Good Question 16: 2018 BC 2(b)

In this post we look at another part of the AP Calculus BC exam. Good Question 15 discussed the unusual units in 2018 BC 2(a). In this post we look at 2018 BC 2(b) where units help us find the correct integral to answer the question.

How do you answer a question of a type you’ve never seen before? I  expect that’s what many of the students taking the 2018 AP Calculus exam were asking when they got to BC 5. If you’ve never done a density question how do you handle this one? 

The question concerns density. Density gives you how much of something exists in a certain length, area, or volume.  Density questions have appeared on the exam now and then, most recently 2008 AB 92 (which really isn’t recent, but then there are a lot of questions we never see). I have a blog post about the density here with several examples. In that post the alternate solution to example 3 explained how I used a unit analysis to find the answer; I used a similar approach here.

2018 BC 2 (b)

The stem of the question tells us that at a depth of meters, 0 < h < 30, the number of plankton in a cubic meter of sea water is modeled by p\left( h \right)=0.2{{h}^{2}}{{e}^{{-0.0025{{h}^{2}}}}} million cells per cubic meter. Part (b) asks for the number of million of plankton in a column of water whose horizontal cross sections have a constant area of 3 square meters.

If the density were constant, then it is just a matter of multiplying the volume of the column times the constant density. Alas, the density is not constant; it varies with the depth. What to do?

Since an amount is asked for, you usually look around for a rate to integrate. Density is a kind of rate: the units are millions of cells per cubic meter. You need to integrate something concerning the density so that you end up with millions of cells; something that will “cancel” the cubic meters.

Consider a horizontal slice thru the column at depth h meters. While I’m not sure plankton is a good topping for pizza, you could picture this as a rather large pizza box whose sides are \sqrt{3} meters long and whose height is  \Delta h meters. This box has a volume of 3 \Delta h cubic meters. For small values of \Delta h the number of million plankton in the box is nearly constant, so at depth hi , there are p(hi) million plankton per cubic meter or {3p\left( {{{h}_{i}}} \right)\Delta h} million plankton in the box.

Notice how the units of the individual quantities combine to assure you the final quantity has the correct units:

\displaystyle (3\text{ square meters)}\cdot \left( {p\left( {{{h}_{i}}} \right)\text{ }\frac{{\text{million plankton}}}{{\text{cubic meters}}}} \right)\left( {\Delta h\text{ meters}} \right)=3p\left( {{{h}_{i}}} \right)\Delta h\text{ million plankton}

Now to find the amount in the column of water we can add up a stack of “pizza boxes.” The sum is \sum\limits_{{i=1}}^{n}{{3p\left( {{{h}_{i}}} \right)\Delta h}}. Now, if we take thinner boxes by letting \Delta h\to 0, we are looking at a Riemann sum. And calculus gives us the answer.

\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{3p\left( {{{h}_{i}}} \right)\Delta h}}=\int_{0}^{{30}}{{3p\left( h \right)dh}}\approx 1,675 million plankton in the column of water (rounded to the nearest million as directed in the question.)


Previous Good Questions can be found under the “Thru the Year” tab on the black navigation bar at the top of  the page, or here.

 

 

 

 


 

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