A) Cross section diameter is y = x/2 (similar triangles), so cross section area is

A(x) = (1/2)π(x/4)^{2} = πx^{2}/32.

B) The volume is

∫_{0}^{2} πx^{2}/32 dx = π/12

II. A) Subdividing the wire into n equal pieces of length Δx = 3/n, the mass of the segment

of the wire between x_{i−1} and x_{i} is approximately

(density) · (length) ≐ (4 − x_{i})\sqrt{1 + (2x)^{2}}Δx

B)

∫_{−1}^{2}(4 − x)\sqrt{1 + 4x^{2}} dx.

C)

\frac{32}{12}\sqrt{17} − ln(\sqrt{17} − 4) + \frac{29}{12}\sqrt{5} − ln(−2 + \sqrt{5}) ≐ 19.6.

III. A) Letting P be the population as a function of time, \frac{dP}{dt} = −k\sqrt{P}. B) All slopes on a horizontal line are equal. For P = 0, slope is 0, P = 1, slope is −1,

P = 2, slope is −\sqrt{2}, etc. C) Solving by separation of variables,

P (t) = (−\frac{k}{2} t + c)^{2}

for some constant c. Letting t = 0, using P (0) = 900, c = 30. Letting t = 6, using P (6) = 441 solving for k, k = 3. So P reaches 0 when t = 20.

IV. A)

\begin{aligned}\frac{dx(t)}{dt} & = 1 + 3 cos(t)/2 = y(t) \\ \frac{dy(t)}{dt} & = −3 sin(t)/2 = −x(t) + t \end{aligned}

B) t solving cos(t) = −2/3, so t ≐ 2.3 radians. C) Three times; once before t = π, once before and once after. (This is a prolate cycloic)