Mathematics 134 – Intensive Calculus for Science 2 Solutions for Exam 3 April 27, 2006

I. A) Series A is not geometric because there is not a constant ratio between successive

terms. For instance \frac{1/2!}{1} = 1/2, but \frac{1/3!}{1/2!} = 2/6 = 1/3. Series B is geometric with a = 3/2, r = 1/2.

B) From the sum formula for infinite geometric series

3 = \frac{a}{1 − r}

= \frac{b/2}{1 − b/2}

⇒ 3 = \frac{b}{2 − b}
6 − 3b = b
6 = 4b
⇒ b = 6/4 = 3/2

II. A) For f(x) = cos(2x), by the chain rule:

f(0) = 1
f′(x) = −2 sin(2x) f′(0) = 0
f′′(x) = −4 cos(2x) f′′(0) = −4
f′′′(x) = +8 sin(2x) f′′′(0) = 0
f^{(4)}(x) = 16 cos(2x) f^{(4)}(0) = 16

So the Taylor polynomial is

1 + 0 · x + \frac{−4}{2!}x^{2} + \frac{0}{3!}x^{3} + \frac{16}{4!}x^{4} = 1 − 2x^{2} + \frac{2}{3}x^{4}

B) Using the general formula for (1+x)^{p}, the first three terms are all nonzero for p = 1/3:

1 + \frac{1}{3}x + \frac{(\frac{1}{3}) (\frac{−2}{3})}{2}x^{2} = 1 + \frac{1}{3}x − \frac{1}{9}x^{2}.

(You would get the same thing by computing the Taylor polynomial of degree 2 using the method of part A too.)

III.

A) It’s slope field 2 for the following reason. \frac{dy}{dx} = (y−1)(y−5) has zero slopes (equilibrium solutions) along the lines y = 1 and y = 5. Slope field 1 doesn’t do that (slope is not zero along y = 1). There are other ways to tell too, from the signs slopes at various y-values.

B) For slope field 1, the solution with y(0) = 3 increases up to a horizontal asymptote at y = 5. For slope field 2, the solution with y(0) = 3 decreases to a horizontal asymptote at y = 1.

C) Separating variables we have

∫ \frac{dy}{y} = ∫ \frac{dx}{x(x + 1)}

The integral on the right can be done with #26 from the table or by the partial fraction method:

\frac{1}{x(x + 1)} = \frac{1}{x} − \frac{1}{x + 1}

so

ln |y| = ln |x| − ln |x + 1| + c

Exponentiating both sides,

y = k (\frac{x}{x + 1})

where k = ±e^{c}.

IV. A) The given information says that if Q is the number of bacteria as a function of time,

then \frac{dQ}{dt} = kQ. This is exponential growth, so

Q(t) = Q(0)e^{kt} = 200e^{kt}

B) We know

360 = Q(.5) = 200e^{(.5)k} ⇒ k = \frac{1}{.5} ln(360/200) ≐ 1.1756

So then we want to solve

10000 = 200e^{(1.1756)t} ⇒ t = \frac{1}{1.1756} ln(10000/200) ≐ 3.33

(hours).