1 ⇒ −4 ≤ −2 ≤ −2 ⇒ 1 ≤ ≤ 2 ⇒ [1, 2] 1
2
2 [cos−1 (3 − 2)] = −
∙ (−2) =
= �1 − (3 − 2)2
�1 − (9 − 12 + 4 2 ) √12 − 4 2 − 8 = −12 ± �144 − 4(−4)(−8) −12 ± √144 − 128 −12 ± √16 −12 ± 4
=
=
=
= 1, 2
−8
−8
−8
−8 Therefore, we have that 12 − 4 2 − 8 > 0 ⇒ −4( − 2)( − 1) > 0 This inequality is only satisfied on the interval (1, 2), so that is the domain of the derivative. 6
Find ′ () and check that your answer is reasonable by comparing the graphs of and ′. 6. ′ () = () = arctan( 2 − ) 2 − 1
2 − 1
1
∙ (2 − 1) =
= 4
2
2
4
3
2
1 + − 2 + − 2 3 + 2 + 1
1 + ( − ) The graphs of and ′ are below, where is the solid curve. 1
2 These graphs seem to correspond since is decreasing on �−∞, �, and that’s precisely where ′ is
1
2 negative. Additionally, is increasing on � , ∞�, which is where ′ is positive. has a relative (and
1
2 1
2 1
2 global) minimum at = , and ′ is zero there. is concave down on �−∞, − � and � , ∞�, where ′ is
1 1
2 2 decreasing, and concave up on �− , �, where ′ is increasing. Lastly, the points of inflection of at
1
2 = ± are where ′ changes direction. From all of this, it seems reasonable ′ is the derivative of .
7. Find the limit
1 + 2
lim arccos �
�
→∞
1 + 2 2 Passing the limit into the rational argument, we have that
lim arccos � →∞ 1
1 + 2 1 + 2
=
arccos
lim
�
�
� = arccos � � =
2
2
→∞ 1 + 2
2
1 + 2
3 7
8. A ladder 10 ft long leans against a vertical wall. If the bottom of the ladder slides away from the
base of the wall at a speed of 2 ft/s, how fast is the angle between the ladder and the wall
changing when the bottom of the ladder is 6 ft from the base of the wall?
Let the angle between the ladder and the wall be . The relationship between and the distance the
ladder is from the base of the wall, say , is given by
sin() = 10 Since the problem asks for the speed at which is changing, solve for . = arcsin � �
10 Implicitly differentiating this equation with respect to time yields When = 6 and = = 2, we have that = 1 2 �1 − � 6 �
10 ∙ 1 2 �1 − � �
10 1
∙2=
10 1 ∙ 1 ∙
10 1
1 1 5 1 1
=
∙ = ∙ =
5
5 4 5 4
9
16
�1 −
�
25
25
∙ Therefore, is increasing at 1/4 of a radian per second.
9. Sketch the curve using calculus. = arctan() Since the domains of and arctan() are ℝ, so is . Furthermore, since both and arctan() are
everywhere continuous, so is .
lim arctan() lim arctan() = →∞ →∞ lim arctan() lim arctan() = →−∞ →−∞ = 2 ≈ 4.810 = − 2 ≈ 0.208 Therefore, the graph of has horizontal asymptotes of = 4.810 and = 0.208.
Therefore, the -intercept is = 1. (0) = arctan(0) = 0 = 1 Since 0 = arctan() has no solution, there are no -intercepts. Since and arctan() are differentiable, so is . 8 arctan()
=
1 + 2 The derivative is never zero nor undefined, so no critical points exist. Therefore, has no relative extrema. Furthermore, since arctan()
2
arctan()
2 ∙ (1 + ) − (2) arctan() (1 − 2) 1
+ =
=
(1 + 2 )2
(1 + 2 )2 2
2 2 2 1
2 > 0 for all , is monotonically increasing. down on the second. 2 2 1
2
1
Furthermore, � , 1.5899� is
2 > 0 on �−∞, � and < 0 on � , ∞�. Therefore, is concave up on the first interval and concave Overall, we have the graph of below. a point of inflection, where 1
arctan� �
2 ≈ 1.5899 10. Evaluate the integral
� 2 √1 − 4
1
2 Let = 2 and = 1. Then, = 2 2 ⇒ = 2 . Therefore, we have that
� 2 √1 − 4 1 1
1 = �
= arcsin() + = arcsin( 2 ) + 2
2 √1 − 2
2 9
11. The region under the curve = 1 √ 2 + 4 from = 0 to = 2 is rotated about the − . Find the volume of the resulting solid.
The bounded region is given below. By the disk method, we have that the volume is
2
2
2
1
1
1 2 = � �
� = � 2 = � arctan � �� = [arctan(1) − arctan(0)]
2
2
2 0 2
0 √ + 4
0 + 4 2
= � − 0� =
2 4
8 10 Section 6.7: Hyperbolic Functions
(Pages 489-491)
1. Find the numerical value of each expression.
sech(0) sech(0) = cosh−1 (1) 0 2
2
=
=1
−0
+ 1+1 cosh−1(1) = ln �1 + �12 − 1� = ln(1) = 0
2. Prove the identity Recall that Therefore, we have that tanh( + ) = tanh() + tanh()
1 + tanh() tanh() tanh() = − − + − − − − −
tanh() + tanh() + − + + −
=
1 + tanh() tanh() 1 + � − − � � − − � + − + −
( + − − − − − − ) + ( + − − − − − − ) + − + − + − −
= − − − − + − −
1 + + − + − + − −
2 − 2 − − + − + − + − − −
−
( + + + − − ) + ( − − − − + − − )
= + − + − + − − − −
2 − 2 + − −(+)
= =
= tanh( + )
2 + 2 − − + + −(+) 11
3. If tanh() = Since tanh() = 12
,
13 12
,
13 find the values of the other hyperbolic functions at . then coth() = 13
.
12 Since 1 − tanh2 () = sech2 (), then we have that
1−� 12 2
144
25
5
� = sech2 () ⇒ 1 −
= sech2 () ⇒
= sech2() ⇒ sech() =
13
169
169
13 Note that sech() = −
Since sech() = Since cosh() = 5
,
13 13
,
5 5
13 is not possible since sech() = then cosh() =
we have that tanh() = Lastly, since sinh() = 12
,
5 13
.
5 1
cosh() and cosh() ≥ 1 for all real . sinh() 12 sinh()
12 13 12
⇒
=
⇒ sinh() =
∙
=
13
cosh() 13
13 5
5
5 then csch() = 5
.
12 4. Use the definitions of the hyperbolic functions to find each of the following limits. − −
lim tanh() = lim =1
→∞
→∞ + − − −
= −1
→−∞ + − lim tanh() = lim →−∞ − −
=∞
→∞
2 lim sinh() = lim →∞ − −
= −∞
lim sinh() = lim
→−∞
→−∞
2
2
=0 →∞ + − lim sech() = lim →∞ + −
lim coth() = lim =1
→∞
→∞ − − lim+ coth() = lim+ →0 →0 lim− coth() = lim− →0 →0 + −
= −∞ − − lim csch() = lim →−∞ + −
=∞ − − →−∞ 2
=0
− − sinh() − − 1
=
lim
=
→∞
→∞ 2 2
lim 12
5. Prove the following derivative formula. Recall that 1 [tanh−1 ()] =
1 − 2 Therefore, we have that 1 + 1
�
tanh−1() = ln �
1 − 2 1 1 + 1 1 − (1)(1 − ) − (−1)(1 + ) 1 1 − 2
[tanh−1()] =
�� = �
�
= �
�
� ln �
2
(1 − ) 2 1 −
2 1 + 2 1 + (1 − )2
1
=
1 − 2
6. Find the derivative. Simplify where possible. = sinh−1 (tan()) 1
sec 2 ()
sec 2()
sec 2 () 2 ()
=
∙ sec
=
=
= |sec()|
= �1 + (tan())2
�1 + tan2 () �sec 2 () |sec()| 13
7. A telephone line hangs between two poles 14 m apart in the shape of the catenary
where and are measured in meters. = 20 cosh � � − 15
20 (a) Find the slope of this curve where it meets the right pole.
(b) Find the angle between the line and pole. 1 = 20 sinh � � ∙
= sinh � � 20 20
20 Since the telephone line meets the pole at = 7, we have that the slope of the line at this point of
intersection is
7 7 20 − −20 7
(7) = sinh � � =
≈ 0.357
2 20 Since 0.357 represents the slope of the tangent line, or the hypotenuse of the right triangle formed, we
have that
tan() = 0.357 where = 90° − . Therefore, we have that = tan−1 (0.357) ≈ 19.66° Therefore, we ultimately have that the angle between the line and the pole is = 90° − 19.66° = 70.34° 14
8. a. Show that any function of the form = sinh() + cosh() satisfies the differential equation ′′ = 2 . b. Find = () such that ′′ = 9, (0) = −4, and ′ (0) = 6. ′ = cosh() + sinh() ′′ = 2 sinh() + 2 cosh() = 2 ( sinh() + cosh()) = 2 Therefore, satisfies the differential equation ′′ = 2 . From the problem statement, we have that 9 = 2 ⇒ = ±3. Since (0) = −4, we have that −4 = (0) = sinh(0) + cosh(0)
−4 = Therefore, we now have that So, we now have that Since ′ (0) = 6, we have that = sinh() − 4 cosh() ′ = cosh() − 4 sinh()
6 = ′ (0) = cosh(0) − 4 sinh(0) Assuming , > 0, one such function is 6 = ⇒ = ±2 = 2 sinh(3) − 4 cosh(3) 15
9. Evaluate the integral.
6 � 4 3 1
3 1 √ 2 −9 6 = � 4 1 2 �9 � 9 − 1� 1 6 = �
3 4 Let = . Then, = ⇒ 3 = . Furthermore, note that = 4 ⇒ = 1 2 �� � − 1
3 2
�� � − 1
3 4
3 = 6 ⇒ = 2 Therefore, we have that
1 6
�
3 4 1 2 4
= [cosh−1 ()]24/3 = cosh−1 (2) − cosh−1 � � ≈ 0.522
2
3
4/3 √ − 1 = � 16 Section 6.8: Indeterminate Forms and l’Hospital’s Rule
(Pages 499-502)
For questions 1 through 4, find the limit. Use l’Hospital’s Rule where appropriate. If there is a more
elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.
1.
lim →0 √1 + 2 − √1 − 4 0
0 Direct substitution produces the indeterminate form . Therefore, we proceed with l’Hospital’s Rule.
1
2
√1 + 2 − √1 − 4
= lim �
+
�=1+2=3
→0
→0 √1 + 2 √1 − 4
lim 2. lim ln() tan � �
2 →1+ Direct substitution produces the indeterminate form 0 ∙ ∞. Therefore, we proceed with l’Hospital’s Rule. 1
sin2 � 2 � 2
2
ln()
= lim+ = − lim+
=−
lim+ ln() tan � � = lim+ →1
→1 cot �
→1 − csc 2 � 2 →1 2�
2
2� 3. 1
1
lim+ � − �
→0 − 1 Direct substitution produces the indeterminate form ∞ − ∞. Therefore, we proceed with l’Hospital’s
Rule.
( − 1) − 1
1
1 − 1 lim+ � − � = lim+
=
lim
=
lim
= +
+
→0
→0
→0 + − 1
→0 + + − 1
2
( − 1) 17
4.
lim 1/ →∞ Direct substitution produces the indeterminate form ∞0 . Therefore, we proceed with l’Hospital’s Rule.
1
ln() = ln()
→∞ lim 1/ = ⇒ ln � 1/ � = ln() ⇒ lim ln� 1/ � = ln() ⇒ lim →∞ →∞ →∞ ln()
1
= ln() ⇒ lim = ln() ⇒ 0 = ln() ⇒ = 0 = 1
→∞ →∞ lim 5. Prove that =∞
→∞ lim for any positive integer . This shows that the exponential function approaches infinity faster than any
power of .
∞
∞ Direct substitution produces the indeterminate form . Therefore, we proceed with l’Hospital’s Rule. Note that since is a positive integer, we have that [ ] = ! = ( − 1)( − 2) ∙ ⋯ ∙ (2)(1) Furthermore, the th derivative of is . Therefore, after applications of l’Hospital’s Rule, we have that =
lim
=∞
→∞ →∞ ( − 1)( − 2) ∙ ⋯ ∙ (2)(1)
lim since the denominator is a constant. This idea would apply for whatever (positive) base of the
exponential function in the numerator. Therefore, the exponential function grows faster than the power
function, in general. 18
6. If an initial amount 0 of money is invested at an interest rate compounded times per year,
the value of the investment after years is = 0 �1 + � If we let → ∞ we refer to the “continuous compounding” of interest. Use l’Hospital’s Rule to show that
if the interest is compounded continuously, then the amount after years is = 0 Notice that 0 is common in both formulas, so we aim to show
where and are assumed fixed. lim �1 + � = →∞ Direct substitution produces the indeterminate form 1∞ . Therefore, we proceed with l’Hospital’s Rule. lim �1 + � = ⇒ ln � lim �1 + � � = ln() ⇒ lim ln ��1 + � � = ln()
→∞
→∞
→∞ 1 ∙ − 2
ln �1 + �
1 + lim ln �1 + � = ln() ⇒ lim
= ln() ⇒ lim
= ln()
1
1
→∞
→∞
→∞ − 2 lim
= ln() ⇒ = ln() ⇒ = →∞ 1 + 19
7. Some populations initially grow exponentially but eventually level off. Equations of the form
() = 1 + − where , , and are positive constants, are called “logistic equations” and are often used to model
such populations. Here, is called the “carrying capacity” and represents the maximum population size that can be supported, and = −0
,
0 where 0 is the initial population. (a) Compute lim (). Explain why your answer is to be expected.
→∞ (b) Compute lim (). (Note that is defined in terms of .) What kind of function is your result?
→∞ = →∞ 1 + − lim () = lim →∞ This answer is to be expected since it says that over time, the population either grows or decays to the
carrying capacity, depending on its initial size. = lim
= lim
−
− − 0 − →∞ →∞ 1 + →∞
+ 0 (1 − − )
1+ 0
0
0
= lim
= 0 →∞ − + 0 (1 − − ) lim () = lim →∞ The last equality comes from the fact that 0 and − are constants. The result is an exponential
function. We will show in Chapter 9 that this function models (unbounded) population growth, which
makes sense if the carrying capacity is made arbitrarily large (unlimited resources). 20
8. Suppose is a positive function. If lim () = 0 and lim () = ∞, show that
→ → lim [()] () = 0 → This shows that 0∞ is not an indeterminate form. lim [()] () = ⇒ ln �lim [()] () � = ln() ⇒ lim ln�[()] () � = ln() → → → lim () ln�()� = ln() → Note that since > 0 and lim () = 0, we have that lim ln�()� = ln �lim ()� = −∞. Combining
→ → this with the other hypothesis says that ln() = lim () ln�()� = −∞
→ Exponentiating both sides of the equation produces Therefore, 0∞ is not an indeterminate form. = −∞ = 0 → 21 Chapter 7: Techniques of Integration
Section 7.1: Integration by Parts
(Pages 516-518)
For questions 1 through 3, evaluate the integral.
1. Let = (ln())2 ⇒ =
1 2 ln()
. �(ln())2 Then, = ⇒ = . Therefore, we have that �(ln())2 = (ln())2 − � 2 ln() Let = ln() ⇒ = . Then, = ⇒ = . Therefore, we have that (ln())2 − � 2 ln() = (ln())2 − 2 � ln() − � � = (ln())2 − 2 ln() + 2 + 2. 2 � 2 sin(2) 0 Using the tabular method, we have the following.
Sign D A + 2 sin(2) − 2 − 0 + 2 1
− cos(2)
2
1
− sin(2)
4
1
cos(2)
8 2
1 2
1
1
1
1
� sin(2) = �− cos(2) + sin(2) + cos(2)� = �−2 2 + � − = −2 2
2
2
4
4
4
0
0
2 2 22
3. 1 Let = 2 ⇒ = 2. Then, =
1 � 0 3 √4 + 2 = � 2� 4+ ≈ 0.1158 1 2 �
0 � √4+ 2 0 3 √4 + 2 ⇒ = √4 + 2 . Therefore, we have the following. 3 1
3
2
2 3
2
2
�
− � 2 4 + = √5 − � (4 + )2 � = √5 − �52 − 42 �
3
3
0
0
1 4. First make a substitution and then use integration by parts to evaluate the integral. � cos() sin(2) 0 Recall that sin(2) = 2 sin() cos(). Therefore, we have that � cos() sin(2) = 2 � cos() sin() cos() 0 0 Let = cos(). Then, = − sin() ⇒ − = sin() . Furthermore, we have that
(0) = cos(0) = 1 () = cos() = −1 So, we have that Sign D A + + 0 − 1 −1 1 2 � cos() sin() cos() = −2 � = 2 � 0 1 −1 From the table above, we have that
1
1 1
4
2 � = 2[ − ]1−1 = 2(1 − 1) − 2 �− − � = ≈ 1.4715 −1 23
5. Use integration by parts to prove the reduction formula.
� = − � −1 Let = ⇒ = −1 . Then, = ⇒ = . Therefore, we have that
� = − � −1 6. Use the method of cylindrical shells to find the volume generated by rotating the region bounded
by the curves about the given axis. = � = − , = 1 about the − The graph of the region bounded by the curves of the functions given above is Note that the distance between the − and the center of a representative vertical rectangle is .
Therefore, by the method of cylindrical shells, we have that the volume generated by rotating the above
enclosed, shaded region about the − is
1 = 2 � ( − − )
0 Let = and = ( − − ). Then, = and = + − . Therefore, we have that 1
1
1 = 2 � ( − − ) = 2 �[( + − )]10 − � ( + − )� = 2 � + − [ − − ]10 � 0
0
1
2
4
1
≈ 4.6229
= 2 � + − � − �� = 2 � = 24 7. A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the
initial mass of the rocket at liftoff (including its fuel) is , the fuel is consumed at rate , and the
exhaust gases are ejected with constant velocity (relative to the rocket). A model for the
velocity of the rocket at time is given by the equation − () = − − ln �
� where is the acceleration due to gravity and is not too large. If = 9.8 m/s2, = 30,000 kg, = 160 kg/s, and = 3000 m/s, find the height of the rocket one minute after liftoff. Since () given above describes the rate of change of height of the rocket, and we can safely assume
the rocket only moves in one direction during liftoff, then the integral of () will provide the total
height climbed by the rocket over some time interval ∆.
60
60
30000 − 160
ℎ = � () = � �−9.8 − 3000 ln �
�� 30000
0
0
60
30000 − 160
2 ]60
[−4.9
=
� 0 − 3000 � ln �
30000
0 30000−160
�
30000 Let = ln � that and = . Then, = 30000
30000−160 ∙− 160 30000 and = . Therefore, we have 60 30000 − 160
� 30000
0
60
30000 − 160 60
160
= −17640 − 3000 �� ln �
�� + �
�
30000
0
0 30000 − 160
60
17
160
= −17640 − 3000 �60 ln � � + �
�
25
0 30000 − 160 [−4.9 2 ]60
0 − 3000 � ln � Let = 30000 − 160. The...

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