Question

5.(a) The percentage error in the radius of a circle is 2%. Calculate the percentage error
in its area and volume?
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Answer 1

Answer:

What is the percentage error in the area if the percentage error in the radius of a circle is 2%?

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Two approaches are possible.

A. Using the calculus.

Since the Area A = (pi)r^2, the differential area dA resulting from a differential change in radius is dA = 2(pi)r dr.

For dr = 0.02 r, then dA= 2(pi)r(0.02r)=(0.04)A

so dA/A=.04, indicating a 2% error in radius corresponds to 4% err in area.

B. Arithmetic Method

The method of the calculus presumes the differential, dr, is infinitesimal compared to r. But a 2% error is actually finite. So the largest possible area of the circle with radius =r(1+.02) is

A*=(pi) r^2 (1. 0 2 )^2

This is greater than the nominal area A=(pi)r ^2 by the amount A*-A or

A*-A =(pi)r ^2 (1+.04+.02^2 )-(pi)r ^2 =(pi)r ^2 [.04+.02^2 ]

so the error, (A*-A)/A = 0.04 +.0004) or 4.04%

NOTE: Here it was assumed the 2 % was positive. By the calculus method, if the error is negative, the error in area is negative 4%. But by the arithmetic method, if will be found to be only -0.04+.0004 or -3.96%. If these calculations are repeated with a much smaller assumed error, say 0.1%, it will be seen that the arithmetic method is converging on the differential method, both from above and below.

Explanation:

It is a correct answer