# Solutions to Homework Assignment#2 by kol12169

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```									                       Solutions to Homework Assignment #2

1. [4 marks] Evaluate each of the following limits.
n−2
i2
(a) lim            .
n→∞         n3
i=1
n     √
( 2i)2
(b) lim                .
n→∞
i=4
n3
n
1    πi
(c) lim           sin .
n→∞
i=1
n    2n
n
i    1
(d) lim          a     +b   , where a and b are constants.
n→∞
i=1
n    n
Solution:
(a) There are 2 ways to do this question. The ﬁrst way is to use a special summation formula:
n−2
i2       (n − 2)(n − 1)(2n − 3)  1
lim              3
= lim             3
= .
n→∞
i=1
n    n→∞          6n             3

n−2             n
i2              i2
The other way is to ﬁrst notice that lim              = lim           , since the last 2 terms in this
n→∞
i=1
n3 n→∞ i=1 n3
latter sum clearly contribute only 0 to the limit. Now recognize the sum in this last limit as
1                               n            1
2                               i2                   1
a Riemann sum for         x dx, and therefore lim           3
=       x2 dx = .
0                         n→∞
i=1
n       0            3
n    √              n     √
( 2i)2               ( 2i)2
(b) First note that lim                = lim                 since the missing 3 terms on the left
n→∞           n3     n→∞           n3
i=4
3   √           i=1
( 2i)2          28
hand side contribute lim                = lim 3 = 0. Therefore
n→∞
i=1
n3     n→∞ n

n         √                  n      √                    n
( 2i)2                    ( 2i)2                     i2  2
lim                    = lim                     = 2 lim                = .
n→∞
i=4
n3     n→∞
i=1
n3      n→∞
i=1
n3  3

n                       1
1     πi                                2                      1       2
(c) lim           sin    =              sin(πx/2) dx = − cos(πx/2)                 =     .
n→∞
i=1
n     2n        0                       π                      0       π
n                                 1
i    1                                       a
(d) lim          a     +b   =                     (ax + b) dx =     + b.
n→∞
i=1
n    n               0                       2

1
2. [4 marks] Evaluate each of the following integrals.

(a)   x2 f (x3 ) dx, where f (u) = eu .

(b)     xf (x2 ) dx, where f (u) = tan u.
(π/4)2    √
f ( x)
(c)             √     dx, where f (u) = sec2 u.
(π/6)2       x
eπ/4
sin(ln x)
(d)          f (x) +              dx, where f (x) = (x − 1)2 (π/4 − ln x)3 + ln x.
1                     x
Solution:
(a) We make the substitution u = x3 , du = 3x2 dx:
3         1 u     1        1 3
x2 f (x3 ) dx = x2 ex dx =        e du = eu + C = ex + C.
3       3        3

(b) We make the substitution u = x2 , du = 2xdx:

1
xf (x2 ) dx =                       x tan(x2 ) dx =
tan u du
2
1                    1
= − ln | cos u| + C = − ln | cos(x2 )| + C.
2                    2

√            dx
(c) We make the substitution u =           x, du = √ :
2 x
(π/4)2         √           (π/4)2      √                                      π/4
f ( x)                (sec x)2
√    dx =              √     dx = 2                                   (sec u)2 du
(π/6)2           x         (π/6)2       x                                     π/6
π/4
= 2 tan u            = 2 (tan(π/4) − tan(π/6))
π/6
1
= 2 1− √
3
(d)
eπ/4                                                   eπ/4                    eπ/4
sin(ln x)                                                              sin(ln x)
f (x) +                              dx =              f (x) dx +                        dx
1                      x                               1                       1              x
eπ/4           eπ/4
sin(ln x)
= f (x)               +                        dx (by the Fundamental Theorem)
1          1              x
π/4
dx
= f (eπ/4 ) − f (1) +                          sin u du (by the substitution u = ln x, du =                  )
0                                                                     x
π                    π/4    π   1
= − cos u                   = − √ + 1.
4                    0      4    2

2
3. [2 marks] Derive the formula for the volume of a cone of radius r and height h by using
the process of integration (not the Fundamental Theorem of Calculus) and then evaluating
a limit.
Solution:
Putting the axis of the cone along the x-axis from x = 0 to x = h and then using the right
hand rule gives
i=n                    2                         i=n
ri       h                           i2  1
V = lim                 π                      = πr 2 h lim               3
= πr 2 h.
n→∞
i=1
n        n         n→∞
i=1
n   3

See the diagram at the end.
The next 3 questions all use the Fundamental Theorem of Calculus in the form
b
f (x) dx = f (b) − f (a).
a

4. [2 marks] Suppose that a tank initially contained 500 gallons of water. Let V (t) denote
dV
the volume of water in the tank at time t. Suppose      = k sin(πt/60) (in units of gallons
dt
per minute), where k is some constant. Determine k if the tank has 400 gallons after 20
minutes.
Solution: On the one hand
20                              20
60k              20
V (20) − V (0) =                   V (t) dt =                      k sin(πt/60) dt = −        cos(πt/60)
0                               0                                π               0
60k                  30k
= −        (cos(π/3) − 1) =     ,
π                    π
10π
and on the other V (20) − V (0) = −100. Comparing these answers we get k = −         .
3
5. [2 marks] The reaction time of a driver (time it takes to notice and react to danger in
the road ahead) is about 0.5 seconds. When the brakes are applied, it then takes the car
some time to decelerate and come to a full stop. If the deceleration rate is a = −8m/sec2 ,
how long would it take the driver to stop from an initial speed of 120 km per hour? (Include
both reaction time and slowing-down time.)
Solution: First we must change units so all measurements are compatible, say
120000         1200         100
120 km/hr =                    m/sec =      m/sec =     m/sec.
3600           36           3
Let v(t) denote the speed at time t and let T be the time the car comes to a full stop. Then
T                       T                             T                                         100
v(T ) − v(0) =           v (t) dt =              −8 dt = −8t                   = −8T, but also v(T ) − v(0) = −          .
0                       0                                 0                                          3
25 1  14
Therefore T = 100/24 = 25/6, and hence the time to stop is                                      + = sec.
6  2 3
3
6. [2 marks] The rate of ﬂow of blood through the heart can be described approximately as
a periodic function of the form F (t) = A(2 + sin(0.15t)), where t is time in seconds and A is
a constant in units of cm3 /sec. Find the total volume of blood that ﬂows through the heart
between t = 0 and t = 1. (Express your answer in terms of A.)
Solution:
Let V (t) be the volume of blood in the heart at time t. Then V (t) = F (t) and so the volume
of blood ﬂowing through the heart between t = 0 and t = 1 is given by
1                      1
V (1) − V (0) =               V (t) dt =             A(2 + sin(0.15t)) dt
0                      0
1              1            1
= A 2−                   cos(0.15t)   =A 2−          (cos(0.15) − 1)
0.15             0           0.15
26      1
= A               −        cos(0.15) ≈ (2.074859481)A.
3     0.15
7. [2 marks] Find the area enclosed by y = f (x) and y = g(x) in the following cases:
(a) f (x) = x2 and g(x) = x + 2.
(b) f (x) = e−|x| and g(x) = 1/e.
Solution:
(a) The curves y = x + 2 and y = x2 intersect at x = 2 and x = −1. Since the line y = x + 2
is above the parabola y = x2 in this range the area enclosed is
2
x2   2           x3   2
A =                   (x + 2 − x2 ) dx =                        +6 −
−1                                   2    −1          3    −1
1              1         9
=             (4 − 1) + 6 − (8 + 1) = .
2              3         2

(b) The graphs intersect at x = ±1 and are symmetric about the y-axis. Therefore the area
is
1
1               1  2        4
A=2         e−x −     dx = −2e−x − = 2 −
0         e               0  e        e.
8. [2 marks] Use MathSheet for the following questions. Hand in the spread sheets.
(a) Use the trapezoidal rule to approximate the area under the graph y = e−x from x = 1
2

to x = 2, accurate to 3 places.
(b) Use the midpoint rule to approximate the length of the ellipse x2 + 2y 2 = 1, accurate to
4 places.
Solution:
2
e−x dx using n = 10, 12, 14 and n = 16. The
2
(a) We ﬁrst apply the trapezoidal rule to
1
T10 = 0.1358096443, T12 = 0.1356407946, T14 = 0.1355390112, T16 = 0.1354729610.

4
On this basis we begin to suspect that the answer to 3 places is 0.135. To see that this is
correct we could either compute the trapezoidal rule Tn for larger and larger values of n or
we could do some error analysis (done in class).
1 √
(b) The ﬁrst quadrant part of the ellipse is given by y = √ 1 − x2 , 0 ≤ x ≤ 1. Therefore
2
the arc length of the total ellipse is given by

1                                    1                                            2
1 1
L = 4                   1 + y (x)2 dx = 4                   1+       √ (1 − x2 )−1/2 (−2x)           dx
0                                    0                      22
1
x2         4               1
2 − x2
= 4                1+                 dx = √                           dx
0                     2(1 − x2)
2         0        1 − x2
π/2
4
= √                         2 − (sin θ)2 dθ (by the substitution x = sin θ).
2         0

Now we use the midpoint rule on this last integral. It turns out that we need only take n = 4
to get 4 places of accuracy. M4 = 5.4026. To verify this one must experiment with larger
values of n to see if increasing the value of n makes any diﬀerence.
n      a        b      c      d      e      f
0   0.1963   3.9618 5.4026 0.0982 3.9904 5.4026
1   0.5890   3.6784        0.2945 3.9148
2   0.9817   3.2356        0.4909 3.7712
3   1.3744   2.8817        0.6872 3.5750
4                          0.8836 3.3496
5                          1.0799 3.1269
6                          1.2763 2.9452
7                          1.4726 2.8420

1
6

5                                                0.8

4
0.6
3                                                 y

2                                                0.4

1
0.2

–2       –1            0       1 x   2
–1     –0.5    0    0.5      1
x
Graph for question 7(a)                                 Graph for question 7(b)

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