A body moves along a straight line according to the law s = and acceleration at the end of 2 seconds. ²-2; hence, when t = 2, u = {(2)² - 2 = 4 ft/sec. V= a = ds 3 dt 2 du = 3r; hence, when t = 2, a = 3(2) = 6 ft/sec². dt ³-2t. Determine its velocity
A body moves along a straight line according to the law s = and acceleration at the end of 2 seconds. ²-2; hence, when t = 2, u = {(2)² - 2 = 4 ft/sec. V= a = ds 3 dt 2 du = 3r; hence, when t = 2, a = 3(2) = 6 ft/sec². dt ³-2t. Determine its velocity
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Hello, can you please answer this given problem using any numerical solution. The exact solution is provided below. You may answer in any of the ff:
Finite Difference Approximation of the Derivative
Forward, backward, and central difference formulas for the first derivative
Trapezoidal Method
Simpson’s 1/3 Method
![Finite Difference Approximation of the Derivative
f(x)+
The derivative f'(x) of a function f(x) at the point x =a is defined by:
Approximated
derivatives
True
derivative
¦S(x)!
X X X X
Point x approaches point a
Figure 8-4: Definition of
derivative.
df (x)
dx x=a
Forward finite difference
True derivative
XI-1 X1 XH-1
approximations of the derivative. In these approximations, illustrated in Fig. 8-5, the derivative at point
(x;) is calculated from the values of two points. The derivative is estimated as the value of the slope of
the line that connects the two points.
Forward Difference
Approximated
derivative
Graphically, the definition is illustrated in Fig. 8-4. The
derivative is the value of the slope of the tangent line to the
function at x =a. The derivative is obtained by taking a point x near
x = a and calculating the slope of the line that connects the two
points. The accuracy of calculating the derivative in this way
increases as point x is closer to point a. In the limit as point x
approaches point a, the derivative is the slope of the line that is
tangent to f(x) at x =a.
= f'(a) = lim
Backward Difference
x
Backward finite difference
Approximated
derivative
f(x) +
f(x) - f(a)
x-a
XLI XI X6+1
df|
dx\x=x
L
Figure 8-5: Finite difference approximation of derivative.
df
dx\x=x
True
derivative
Is the slope that connects points (x₁, f(x₁)) and (x₁+1 f (x₁+1));
dfl
dx\x=x₁
=
X
Is the slope that connects points (x₁-1₁, f(x₁-1)) and (x₁, f(x));
f(x₁+1) - f(x₁)
Xi+1 - Xi
f(x) +
f(x₁)-f(x₁-1)
xi-xi-1
Central Difference
Is the slope that connects points (x₁-1, f(x₁-1)) and (x₁+1, f(xi+1)):
f(x₁+1) -f(x₁-1)
Xi+1-XI-1
Central finite difference
Truc
derivative-
đại
Approximated
derivative
Xfol
X
УА
y=f(x)
L
Gives:
Integration is frequently encountered when solving
problems and calculating quantities in engineering and science.
Integration and integrals are also used when solving differential
equations. One of the simplest examples for the application of
integration is the calculation of the length of a curve. When a
curve in the x-y plane is given by the equation y = f (x), the length
L of the curve between the points a and x = b is given by:
Trapezoidal Method
A refinement over the simple rectangle and midpoint
methods is to use a linear function to approximate the
integrand over the interval of integration (Fig. 9-12).
Newton's form of interpolating polynomials with two points
x = a and x = b , yields:
b-a
f(x) = f(a)+(x-a)ƒ [a, b] = f(a) + (x −a)[ƒ(b)−ƒ(a)]
Substituting the above formula to:
The integrand can be an analytical function or a set of discrete points (tabulated data). When the
integrand is a mathematical expression for which the antiderivative can be found easily, the value of
the definite integral can be determined analytically. Numerical integration is needed when analytical
integration is difficult or not possible, and when the integrand is given as a set of discrete points.
y₁y=f(x)
Then simplifying it gives:
~√(√(a) + F(b)(b-
L =
= [*√/1 + [f'(x)]³dx
1(f) ~
f(b)
f(a)
((S)-° f(x) dx
1(ƒ ) = f* (ƒ(a) + (x − a)\ƒ(b)-f(a)]) dx
· ƒ(a)(b −a) + } [ƒ(b)− ƒ(a)](b.
b)) (b-a)/
Alea
f(b-a) b) - Ma))
rea
2
xa
x=b
Figure 9-12: The trapezoidal method.
a)(b-a)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07c69857-616f-4104-9533-0d686edb08a9%2F54d1e599-5a22-414c-be3c-f42d398cb3c5%2Fvyn561_processed.png&w=3840&q=75)
Transcribed Image Text:Finite Difference Approximation of the Derivative
f(x)+
The derivative f'(x) of a function f(x) at the point x =a is defined by:
Approximated
derivatives
True
derivative
¦S(x)!
X X X X
Point x approaches point a
Figure 8-4: Definition of
derivative.
df (x)
dx x=a
Forward finite difference
True derivative
XI-1 X1 XH-1
approximations of the derivative. In these approximations, illustrated in Fig. 8-5, the derivative at point
(x;) is calculated from the values of two points. The derivative is estimated as the value of the slope of
the line that connects the two points.
Forward Difference
Approximated
derivative
Graphically, the definition is illustrated in Fig. 8-4. The
derivative is the value of the slope of the tangent line to the
function at x =a. The derivative is obtained by taking a point x near
x = a and calculating the slope of the line that connects the two
points. The accuracy of calculating the derivative in this way
increases as point x is closer to point a. In the limit as point x
approaches point a, the derivative is the slope of the line that is
tangent to f(x) at x =a.
= f'(a) = lim
Backward Difference
x
Backward finite difference
Approximated
derivative
f(x) +
f(x) - f(a)
x-a
XLI XI X6+1
df|
dx\x=x
L
Figure 8-5: Finite difference approximation of derivative.
df
dx\x=x
True
derivative
Is the slope that connects points (x₁, f(x₁)) and (x₁+1 f (x₁+1));
dfl
dx\x=x₁
=
X
Is the slope that connects points (x₁-1₁, f(x₁-1)) and (x₁, f(x));
f(x₁+1) - f(x₁)
Xi+1 - Xi
f(x) +
f(x₁)-f(x₁-1)
xi-xi-1
Central Difference
Is the slope that connects points (x₁-1, f(x₁-1)) and (x₁+1, f(xi+1)):
f(x₁+1) -f(x₁-1)
Xi+1-XI-1
Central finite difference
Truc
derivative-
đại
Approximated
derivative
Xfol
X
УА
y=f(x)
L
Gives:
Integration is frequently encountered when solving
problems and calculating quantities in engineering and science.
Integration and integrals are also used when solving differential
equations. One of the simplest examples for the application of
integration is the calculation of the length of a curve. When a
curve in the x-y plane is given by the equation y = f (x), the length
L of the curve between the points a and x = b is given by:
Trapezoidal Method
A refinement over the simple rectangle and midpoint
methods is to use a linear function to approximate the
integrand over the interval of integration (Fig. 9-12).
Newton's form of interpolating polynomials with two points
x = a and x = b , yields:
b-a
f(x) = f(a)+(x-a)ƒ [a, b] = f(a) + (x −a)[ƒ(b)−ƒ(a)]
Substituting the above formula to:
The integrand can be an analytical function or a set of discrete points (tabulated data). When the
integrand is a mathematical expression for which the antiderivative can be found easily, the value of
the definite integral can be determined analytically. Numerical integration is needed when analytical
integration is difficult or not possible, and when the integrand is given as a set of discrete points.
y₁y=f(x)
Then simplifying it gives:
~√(√(a) + F(b)(b-
L =
= [*√/1 + [f'(x)]³dx
1(f) ~
f(b)
f(a)
((S)-° f(x) dx
1(ƒ ) = f* (ƒ(a) + (x − a)\ƒ(b)-f(a)]) dx
· ƒ(a)(b −a) + } [ƒ(b)− ƒ(a)](b.
b)) (b-a)/
Alea
f(b-a) b) - Ma))
rea
2
xa
x=b
Figure 9-12: The trapezoidal method.
a)(b-a)

Transcribed Image Text:A body moves along a straight line according to the law s =
and acceleration at the end of 2 seconds.
²-2; hence, when t = 2, u = (2)² -2= 4 ft/sec.
U=
a =
ds
dt
du
dt
=
3
3r; hence, when t = 2, a = 3(2) = 6 ft/sec².
³-2t. Determine its velocity
Expert Solution
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