MATH 180 FALL 1995

ASSIGNMENT #33

5.5, #7(a). The reason for the rise in the graph: flying at higher velocities requires correspondingly more energy. The reason for the dip in the graph: taking off from the ground requires more energy than staying airborn and flying at low velocities.

(b) A dimension analysis shows that a(v) = f(v)/v, since the units of a(v), f(v), v are respectively joules/meter, joules/second, meters/second

(c) a(v) is related to the graph of f(v) as follows: Draw the line ℓ_{v} from the origin to the point (v, f(v)) on the graph. The slope of ℓ_{v} is a(v) = f(v)/v. This has a local extremum when a′(v) = (vf′(v) − f(v))/v^{2} = 0, that is, when f′(v) = f(v)/v = a(v). At that point ℓ_{v} is the tangent line to the graph at (v, f(v)).

(d) The bird should minimize a(v), since conserving energy with respect to distance rather velocity is more important.

5.5, #11. f(x) = e^{−x^{2}} is an even function, that is, f(−x) = f(x). So the best possible bounds on f(x) on [−0.3, 0.3] are the same as the best possible bounds on [0, 0.3]. As x increases from 0 to 0.3, x^{2} increases from 0 to 0.09, while f(x) decreases from e^{0} = 1 to e^{−0.09} = .914. Thus 0 ≤ f(x) ≤ .914.

5.5, #14. f(x) = x^{3} −4x^{2} +4x, f′(x) = 3x^{2} −8x+4, f′′(x) = 6x−8. The roots of f′(x) are x = 1/3 and x = 2. Since f′′(1/3) < 0 and f′′(2) > 0, it follows by the second-derivative test that x = 1/3 is a local maximum, x = 2 is a local minimum. The domain in the actual problem is restricted to 0 ≤ x ≤ 4. Both critical points are in this domain. We calculate f at the critical points and at the end-points of the domain:

f(0) = 0, f(1/3) = .926, f(2) = 0, f(4) = 16.

So the smallest and the largest value of f(x) on [0, 4] are 0 and 16, and thus 0 ≤ f(x) ≤ 16.