Advanced Search

Maple 2019 Questions and Posts

These are Posts and Questions associated with the product, Maple 2019
View unanswered posts
Maple 2019 Questions and Posts Feed

Calculating deviation with derivatives ...

Asked by: The function 105
optimization calculus
May 12 2023
0  5

Hello, 

Ive followed the examples, and yet not the result the book gave. The question before this one, same type, and i did get the right answer there. 

Would you have a clue why it is not going as planned? 

Translation: "Through a resistancewire with the measured resistance R= (1.5+-0.5)Ohm goes an electric current with the measured strength I= (2.5+-0.05)A, During a timespan of t=(10+-0.5)s develops an amount of heat Q=RI^2*t. Give the estimate of the maximum relative deviation in Q."

Here i got the right answer of .42, and yet when doing the exact same thing in assignment 4, i got the wrong answers. 

Translation: "Two bodies with mass M and m have a massmidpointdistance r. For the gravitational force between these bodies counts: F=y*M*m/r^2, at which y is the gravitational constant. Give an estimation of the maximum realative deviation in F on ground of the following measurementdata: M=(7.0+-0.05)kg, m=(2.0+-0.02)kg and r=(5.0)+-0.05)m."

I had the answer -0.0029. The books says its wrong, but i cant see what i did wrong. 

Translation of the example: "According to the law of Ohm the currentstrength in the stream(/current)circle(circuit) with tension(voltage) U and resistance R is equal to I=U/R. For a given currentcircle((part of a)circuit) is given: U=(30.0+-0.05)Volt and R=(2.5+-0.03)Ohm. Determine the maximum absolute and relative deviation in I. 

Solution

The maximum absolute deviation in I is

This gives the currentstrength:

The maximum realtive deviation in I is: .... or about 1.4%"

Could somebody tell me what i am doing wrong?

Thank you!

Greetings,

The Function

> 

#opdracht 3
> 

R := 1.5; A := 2.5; t := 10; Q := R*A^2*t

1.5

  

2.5

  

10

  

93.750

 (1)
> 

NULL
> 

restart
> 

Q := R*A^2*t

R*A^2*t

 (2)
> 

diff(R*A^2*t, R)

A^2*t

 (3)
> 

diff(R*A^2*t, A)

2*R*A*t

 (4)
> 

diff(R*A^2*t, t)

R*A^2

 (5)
> 

NULL
> 

R := 1.5; A := 2.5; t := 10

1.5

  

2.5

  

10

 (6)
> 

.5*A^2*t+2*R*A*t*0.5e-1+.5*R*A^2

39.6875

 (7)
> 

Q := R*A^2*t

93.750

 (8)
> 

39.6875/(93.750)

.4233333333

 (9)
> 

#Opdracht 4
> 

restart
> 

M := 7; m := 2; r := 5; F := y*M*m/r^2

7

  

2

  

5

  

(14/25)*y

 (10)
> 

restart
> 

diff(y*M*m/r^2, M)

y*m/r^2

 (11)
> 

diff(y*M*m/r^2, m)

y*M/r^2

 (12)
> 

diff(y*M*m/r^2, r)

-2*y*M*m/r^3

 (13)
> 

M := 7; m := 2; r := 5; y := 9.81

7

  

2

  

5

  

9.81

 (14)
> 

0.5e-1*y*m/r^2+0.2e-1*y*M/r^2+(-2*y*M*m/r^3)*0.5e-1

-0.1569600000e-1

 (15)
> 

F := y*M*m/r^2

5.493600000

 (16)
> 

-0.1569600000e-1/(5.493600000)

-0.2857142857e-2

 (17)
> 

NULL

Download Mapleprimes_Question_Book_2_Paragraph_5.13_Question_4.mw

The book didnt even gave the right answer...

Asked by: The function 105
leastsquares
May 10 2023
0  4

The answers from the book dont make any sense after using their solution. Its not producing the expected results. 

This is the translation by the way: "A battery that has a voltage U, is in series with a resistor R, and a condenser with capacity C. The current strength on a given time "t" in this circuit is given by i=i(t)=U/R*e^-(t/R*C). Determine R and C based on the data given in table 5.6, the current is U=100V."

Greetings,

The Function


 

> 

restart
> 

U := 100

100

 (1)
> 

R := 4900

4900

 (2)
> 

C := 20*(1/1000)

1/50

 (3)
> 

"v(t):=U/(R)*(e)^(-(t/(R*C)))"

proc (t) options operator, arrow, function_assign; U*exp(-t/(R*C))/R end proc

 (4)
> 

v(.1)

0.2038734923e-1

 (5)

NULL

> 

data := LinearAlgebra:-Transpose(`<,>`(`<|>`(.1, .2, .3, .4, .5), `<|>`(7.36, 2.7, .99, .37, .13)))

Matrix(%id = 18446746712187741238)

 (6)
> 

plot(data, style = point)

 
> 

X := data[() .. (), 1]; Y := data[() .. (), 2]

Vector(5, {(1) = .1, (2) = .2, (3) = .3, (4) = .4, (5) = .5})

  

Vector[column](%id = 18446746712187761598)

 (7)
> 

sumX := add(X); sumY := add(Y); sumX2 := add(x^2, `in`(x, X)); sumXY := X.Y

1.5

  

11.55

  

.55

  

1.78600000000000003

 (8)
> 

eq1 := numelems(X)*a+b*sumX = sumY; eq2 := a*sumX+b*sumX2 = sumXY

5*a+1.5*b = 11.55

  

1.5*a+.55*b = 1.78600000000000003

 (9)
> 

solve({eq1, eq2}, {a, b})

{a = 7.347000000, b = -16.79000000}

 (10)

``

Download Mapleprimes_Question_Book_2_Paragraph_5.12_Question_5.mw

Undefined integral of zero....

Asked by: Bart 117
integration undefined
April 19 2023
1  7

The following 2D integrals of 0 are seemingly trivial and one would expect them to evaluate to zero, but Maple evaluates them to undefined

int(0, x=0..infinity, y=0..1) # undefined
int(0, x=0..1, y=0..infinity) # undefined

When the 2D integral is split into two 1D integrals, it does evaluate to zero, as the following examples show

int(0,x=0..infinity) # 0
int(int(0,x=0..infinity), y=0..1) # 0
int(int(0,x=0..1),y=0..infinity) # 0

If infinity is replaced by a variable (say 'c'), the first two integrals are also evaluated to zero.

It may be connected by the following

int(a, x=0..infinity, y=0..1) # a*infinity
int(a, x=0..1, y=0..infinity) # a*infinity
int(a,x=0..infinity) # signum(a)*infinity

So for the 1D integrals the signum is applied to 'a' when the interval is infinite, but not for the 2D integrals. I'm not sure about this difference.

Problem in solving a differential equation...

Asked by: pooyan1990 30
syntax dsolve differential-equations
April 05 2023
0  2

I receive the following error when I run maple to solve a diff equation.
Thank you in advance for your help.

> 

eq1:=A__n=[X__n+[(C-z)*(tan(theta1)+tan(theta2))]] * [B + (2*(C-z)*tan(theta3))];
eq2:=P__n=[2*(X__n+B)] + 2 * (C-z) * [(tan(theta1)+tan(theta2)) + (2*tan(theta3))];
eq3:=gamma - [cu*(P__n/A__n)] = diff(sigma__v(z),z);
eq4:=subs([eq1,eq2],eq3);
ics:=sigma__v(0)=sigma__s;
sol:=simplify(dsolve(([eq4,ics]), sigma__v(z)));
NULL

A__n = [X__n+[(C-z)*(tan(theta1)+tan(theta2))]]*[B+2*(C-z)*tan(theta3)]

  

P__n = [2*X__n+2*B]+2*(C-z)*[tan(theta1)+tan(theta2)+2*tan(theta3)]

  

gamma+[-cu*P__n/A__n] = diff(sigma__v(z), z)

  

gamma+[-cu*([2*X__n+2*B]+2*(C-z)*[tan(theta1)+tan(theta2)+2*tan(theta3)])/([X__n+[(C-z)*(tan(theta1)+tan(theta2))]]*[B+2*(C-z)*tan(theta3)])] = diff(sigma__v(z), z)

  

sigma__v(0) = sigma__s

  

Error, (in PDEtools/NumerDenom) invalid input: `PDEtools/NumerDenom` expects its 1st argument, ee, to be of type algebraic, but received [tan(theta1)+tan(theta2)+2*tan(theta3)]

  
> 

``

Download diffeq.mw

How to import this data file?...

Asked by: FDS 115
import
April 03 2023
1  7

Dear power users, I have a probably trivial question. How can I import the attached file in Maple? The file consists of a header part with below the header two columns with data. The file coming from a measuring device has no extension and I could not upload it to the forum as it was considered a null file. In order to solve this I gave it a .txt extension but keep in mind that this normally is not present.

Additional question: if the file would have several headers each time followed by 2 data columns, is there an easy way to separate them in a sort of automatic manner?

Spiking2011622.txt

How to solve this question?...

Asked by: Harry Xu 20
rootfinding homework do-my-homework-for-me
April 03 2023
0  1