Nonlinear EquationsYour nonlinearity confuses me

02345 fexdxcxbxax

“The problem of not knowing what we missed is that we believe we haven't missed anything” – Stephen Chew on Multitasking

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2

Example – General EngineeringYou are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC commodes. The floating ball has a specific gravity of 0.6 and has a radius of 5.5 cm. You are asked to find the depth to which the ball is submerged when floating in water.

Figure Diagram of the floating ball

010993.3165.0 423 xx

For the trunnion-hub problem discussed on first day of class where we were seeking contraction of 0.015”, did the trunnion shrink enough when dipped in dry-ice/alcohol mixture?

1. 2.

0%0%

1. Yes2. No

30

4

Example – Mechanical Engineering

Since the answer was a resounding NO, a logical question to ask would be:

If the temperature of -108oF is not enough

for the contraction, what is?

5

Finding The Temperature of the Fluid

dTTDDc

a

T

T

)(

Ta = 80oFTc = ???oFD = 12.363"∆D = -0.015" -350 -300 -250 -200 -150 -100 -50 0 50 100

2

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

T

3528 10349.610457.710992.5015.0 cc TT

TT 009696.0033.6

010651.810457.710992.5)( 3528 ccc TTTf

6

Finding The Temperature of the Fluid

dTTDDc

a

T

T

)(

Ta = 80oFTc = ???oFD = 12.363"∆D = -0.015"

352639 10166.610435.710829.310059.5015.0 ccc TTT

015.610195.610228.1 325 TT

010834.810435.710829.310059.5)( 352639 cccc TTTTf

Nonlinear Equations(Background)

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How many roots can a nonlinear equation have?

-20 -10 0 10 20-3

-2

-1

0

1

2

3sin(x)=2 has no roots

How many roots can a nonlinear equation have?

-20 -10 0 10 20-3

-2

-1

0

1

2

3sin(x)=0.75 has infinite roots

How many roots can a nonlinear equation have?

-20 -10 0 10 20-3

-2

-1

0

1

2

3sin(x)=x has one root

How many roots can a nonlinear equation have?

-20 -10 0 10 20-3

-2

-1

0

1

2

3sin(x)=x/2 has finite number of roots

The value of x that satisfies f (x)=0 is called the

1. 2. 3. 4.

0% 0%0%0%

1. root of equation f (x)=0 2. root of function f (x) 3. zero of equation f (x)=0 4. none of the above

A quadratic equation has ______ root(s)

1. 2. 3. 4.

0% 0%0%0%

1. one2. two3. three4. cannot be determined

For a certain cubic equation, at least one of the roots is known to be a complex root. The total number of complex roots the cubic equation has is

1. 2. 3. 4.

0% 0%0%0%

1. one2. two3. three4. cannot be

determined

Equation such as tan (x)=x has __ root(s)

1. 2. 3. 4.

0% 0%0%0%

1. zero2. one3. two4. infinite

A polynomial of order n has zeros

1. 2. 3. 4.

0% 0%0%0%

1. n -12. n3. n +14. n +2

END

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Bisection Method

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Bisection method of finding roots of nonlinear equations falls under the category of a (an) method.

1. 2. 3. 4.

0% 0%0%0%

1. open2. bracketing3. random4. graphical

If for a real continuous function f(x),f (a) f (b)<0, then in the range [a,b] for f(x)=0, there is (are)

1. 2. 3. 4.

0% 0%0%0%

1. one root2. undeterminable number of roots3. no root4. at least one root

The velocity of a body is given by v (t)=5e-t+4, where t is in seconds and v is in m/s. We want to find the time when the velocity of the body is 6 m/s. The equation form needed for bisection and Newton-Raphson methods is

1. 2. 3. 4.

25% 25%25%25%1. f (t)= 5e-t+4=02. f (t)= 5e-t+4=63. f (t)= 5e-t=24. f (t)= 5e-t-2=0

To find the root of an equation f (x)=0, a student started using the bisection method with a valid bracket of [20,40]. The smallest range for the absolute true error at the end of the 2nd iteration is

1. 2. 3. 4.

25% 25%25%25%1. 0 ≤ |Et|≤2.5

2. 0 ≤ |Et| ≤ 5

3. 0 ≤ |Et| ≤ 10

4. 0 ≤ |Et| ≤ 20

For an equation like x2=0, a root exists at x=0. The bisection method cannot be adopted to solve this equation in spite of the root existing at x=0 because the function f(x)=x 2

1. 2. 3. 4.

0% 0%0%0%

1. is a polynomial2. has repeated zeros at

x=03. is always non-negative4. slope is zero at x=0

END

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Newton Raphson Method

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Newton-Raphson method of finding roots of nonlinear equations falls under the category of __________ method.

1. 2. 3. 4.

0% 0%0%0%

1. bracketing2. open3. random4. graphical

The next iterative value of the root of the equation x 2=4 using Newton-Raphson method, if the initial guess is 3 is

1. 2. 3. 4.

0% 0%0%0%

1. 1.5002. 2.0663. 2.1664. 3.000

The root of equation f (x)=0 is found by using Newton-Raphson method. The initial estimate of the root is xo=3, f (3)=5. The angle the tangent to the function f (x) makes at x=3 is 57o. The next estimate of the root, x1 most nearly is

1. 2. 3. 4.

0% 0%0%0%

1. -3.24702. -0.24703. 3.24704. 6.2470

The Newton-Raphson method formula for finding the square root of a real number R from the equation x2-R=0 is,

1. 2. 3. 4.

25% 25%25%25%

21i

i

xx

2

31

ii

xx

iii x

Rxx

2

11

iii x

Rxx 3

2

11

1.

2.

3.

4.

END

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