Which area's natural resources are LEAST LIKELY to be able to support a high population density?
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A. Depending on which variable you choose to integrate with, you can capture the total bounded region with either -2 ≤ x ≤ (-1 + √5)/2 or 1 ≤ y ≤ (5 + √5)/2, where the upper endpoints correspond to the coordinates of the appropriate intersections:
y = x² + 1
⇒ x = (x² - 2)² - 2
⇒ x⁴ - 4x² - x + 2 = 0
⇒ (x - 2) (x + 1) (x² + x - 1) = 0
⇒ x = 2, x = -1, x = -1/2 ± √5/2
⇒ y = 5, y = 2, y = (5 ± √5)/2
On the other hand, we can compute the areas of A and B separately, then sum those integrals. Area A is easier to compute by integrating with respect to y over 2 ≤ y ≤ (5 + √5)/2, while area B is easier to find by integrating x over -1 ≤ x ≤ (-1 + √5)/2.
B. I'll stick to the split-region approach.
First, we find equations for the appropriates halves of either parabola:
• y = x² + 1 ⇒ x = ± √(y - 1)
and x = -√(y - 1) describes the left half of the blue parabola;
• x = (y - 3)² - 2 ⇒ y = 3 ± √(x + 2)
and y = 3 - √(x + 2) describe the bottom half of the red parabola.
Now we can set up the integrals.
Area of A:
Area of B:
Alternatively, one can prove that the regions A and B are symmetric across the line y = x + 3, so we can simply pick one of these integrals and double it.
C. Computing the integrals, we find
area of A = 2/3
area of B = 2/3
and so the total area is 2/3 + 2/3 = 4/3.
Each of these ODEs is linear and homogeneous with constant coefficients, so we only need to find the roots to their respective characteristic equations.
(a) The characteristic equation for
is
which arises from the ansatz .
The characteristic roots are and
. Then the general solution is
where are arbitrary constants.
(b) The characteristic equation here is
with a root at of multiplicity 2. Then the general solution is
(c) The characteristic equation is
with roots at , where
. Then the general solution is
Recall Euler's identity,
Then we can rewrite the solution as
or even more simply as