Question: 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

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

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