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Math  /  Calculus

QuestionProblem 1: Multi-species Flocks Ph.D. candidate Jenny Muñez studies multi-species bird flocks in the Colombian Andes. She surveys birds from two species (species 1 and species 2) counting the number of birds every 50 m of elevation gain of along an elevational gradient starting at an altitude of 100 m and ending at 400 m . Her data is given in the following table and plot: \begin{tabular}{c|c|c} \begin{tabular}{c} Elevation \\ ee \end{tabular} & \begin{tabular}{c} \#Species 1 \\ N1(e)N_{1}(e) \end{tabular} & \begin{tabular}{c} \# Species 2 \\ N2(e)N_{2}(e) \end{tabular} \\ \hline 100 & 2.32 & 0.64 \\ 150 & 30.84 & 7.37 \\ 200 & 15.88 & 13.95 \\ 250 & 7.21 & 14.61 \\ 300 & 3.88 & 18.53 \\ 350 & 1.02 & 17.22 \\ 400 & 1.94 & 18.14 \\ \multicolumn{3}{c}{Δe=400−1002=50\Delta e=\frac{400-100}{2}=50} \end{tabular} A. Use Riemann Sums to approximate the Niche size of species 1 (Blue) and species 2 (Orange) by approximating the following two definite integrals (show your calculations):
Niche Size Species 1: ∫100400N1(e)de≈∑bN1(ei)Δe\quad \int_{100}^{400} N_{1}(e) d e \approx \sum^{b} N_{1}\left(e_{i}\right) \Delta e Δe=50\Delta e=50

Studdy Solution

STEP 1

What is this asking? We need to estimate the *niche size* of two bird species by calculating the area under their population curves using something called Riemann sums! Watch out! Don't forget to multiply each population count by the elevation step (Δe\Delta e), and make sure you're using the right population numbers for each species!

STEP 2

1. Calculate the niche size of Species 1.
2. Calculate the niche size of Species 2.

STEP 3

We're going to **approximate** the area under the curve for Species 1 using a Riemann sum.
Think of it like drawing a bunch of rectangles under the curve and adding up their areas.
The width of each rectangle is Δe=50\Delta e = \mathbf{50} meters.

STEP 4

The Riemann sum for Species 1 is given by: ∫100400N1(e)de≈∑N1(ei)⋅Δe \int_{100}^{400} N_{1}(e) de \approx \sum N_{1}(e_i) \cdot \Delta e Let's break it down!
We **multiply** each population count N1(e)N_1(e) by Δe\Delta e and then **add** them all up:
Niche Size Species 1≈(2.32⋅50)+(30.84⋅50)+(15.88⋅50)+(7.21⋅50)+(3.88⋅50)+(1.02⋅50)+(1.94⋅50)=116+1542+794+360.5+194+51+97=3154.5\begin{aligned} \text{Niche Size Species 1} &\approx (2.32 \cdot 50) + (30.84 \cdot 50) + (15.88 \cdot 50) + (7.21 \cdot 50) \\ &+ (3.88 \cdot 50) + (1.02 \cdot 50) + (1.94 \cdot 50) \\ &= 116 + 1542 + 794 + 360.5 + 194 + 51 + 97 \\ &= \mathbf{3154.5} \end{aligned}So, the **approximate niche size** of Species 1 is 3154.5\mathbf{3154.5}.

STEP 5

Now, we do the same thing for Species 2!
We'll use the same Δe=50\Delta e = \mathbf{50} meters.

STEP 6

∫100400N2(e)de≈∑N2(ei)⋅Δe \int_{100}^{400} N_{2}(e) de \approx \sum N_{2}(e_i) \cdot \Delta e Let's **multiply** each population count N2(e)N_2(e) by Δe\Delta e and then **add** them all up: Niche Size Species 2≈(0.64⋅50)+(7.37⋅50)+(13.95⋅50)+(14.61⋅50)+(18.53⋅50)+(17.22⋅50)+(18.14⋅50)=32+368.5+697.5+730.5+926.5+861+907=4523.5\begin{aligned} \text{Niche Size Species 2} &\approx (0.64 \cdot 50) + (7.37 \cdot 50) + (13.95 \cdot 50) + (14.61 \cdot 50) \\ &+ (18.53 \cdot 50) + (17.22 \cdot 50) + (18.14 \cdot 50) \\ &= 32 + 368.5 + 697.5 + 730.5 + 926.5 + 861 + 907 \\ &= \mathbf{4523.5} \end{aligned} The **approximate niche size** of Species 2 is 4523.5\mathbf{4523.5}.

STEP 7

The approximate niche size of Species 1 is 3154.5\mathbf{3154.5}, and the approximate niche size of Species 2 is 4523.5\mathbf{4523.5}.

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