Approximating error in area of a circle from its radius
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Given a circle with area 15 and an error in its area of .2, what would the maximum error in its radius be?
Given that $$A=pi r^2$$ $$A'=2pi r$$ it can be found that the radius when A = 15 must be $$r=sqrt{frac{15}{pi}} $$ so using differentials $$A(r+Delta r)approx A(r)+A'(r)Delta r $$ $$15approxfrac{sqrt{15}}{pi}+2pisqrt{frac{15}{pi}}Delta r$$
so $$Delta rapproxfrac{15-sqrt{frac{15}{pi}}}{2pi sqrt{frac{15}{pi}}}$$
Is this correct?
calculus ordinary-differential-equations
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add a comment |
$begingroup$
Given a circle with area 15 and an error in its area of .2, what would the maximum error in its radius be?
Given that $$A=pi r^2$$ $$A'=2pi r$$ it can be found that the radius when A = 15 must be $$r=sqrt{frac{15}{pi}} $$ so using differentials $$A(r+Delta r)approx A(r)+A'(r)Delta r $$ $$15approxfrac{sqrt{15}}{pi}+2pisqrt{frac{15}{pi}}Delta r$$
so $$Delta rapproxfrac{15-sqrt{frac{15}{pi}}}{2pi sqrt{frac{15}{pi}}}$$
Is this correct?
calculus ordinary-differential-equations
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1
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$A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
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– Dando18
Dec 11 '18 at 23:47
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Yes, edited to show that.
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– ovil101
Dec 11 '18 at 23:51
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Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
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– LutzL
Dec 12 '18 at 8:56
add a comment |
$begingroup$
Given a circle with area 15 and an error in its area of .2, what would the maximum error in its radius be?
Given that $$A=pi r^2$$ $$A'=2pi r$$ it can be found that the radius when A = 15 must be $$r=sqrt{frac{15}{pi}} $$ so using differentials $$A(r+Delta r)approx A(r)+A'(r)Delta r $$ $$15approxfrac{sqrt{15}}{pi}+2pisqrt{frac{15}{pi}}Delta r$$
so $$Delta rapproxfrac{15-sqrt{frac{15}{pi}}}{2pi sqrt{frac{15}{pi}}}$$
Is this correct?
calculus ordinary-differential-equations
$endgroup$
Given a circle with area 15 and an error in its area of .2, what would the maximum error in its radius be?
Given that $$A=pi r^2$$ $$A'=2pi r$$ it can be found that the radius when A = 15 must be $$r=sqrt{frac{15}{pi}} $$ so using differentials $$A(r+Delta r)approx A(r)+A'(r)Delta r $$ $$15approxfrac{sqrt{15}}{pi}+2pisqrt{frac{15}{pi}}Delta r$$
so $$Delta rapproxfrac{15-sqrt{frac{15}{pi}}}{2pi sqrt{frac{15}{pi}}}$$
Is this correct?
calculus ordinary-differential-equations
calculus ordinary-differential-equations
edited Dec 11 '18 at 23:50
ovil101
asked Dec 11 '18 at 23:44
ovil101ovil101
373
373
1
$begingroup$
$A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
$endgroup$
– Dando18
Dec 11 '18 at 23:47
$begingroup$
Yes, edited to show that.
$endgroup$
– ovil101
Dec 11 '18 at 23:51
$begingroup$
Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
$endgroup$
– LutzL
Dec 12 '18 at 8:56
add a comment |
1
$begingroup$
$A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
$endgroup$
– Dando18
Dec 11 '18 at 23:47
$begingroup$
Yes, edited to show that.
$endgroup$
– ovil101
Dec 11 '18 at 23:51
$begingroup$
Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
$endgroup$
– LutzL
Dec 12 '18 at 8:56
1
1
$begingroup$
$A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
$endgroup$
– Dando18
Dec 11 '18 at 23:47
$begingroup$
$A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
$endgroup$
– Dando18
Dec 11 '18 at 23:47
$begingroup$
Yes, edited to show that.
$endgroup$
– ovil101
Dec 11 '18 at 23:51
$begingroup$
Yes, edited to show that.
$endgroup$
– ovil101
Dec 11 '18 at 23:51
$begingroup$
Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
$endgroup$
– LutzL
Dec 12 '18 at 8:56
$begingroup$
Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
$endgroup$
– LutzL
Dec 12 '18 at 8:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your approach is correct but likely an overkill. The final result is also not really pretty.
If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.
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add a comment |
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The errors are related by
$$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$
Therefore
$$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$
Your formula is incorrect because you had $Delta A = A - r$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Your approach is correct but likely an overkill. The final result is also not really pretty.
If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.
$endgroup$
add a comment |
$begingroup$
Your approach is correct but likely an overkill. The final result is also not really pretty.
If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.
$endgroup$
add a comment |
$begingroup$
Your approach is correct but likely an overkill. The final result is also not really pretty.
If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.
$endgroup$
Your approach is correct but likely an overkill. The final result is also not really pretty.
If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.
answered Dec 12 '18 at 2:07
Lucas HenriqueLucas Henrique
1,059414
1,059414
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The errors are related by
$$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$
Therefore
$$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$
Your formula is incorrect because you had $Delta A = A - r$
$endgroup$
add a comment |
$begingroup$
The errors are related by
$$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$
Therefore
$$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$
Your formula is incorrect because you had $Delta A = A - r$
$endgroup$
add a comment |
$begingroup$
The errors are related by
$$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$
Therefore
$$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$
Your formula is incorrect because you had $Delta A = A - r$
$endgroup$
The errors are related by
$$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$
Therefore
$$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$
Your formula is incorrect because you had $Delta A = A - r$
answered Dec 13 '18 at 6:13
DylanDylan
12.9k31027
12.9k31027
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1
$begingroup$
$A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
$endgroup$
– Dando18
Dec 11 '18 at 23:47
$begingroup$
Yes, edited to show that.
$endgroup$
– ovil101
Dec 11 '18 at 23:51
$begingroup$
Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
$endgroup$
– LutzL
Dec 12 '18 at 8:56