## Calculating deviation with derivatives ...

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
 >
 (1)
 >
 >
 >
 (2)
 >
 (3)
 >
 (4)
 >
 (5)
 >
 >
 (6)
 >
 (7)
 >
 (8)
 >
 (9)
 > #Opdracht 4
 >
 >
 (10)
 >
 >
 (11)
 >
 (12)
 >
 (13)
 >
 (14)
 >
 (15)
 >
 (16)
 >
 (17)
 >

## The book didnt even gave the right answer...

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

 >
 >
 (1)
 >
 (2)
 >
 (3)
 >
 (4)
 >
 (5)

 >
 (6)
 >
 >
 (7)
 >
 (8)
 >
 (9)
 >
 (10)

## Undefined integral of zero....

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...

I receive the following error when I run maple to solve a diff equation.

 > 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)));
 >