child's toy: rect. with semicirc. ends: find error of area

[Sophie]

A child toy is a flat disk shaped, as in the diagram (I could not upload the diagram but I drew one by hand and took a picture, I hope this helps), in a rectangle with a semi-circular ends.



The width of the rectangle is measured to be 8cm with an error of 0.05cm and the length is three times the width. Use differentials to find the percentage error in the area.

I feel 80% confidant I have done the right thing, however I always have problems interpreting written questions. If someone could check I am right I would be very grateful…

r=4
x=8
dx=0.05

Area = (Pi) (r^2) + x * 3x
Area = (Pi) (r^2) + 3(x^2)
Area = 50.24 + 3(x^2)

dA = (6x) dx
dA = (6*8) (0.05)
dA = 2.4cm squ This is the max error

Relative error

((6x) dx) / (50.24 + 3(x^2))

2.4/242.24 = 0.009907 is the relative error

% error = relative error * 100%

%error = 0.009907 * 100 = 0.99% error


Thanks for your time, Sophie

 

[galactus]

Use ImageShack to host and display your diagram.

http://www.imageshack.us/

 

[Sophie]

I think I solved my mistake, yippeee, but I can not make mathType cut and paste anymore, sorry...

x=8

r=x/2

dx=0.05


Area = (3(x^2)) + ((Pi. (x^2))/4)

dA = (6x + ((Pi.x)/2)) dx

dA = 3.03cm sq, which is the max error

Relative error = 3.03/(3(x^2)) + ((Pi. (x^2))/4) = 3.03/242.27 = 0.0125

%error = relative error * 100%

% error = 0.0125 * 100 = 1.25%