The radius of a circular disk is given as 19 cm with a maximum error in measurement of 0.2 cm.

Question

(a) Use differentials to estimate the maximum error in the calculated area of the disk. (Round your answer to two decimal places.)

 

(b) What is the relative error? (Round your answer to four decimal places.)

What is the percentage error? (Round your answer to two decimal places.)

 

Answer 1

Ok, let us take a lightning tour of 'Error Analysis', as it is usually called, or 'Analysis of Uncertainties', which is conceptually less misleading than the popular name. The term 'Error' suggests that it is something that comes about from the experimenter being careless (which may well be the case, but not always). In reality, any measuring apparatus measuring any physical quantity (like r) always has some uncertainty; no measuring device is infinitely resolving. So, when we measure r, we always estimate the mean value to be within some neighborhood δr>0 of an estimate, r₀. In other words, the estimated value of r lies in the range (r₀-δr, r₀+δr), or, more elegantly, r ∈ (r₀ ± δr), with possible values of r lying in an interval of length 2*δr

First, a little point regarding the phrasing of your question. You say that we have measured the radius r of a disk to be (19 ± 0.1) cm. The mean value should be specified till the least significant digit (so, 19.x instead of 19). The least (or last) significant digit is the one whose place value is comparable to the uncertainty of the measuring apparatus, which in this case is δr = 0.1 cm. That is, when we measured a distance using a mm scale, we record a value upto the mm value, and then take into account the fact that the measurement is uncertain due to the fact that the scale has a smallest marking (1mm), and any point lying between two markings has an uncertain position. The measured r could be (19.1 ± 0.1) cm, for example.

Now, coming to the question, we wish to use the uncertainty in the measured value of r in order to estimate the uncertainty in the calculated value of area A. In order to do this, we need to perform what is known as propagation of uncertainty. The basic idea goes as follows:

Let us say that we wish to compute some R that is a function of r : R = R(r). Obviously, if we could only estimate the value of r upto some uncertainty, we can only calculate the value of R(r) upto some uncertainty. as well. The simplest way one could go about doing so is to use the range of values estimated for r , (r_min,r_max) = (r₀-δr, r₀+δr) to find the range of possible values of R. If, for example, R ∼ rⁿ where n > 0 (for example, R(r) = A = πr²), then R_max = r_max ⁿ and R_min = r_min ⁿ .

So, δR = ( r_max ⁿ - r_min ⁿ )/2 = (r_max - r_min)(r_max^n-1 + r_max^n-2r_min + ... r_min^n-1)/2

= δr (r_max^n-1 + r_max^n-2r_min + ... r_min^n-1). So, δR ∝ δr.

Now comes the crucial insight. In case we have a sensible measuring setup, the absolute value of the estimated mean value should be much greater than the uncertainty, that is |r₀| >> δr (If we wish to measure something a few mm in length using a mm scale, we don't have a very sensible experimental setup). When this happens, we can expand any function of the two variables, δR(r₀,δr), into a power series of terms of decreasing importance: δR(r₀,δr) = δR₀ + δR₁(δr/r₀) + δR₂(δr/r₀)² + ... .

It is clear that the constant δR₀ = δR(r₀, δr=0) = 0.

The terms decrease in importance as |δr/r₀| << 1, so |δr/r₀|² <<<< 1 etc.

If we only wish to estimate the most dominant contribution of the measurement uncertainty δr to the calculation uncertainty δR, we should only retain the δR₁ term.

So, δR ≅ δr(δR₁/r₀ ) ~ # δr ~ δr (r_max^n-1 + r_max^n-2r_min + ... r_min^n-1) from above.

Since we are only interested in the dominant contribution of δr (i.e. the δr¹ contribution), we should estimate both r_max and r_min in the bracket to be equal to r₀; the dependence of r_max and r_min on δr introduces additional powers of δr. Then, we have : δR ≅ (n r₀^n-1)δr. If we have R = crⁿ for some constant c, then δR ≅ c(nr^n-1) δr.

Note that this is very suggestive, as it should be. Actually, we can derive the above result in a much more elegant (and illuminative) method. Essentially, when |r₀| >> δr, the problem of propagating uncertainties reduces to the mathematical problem of finding the Taylor series expansion for the function R of r about some value r₀ . This is given as:

R(r₀+δr) = R(r₀) + R'(r₀) δr + R''(r₀) (δr)²/2 +....+ R^(m)(δr)^m/(m)! + ... ≅ R₀ + R'(r₀) δr ,

where R'(r₀) is the first derivative of the fn R with respect to r evaluated at r₀ (and R^(m) the m^th derivative). When R = crⁿ, R' = cnr^n-1, giving the above result. The relative uncertainty is then δR/R₀ ≅ n δr/r₀.

Note that the constant c is unimportant when evaluating the relative uncertainty.

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BONUS: This second method actually allows us to estimate the uncertainty in a calculated quantity R that depends on several measured quantities a,b,c,.... By employing the same logic, we can perform a power series expansion on R. Now, the power series expansion will involve partial derivatives.

R(a+δa , b+δb , c+δc , ...) ≅ R(a,b,c,...) + (∂R/∂a (a,b,c) δa + ∂R/∂b δb + ∂R/∂c δc +...).

If R = # aⁿ b^m c^o ... , then it is easy to show that this method yields

δR / R₀ = δR / R(a,b,c) = n(δa / a) + m(δb / b) + o(δc / c) + ...

Actually, this is an overestimate unless a, b , c are correlated variables. That is, if a, b , c are physical quantities that are connected in a manner that the observed deviation from the true mean is in the same direction for a , b, and c. Typically though, a , b , c will be uncorrelated variables (like distance, time, mass etc.) in this case, a better estimate for the relative uncertainty in R is

δR / R₀ = ((n(δa / a))² + (m(δb / b))² + (o(δc / c))² + ...)^1/2.

But how we get there is a whole other story.

Hope this helps you with your question.

Answer 2

Hi Natalie!

For part (a), to calculate the maximum error, we will want to utilize the concept of differentials as it states in the instruction. Why is that? Because if there was a maximum error of 0.2 cm, this means that the radius could have been as large as 19.2 cm, which means that we are attempting to locate the difference in area of the disk when the radius changes from 19 cm to 19.2 cm. To calculate this most accurately, we want to use the derivative equation for the area.

First, we know that the area of a circle is found by:

A = π r²

Therefore, the area of the disk with the radius of 19 cm is A = π (19)² = 361π cm²

To estimate the maximum error as a result, we can take the derivative of this equation:

A = π r²

dA/dr = 2πr (Using the Power Rule of derivatives)

dA = 2πr ⋅dr (Change in area, ΔA, with respect to the change in radius)

If we substitute the given values, r = 19 cm, and the maximum error in the measurement of the radius (change in r, Δr or dr) equals 0.2 cm, we will find that:

dA = 2π (19)(0.2) = 7.6π ≈ 23.88

Therefore, the maximum error of the area of the disk, rounded to two decimal places is 23.88 cm²

The relative error is going to take this maximum error of the area of the disk and divide that by what the actual area is with the radius measurement of 19 cm:

ΔA / A

Now, we can answer part (b) by taking (7.6π )/(361π), which would get us approximately 0.0211 rounded to 4 decimal places for the relative error.

The percentage error, part (c), would simply be this 0.0211 multiplied by 100, which is 2.11% rounded to two decimal places!

If you have any additional questions, feel free to reach out to let me know!

Thanks!

Answer 3

See the video.

https://www.youtube.com/watch?v=wsSOx4uxURU