Mathematics 36, section 1 -- AP Analysis

Practice Final Exam -- Solutions

December 12, 1997

A) In each case, the line is the tangent line to the graph at $x = 0$ -- the line with equation $y = f_i'(0)x + 1$. Computing the derivatives, you should get:
1. f_1: $y = 3x + 1$
2. f_2$: $y = 4x + 1$
3. $f_3$: $y = 2x + 1$.
B) For $y = f₁(x) + f₃(x)$, at $x = 0$, $y = 2$, and $y' = f₁'(0) + f₃'(0) = 5$. {\it Answer:} $y = 5x+2$.

II.

A) $f'$ is negative and decreasing asymptotically from $0$ until about $x = -4$, negative but increasing on $(-4,-2.1)$ (approximately), positive and increasing on $(-2.1,0)$, positive but decreasing on $(0,2.1)$, then negative and decreasing on $(-2.1,4)$, then negative but increasing asymptotically to 0 from $x = 4$ on.
B) $(0,0)$ must be a local minimum on $y = F(x)$ where $F' = f$. There are no other critical points for $F$.

III. The height function is increasing up to the time $t = a$ that the left hand portion of the "W" is filled up to the level of the middle segment. It is concave {\it down\/} since the walls of the container slope outward -- the water is being poured at a constant volume per unit time so as the water level rises, more water is needed to produce the same increase in level. From time $t = a$ to $t = 2a$, water will spill over into the right hand portion of the "W" and the water level on the left will not change. Then from that point the water level will be increasing and concave down as before.

IV.

A) Taylor polynomial:
p_3(x) = 1 + {1\over 3} x - {1\over 9} x^2 + {5\over 81} x^3 Approximation: $$(1.1)^{1/3} \doteq p_3(.1) = 1.03228395$$ ($1.0323$, rounded to 4 decimal places). \item{B)} From the error bound for Taylor polynomials, $$\eqalign{|\hbox{error}| &\le {\max_{0 \le z \le .1} f^{(4)}(z) \over 4!}(.1)^4\cr & = {1\over 24} \cdot \max_{0 \le z \le .1} {80\over 81 (1 + z)^{11/3}} (.1)^4\cr &= {1\over 24} {80 \over 81} (.1)^4 \hbox{ (max is at $z = 0$)}\cr &\doteq .00004\cr}$$ The actual error (comparing answer from A with calulator value) is approx. $$|1.032280115 - 1.03228395| = .00000383 = .383 \times 10^{-6}$$

V. \item{A)} Half of a circular ring with inner radius $a$ and outer radius $a + \Delta r$, with area approx. $\pi a \Delta r$. \item{B)} Using a right-hand Riemann sum with $\Delta r = 1$: $$\hbox{total pop.} \doteq \pi(70 + 61.5\cdot 2 + 58\cdot 3 + 51\cdot 4 + 45\cdot 5 + 29\cdot 6 + 25\cdot 7 + 19.5\cdot 8)$$ which is approx. $4.087 \times 10^6$. \bigskip \noindent VI. \item{A)} (Using the substitution $u = \sin 2x$) ${1\over 2} e^{\sin 2x} + C$. \item{B)} (Using parts, or number 27 in the table) ${x^8 \over 8}\ln 5x - {x^8\over 64} + C$. \item{C)} (Using number 24 in the table, with $a = 3$, $b = 4$, $c = 1$, $d = 3$: $$7\ln|x - 4| - 6\ln|x-3| \vert_5^6 \doteq 2.42$$ \bigskip \noindent VII. \item{A)} $\int_0^\pi \sqrt{x}\, dx = {2\over 3}\pi^{3/2}$. \item{B)} $M = \int_0^\pi x \sin(x/2)\, dx$. Using parts (or the table) $$\eqalign{\int_0^\pi x \sin(x/2)\, dx &= -2x\cos(x/2) + 4\sin(x/2)\vert_0^\pi\cr &= 4\cr}$$ \bigskip \noindent VIII. \item{A)} ${dT\over dt} = k(T - A)$. Separating variables and integrating: $$T = A + (T(0) - A)e^{kt}$$ \item{B)} From given information $T(0) = 0$, $A = 350$ (temp of oven). So $T = 350(1 - e^{kt})$. At $t = 10$, $T = 50$, so $$\eqalign{50 &= 350(1 - e^{10k})\cr 1/7 &= (1 - e^{10k})\cr e^{10k} &= 6/7\cr k &= \ln(6/7)/10 \doteq -.0154}$$ When $T = 180$, $$180 = 350(1 - e^{(-.0154)t})$$ Solving for $t$, $t \doteq 46.9$. So you'll need to wait another $36.9$ minutes (get a snack to tide you over!)