What Should Be Included With Your Computations

Many think that a computation can be presented as a few lines of mathematical expressions. In rigorous mathematics, much more is expected of your write-up. What we are really looking for are computations with explanations.

Computations With Explanations are mathematical expressions which are continually being justified and clarified by prose.

Needless to say the prose should always be in complete sentences.

Warning: Do not get carried away with your prose. A longer solution is not necessarily a better solution. Only include prose which justifies or clarifies. Leave out anything superfluous.

Video Tutorial

Below is a video tutorial discussing how to present computations.

All problems are from:

Deborah Hughes-Hallett, Andrew Gleason, et al.: Calculus: Single Variable, Fourth Edition, Wiley, 2004

Test Yourself

Instructions

Read the following team homework problem. Below it are four possible presentations of computations. Think about which one best fits the criteria described above and then check your answer.

The Question

One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a continuous rate of approximately 2.47% per year. After the Chernobyl disaster, it was suggested that it would be about 100 years before the region would again be safe for human habitation. What percent of the original strontium-90 would still remain then?

Possible Presentations Of Computations

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(a) 8.46%

Click here if you think (a) is the best possible choice.

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(b) S(t)=A e^{-0.0247 t}

S(100)= A e^{-0.0247*(100)}= A (0.08458)

8.46%

Click here if you think (b) is the best possible choice.

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(c) Since strontium-90 decays exponentially at a continuous rate of approximately 2.47% per year, the amount of strontium-90 left after t years is given by S(t)=A e^{-0.0247 t} where A is the initial amount of strontium-90. To figure out the amount of strontium-90 left after 100 years we plug in t=100.

S(100)= A e^{-0.0247 (100)}= A (0.08458)

So 8.46% the original strontium-90 is left.

Click here if you think (c) is the best possible choice.

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(d) We have exponential decay so we have P₀ a^t. Continuous decay means we use e instead. So P₀ e^-kt. Here k is 2.47% and t is 100. So

P₀ e^{-0.0247 * 100}=P₀(0.08458).

So 8.46%.

Click here if you think (d) is the best possible choice.

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