MA 104 Graded Homework 2 Solutions

[Pages:4]MA 104 Graded Homework 2 Solutions

Dr. L. Zhao, Sections B7, C7, D7 Issued: Monday 23 February 2004 Due: Beginning of class, Monday 8 March 2004

0

(1) (5 Points) Evaluate the integral by interpreting it in terms of areas: (2 +

-3

Consider the graph of y = f (x) = 2 + 9 - x2. We have

9 - x2)dx.

y = 2 + 9 - x2 = (y - 2)2 = 9 - x2 = x2 + (y - 2)2 = 32,

which is the equation of a circle of radius 3 centered at (0, 2). The integral can be interpreted as the area of a quarter of the said circle and the area of the rectangle whose vertices are (0, 0), (0, 2), (-3, 2) and (-3, 0). Therefore, we have

0

(2 +

-3

9

-

x2)dx

=

1 4

?

?

32

+

2

?

3

=

9 4

+

6.

A plot of the figure and the numerical computations can be done on Mathematica with the

following commands.

Plot[2 + Sqrt[9 - x^2], {x, -3, 0}]

Integrate[2 + Sqrt[9 - x^2], {x, -3, 0}]

(2) (5 Points) Evaluate the following definite integral

4 0

ex x

dx.

This problem is done with Mathematica using the following command.

Integrate[Exp[Sqrt[x]]/Sqrt[x], {x, 0, 4}]

Manually,

we

set

u

=

x

and

get

du

=

dx 2x

or

2du

=

dx x

.

We

have

4 0

ex x

dx

=

x=4

2eudu

=

2eu

x=4

=

2ex

4

=

2(e2

- 1).

x=0

x=0

0

(3) The following table gives the values of electric intensity I(t) at a point on a circuit at time t seconds.

t(seconds)

0 0.5 1.0 1.5 2.0 2.5 3.0

I(t) (Amperes) 19.3 19.0 18.3 16.5 12.3 3.3 2.6

(a) (5 Points) Estimate the amount of electric charge passed through the point from t = 0 to t = 3 with the right endpoints as the sample points. The total amount of electric charging passing through the point is

0.5 ? 19.0 + 0.5 ? 18.3 + 0.5 ? 16.5 + 0.5 ? 12.3 + 0.5 ? 3.3 + 0.5 ? 2.6 = 36 columb.

(b) (3 Points) Suppose it is known that I(t) is a decreasing function, is our estimate in part (a) an over-estimate or under-estimate? Why? If I(t) is a decreasing function, then for each of the subintervals from 0 to 3, I(t) at the right-hand end point is less than all values of I(t) in the sub-interval. Hence, the approximating rectangles lie below the graph of I(t) and that is to say the approximation we have in part (a) is an under-estimate.

1

2

MA 104 HOMEWORK 2 SOLUTIONS

(c) (2 Points) Had we used the left end point as the sample points and supposed that I(t)

is a decreasing function, would our estimate be an over-estimate?

Had we use the left-hand end points as the sample points, we would have gotten an

over-estimate via similar reasoning as in part (b).

2

(4) (10 Points) Use the mid-point and trapezoidal rules with n = 8 to approximate

4 - x2dx.

-2

The interval of interest

is [-2, 2].

We have x =

2 - (-2) 8

=

1 2

.

The end points of the

subintervals

are

xi

=

-2 + ix

=

-2 +

i 2

.

The

mid-point

of

the

i-th

subinterval

is

x?i

=

1 2

(xi

+

xi-1)

=

1 2

-2

+

i 2

+

-2

+

i

- 2

1

=

-2

+

2i

- 4

1.

Hence, for the mid-point rule

2 -2

4 - x2dx

8

xf (x?i) =

8

1 2

i=1

i=1

4-

-2

+

2i

- 4

1

2

= 6.36786.

This can be done on Mathematica with the following command.

Sum[1/2*Sqrt[4 - (-2 + (2i - 1)/4)^2], {i, 1, 8}] // N

For the trapezoidal rule, we have

2 -2

4 - x2dx

x 2

[f

(x0)

+

2f

(x1)

+

2f

(x2)

+

?

?

?

+

2f

(x6)

+

2f

(x7)

+

f

(x8)]

=

1 4

4 - (-2)2 +

7

4 - 22 + 2 ?

4-

-2

+

i 2

2

i=1

= 5.99142.

The above can be done on Mathematica with the following command.

1/4*(Sqrt[4-(-2)^2]+Sqrt[4-2^2]+Sum[2*Sqrt[4 - (-2 + i/2)^2], {i, 1, 7}]) // N

For the cognoscenti, this integral can be viewed as the area of a semi-circle centered at the

origin with radius 2. Hence the exact value of this integral is 2. The value of this integral

and that of the integral in Problem (1) can be computed algebraically without interpreting

them as areas via a somewhat involved substitution.

(5) (5 Points) A uniform cable hanging over the edge of a tall building is 40 meters long and

has mass 60 kilograms. How much work is require to pull the cable completely to the top?

(You may assume gravitational acceleration is 9.8m/s2 = 9.8N/Kg.)

The

rope

weighs

60?9.8

=

588

Newtons.

Hence,

the

weight

density

of

the

rope

is

588 40

=

14.7

Newtons per meter. Divide the rope into n equal pieces, each of length x. The weight of

each piece is 14.7x and the work needed to lift that piece to the top is 14.7xi, where xi

is the position of the section in the rope. The total work needed to lift the rope to the roof

is

n

Total Work = lim x14.7xi =

40 14.7xdx = 7.35x2 40 = 7.35 ? 402 = 11760 J.

n i=1

0

0

SECTIONS B14, C14, D14

3

In Mathematica, this is done via the following command.

Integrate[14.7x, {x, 0, 40}]

(6) (5 Points) If 10 J of work is needed to stretch a spring from 10 cm to 11 cm and another 15 J to stretch it from 11 cm to 12 cm, what is the natural length of the spring? Let L meters be the natural length of the spring and k be its spring constant. We have

0.11-L

0.12-L

(i)

kxdx = 14 and

kxdx = 15.

0.1-L

0.11-L

Note that 10 centimeters is 0.1 meters and the same goes for the other quantities.This problem can now be solved using Mathematica via the following command.

Solve[{Integrate[k*x, {x, 0.1-L, 0.11-L}] == 14, Integrate[k*x, {x, 0.11-L, 0.12-L}] == 15}, {L, k}]

The answer is that L = 0.035 meters and the natural length of the spring is 3.5 centimeters. Manually, the equations in (i) can be re-written as

k 2

(0.11 - L)2 - (0.10 - L)2

=

14,

and

k 2

(0.12 - L)2 - (0.11 - L)2

= 15.

Expanding the left-hand sides of both the above equations and simplify, we get

k 2

(0.0021

-

0.02L)

=

14

and

k 2

(0.0023

-

0.02L)

=

15.

Divide the first equation by the second, we get

0.0021 0.0023

- -

0.02L 0.02L

=

14 15

=

15(0.0021

-

0.02L)

=

14(0.0023

-

0.02L)

= -2L = 14 ? 0.23 - 15 ? 0.21 = L = -0.035 meters.

It is of course absurd to have a negative length, but the my negligence has caused this

(7)

mistake to happen. (5 Points) A supply

curve

is

given

by

p=

15 +

x 5

.

Find

the

producer

surplus

when

the

selling price is $10.

First, setting the price to $16, we have

16

=

15

+

x 5

=

1

=

x 5

=

5

=

x

=

x

=

25.

Hence, the producer surplus is

25

16 -

0

15

+

x 5

dx =

x

-

2 15

x

3 2

25 0

=

25 3

=

8.33333.

In Mathematica, this can be done via the following commands.

price[x_] := 15 + Sqrt[x]/5 Solve[price[x] == 16, x] Integrate[Abs[16 - price[x]], {x, 0, 25}] // N

4

MA 104 HOMEWORK 2 SOLUTIONS

(8) (5 Points) For a constant c given, find all functions f (x) that satisfy the equation

x

(ii)

f (t)dt = [f (x)]2 + c.

0

By the fundamental theorem of calculus, the left-hand side of (ii) can be re-written as

x

(iii)

f (t)dt = F (x) - F (0) = F (x) - F (0) = [f (x)]2 + c,

0

where F (x) is an anti-derivative of f (x). Differentiating with respect to x on both sides of

the last equality (iii), we have

d dx

(F (x)

-

F (0))

=

d dx

[f (x)]2 + c

=

f

(x)

=

2f

(x)

df (x) dx

.

Hence

it

must

be

that

df (x) dx

=

1 2

or

f (x) =

1 2

x

+

c1

,

for

some

constant

c1.

Inserting

this

back into (ii) in order to solve for c1, we have

x 0

t 2

+

c1dt

=

x 2

+

c1

2

+ c.

The left-hand side of this above is

(iv)

t2 4

+

c1t

x 0

=

x2 4

+

c1x,

while expanding the right-hand side yields

(v)

If

the

expressions

in

(iv)

and

(v)

x2 4

+

c1x

are to be

+ c21 + c. the same,

we

must

have

c21

+c

=

0

or

c1

=

? -c.

Therefore, if c < 0, then the functions that satisfy (ii) are

f (x)

=

1 2

x

?

-c.

Note that, there is no real-valued function satisfying (ii) if c > 0. Finally, for the special

case of c = 0, there are two functions that satisfy the equation in question, namely

f (x)

=

1 2

x,

and

f (x)

=

0.

For the cognoscenti, the statement regarding c > 0 can be seen more readily by setting

x = 0 in (ii). In that case, the left-hand side of (ii) is zero, while and right-hand side of (ii)

is positive due to the fact that c > 0 and a square of a real number is always non-negative.

Moreover, I doubt one can incorporate the use of Mathematica into this problem.

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download