The measurement of the circumference of a circle is found to be 64 centimeters, with a possible error of 0.9 centimeter. Approximate the percent error in computing the area of the circle.
MY ANSWER 8.83%
11 answers
Also, it says to estimate the maximum allowable percent error in measuring the circumference if the error in computing the area cannot exceed 3%. How do I do this?
C = 2 pi r = 64
so r = 64/(2pi) = 10.19
dC/dr = 2 pi
so dC = 2 pi dr
.64 = 2 pi dr
so
dr = .10186
A = pi r^2 = pi (10.19)^2 = 325.95
dA/dr = 2 pi r (of course:)
dA = 2 pi r dr
dA = 2 pi (10.19)(.10186)
dA = 6.52
percent error = (6.52/325.95)100 = 2%
so r = 64/(2pi) = 10.19
dC/dr = 2 pi
so dC = 2 pi dr
.64 = 2 pi dr
so
dr = .10186
A = pi r^2 = pi (10.19)^2 = 325.95
dA/dr = 2 pi r (of course:)
dA = 2 pi r dr
dA = 2 pi (10.19)(.10186)
dA = 6.52
percent error = (6.52/325.95)100 = 2%
part 2
(dA/A)100 = 3
(2 pi r dr /pi r^2)100 = 3
or
dr/r = .015
but dr = dC/(2pi)
so
dC/2 pi r = .015
or
dC/C = .015
so 1.5 % error in C gives 3% error in A
(dA/A)100 = 3
(2 pi r dr /pi r^2)100 = 3
or
dr/r = .015
but dr = dC/(2pi)
so
dC/2 pi r = .015
or
dC/C = .015
so 1.5 % error in C gives 3% error in A
Okay, I really don't understand how you did any of that. What's the point of half of it? I tried to do some of what you did, but I got different numbers for my answers.
I made that a lot harder than it needed to be but follow it and you will be able to shorten it.
well, let me do the second part slowly
C = 2 pi r
dC = 2 pi dr
so
dC/(2 pi r) = dr/r
or
dC/C = dr/r which any architect will tell you
dA = C dr
dA/A = C dr/A
dA/A = C dr/(pir^2)
dA/A = 2 pi r dr/(pi r^2)
dA/A = (2)dr/r
SO
dr/r = (1/2) dA/A Important
if dA/A = 3%
then dr/r = half that or 1.5 %
C = 2 pi r
dC = 2 pi dr
so
dC/(2 pi r) = dr/r
or
dC/C = dr/r which any architect will tell you
dA = C dr
dA/A = C dr/A
dA/A = C dr/(pir^2)
dA/A = 2 pi r dr/(pi r^2)
dA/A = (2)dr/r
SO
dr/r = (1/2) dA/A Important
if dA/A = 3%
then dr/r = half that or 1.5 %
You realize I'm a terrible Calculus student? I don't understand the concepts that the other students can pick up on right away. I don't really get why you did any of what you did, and there aren't any examples like this one in our book for me to look back on.
C = 2 pi r = 64
so r = 64/(2pi) = 10.19
dC/dr = 2 pi
so dC = 2 pi dr
.9 = 2 pi dr (used .64 by mistake)
so
dr = .1433
A = pi r^2 = pi (10.19)^2 = 325.95
dA/dr = 2 pi r (of course:)
percent error = (9.17/325.95)100 = 2.81%
dA = 2 pi r dr
dA = 2 pi (10.19)(.1433)
dA = 9.17
so r = 64/(2pi) = 10.19
dC/dr = 2 pi
so dC = 2 pi dr
.9 = 2 pi dr (used .64 by mistake)
so
dr = .1433
A = pi r^2 = pi (10.19)^2 = 325.95
dA/dr = 2 pi r (of course:)
percent error = (9.17/325.95)100 = 2.81%
dA = 2 pi r dr
dA = 2 pi (10.19)(.1433)
dA = 9.17
note that if I had done the second part first I would have known that
dA/A = 2 (dC/C)
dA/A = 2(.9/64)
= .028
which is 2.8 percent which I got after much algebra
dA/A = 2 (dC/C)
dA/A = 2(.9/64)
= .028
which is 2.8 percent which I got after much algebra
C = 2 pi r = 64 {circumference is 2 pi r}
so r = 64/(2pi) = 10.19
dC/dr = 2 pi { just taking the derivative of 2 pi r}
so dC = 2 pi dr {multiplied both sides by r}
.9 = 2 pi dr (used .64 by mistake) {the .9 cm was given in the problem statement for dC}
so
dr = .1433 {divided by 2 pi}
A = pi r^2 = pi (10.19)^2 = 325.95
{because area = pi r^2 of circle}
dA/dr = 2 pi r (of course:) { taking derivative but if we sketch we see that increase in area is circumference times dr}
then I inverted some lines
dA = 2 pi r dr {multiplying both sides of above by dr
dA = 2 pi (10.19)(.1433) { putting in the numbers for r and dr }
dA = 9.17
percent error = (9.17/325.95)100 = 2.81%
{because we want dA/A * 100}
so r = 64/(2pi) = 10.19
dC/dr = 2 pi { just taking the derivative of 2 pi r}
so dC = 2 pi dr {multiplied both sides by r}
.9 = 2 pi dr (used .64 by mistake) {the .9 cm was given in the problem statement for dC}
so
dr = .1433 {divided by 2 pi}
A = pi r^2 = pi (10.19)^2 = 325.95
{because area = pi r^2 of circle}
dA/dr = 2 pi r (of course:) { taking derivative but if we sketch we see that increase in area is circumference times dr}
then I inverted some lines
dA = 2 pi r dr {multiplying both sides of above by dr
dA = 2 pi (10.19)(.1433) { putting in the numbers for r and dr }
dA = 9.17
percent error = (9.17/325.95)100 = 2.81%
{because we want dA/A * 100}
Honestly, I just waited till he explained it on board. i just put random stuff. :)