Mathematics 126 – Calculus for Social Science 2 Solutions – Makeup Exam 3 December 4, 2001

I. Let Q be the inside temperature of the house as a function of time. The statement of the problem gives that

\frac{dQ}{dt} = k(Q − 10)

for some constant k. This is a separable differential equation. To solve it, we can proceed as follows.

\frac{dP}{dt} = k(Q − 10)

Then divide by Q − 10 and multiply by dt to separate. Integrating, then exponentiating both sides, we have:

\begin{aligned}∫ \frac{dQ}{Q − 10} & = ∫ k dt \\ ln |Q − 10| & = kt + c \\ Q − 10 & = be^{kt} (where b = ±e^{c}) \\ P & = 10 + be^{kt} \end{aligned}

When t = 0 (10 am), Q = 68, so b = 68 − 10 = 58. When t = 2 (noon), Q = 58, so 58 = 10 + 58e^{2k}, or k = \frac{1}{2} ln(48/58) ≐ −.0946. Then the temperature at 3pm (t = 5) is

Q(5) = 10 + 58e^{(−.0946)(3)} ≐ 46.1

(Brr!! Where is that furnace repair guy?!!)

II. A) The probability that t is in the range 0 ≤ t ≤ 1 is

P (0 ≤ t ≤ 1) = ∫_{0}^{1} \frac{9}{(9 + t^{2})^{3/2}} dt

Because we do not have du = 2t in the numerator for u = 9 + t^{2}, this is not of the form ∫ u^{−3/2} du. But we can use entry #12 from the table of integrals with a = 3:

∫_{0}^{1} \frac{9}{(9 + t^{2})^{3/2}} dt = \frac{9}{9} · \frac{t}{(9 + t^{2})^{1/2}} |_{0}^{1}

= \frac{1}{\sqrt{10}} − 0

≐ .3162 B) The expected value is

\begin{aligned}E(t) & = ∫_{0}^{∞} tf(t) dt \\ & = ∫_{0}^{∞} \frac{9t}{(9 + t^{2})^{3/2}} dt \end{aligned}

With the factor of t from the expected value formula, this is of the form ∫ u^{−3/2} du. We have

∫ \frac{9t}{(9 + t^{2})^{3/2}} dt = \frac{−9}{(9 + t^{2})^{1/2}}

So

∫_{0}^{∞} \frac{9t}{(9 + t^{2})^{3/2}} dt = lim_{b→∞} \frac{−9}{(9 + t^{2})^{1/2}} |_{0}^{b} = 3

(planned obsolescence!) C) The variance is

Var(t) ∫_{0}^{∞}(t − 3)^{2} \frac{9}{(9 + t^{2})^{3/2}} dt

III. The conversion from the car speed to the standard normal is z = \frac{x−36}{7} A) The probability that a randomly selected car has speed x ≤ 35 is:

P (x ≤ 35) = P (z ≤ \frac{35 − 36}{7})

= P (z ≤ −.14)

≐ .4443 (from the table)

B) The probability that a randomly selected car has speed 40 ≤ x ≤ 45 is:

P (40 ≤ x ≤ 45) = P (.57 ≤ z ≤ 1.29)

= P (z ≤ 1.29) − P (z ≤ .57) ≐ .9015 − .7157 (from the table) = .1840

IV.

A) Using the quotient rule:

f_{x} = \frac{1 − x^{2} + y^{2}}{(1 + x^{2} + y^{2})^{2}}
f_{y} = \frac{−2xy}{(1 + x^{2} + y^{2})^{2}}
f_{xx} = \frac{(1 + x^{2} + y^{2})^{2}(−2x) − (1 − x^{2} + y^{2}) · 2(1 + x^{2} + y^{2})(2x)}{(1 + x^{2} + y^{2})^{4}}

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

f_{xy} = \frac{(2y)(3x^{2} − y^{2} − 1)}{(1 + x^{2} + y^{2})^{3}}
f_{yy} = \frac{(2x)(3y^{2} − x^{2} − 1)}{(1 + x^{2} + y^{2})^{3}}

B) The critical points occur where f_{x} = f_{y} = 0. Look at f_{y} first. This is zero only when the numerator is zero, so either x = 0 or y = 0. If x = 0, then f_{x}(0, y) = \frac{1}{1+y^{2}} ≠ 0,

so there are no critical points with x = 0. But if y = 0, then f_{x}(x, 0) = \frac{1−x^{2}}{(1+x^{2})^{2}} = 0 when x = ±1. The two critical points are (1, 0) and (−1, 0).

point H(x, y) f_{xx}  type
(1, 0)  1/4 −1/2 local max.
(−1, 0) 1/4  1/2 local min.

(For example,

H(1, 0) = \frac{(2)(1 − 3)}{8} \frac{(2)(−2)}{8} − 0^{2} = 1/4)