Math 114, Spring 1998

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SOLUTIONS TO THE FINAL EXAM

1.i) a_n=n!/(n^n). Ratio test converges to 1/e<1, so the series is convergent.

1.ii) a_n=sin(n)/n. Use Dirichlet's test; |sum(sin(n),n=1..N)| <= |cot(1/2)|+1/2, and 1/n decreases to 0. So sum(sin(n) *(1/n), n=0..infinity) converges.

1.iii) Observe that a_{2n}=(-1)^n(3^{2n}+3)/(4*(2n)!). Now use ratio test with |a_{2n+2}/a_{2n}|. Convergent.

1) In the ratio test you take the limit. Without the limit your conclusions will be wrong. For example for the harmonic series a_{n+1}/a_n=n/(n+1)<1 for all n but the series diverges.

2) A surprising number of students found a surprising number of different limits for sin(n)/n as n goes to infinity... (For the confused: this limit is zero.)

3) If you attempt to apply the root test for 1.i you should be prepared for a detailed analysis of the area under y=1/x curve in order to find the behaviour of (n-th root of n!), or you could just quote it. (It behaves like n/e for large n.)

4) Finally applying non-standard tests, i.e. criteria you invent on the spot, is not acceptable without proper evidence for their credibility.

2) Set z=(1/2)(3-x^2-y^2). Gradient is (x,y,1), its length is sqrt(1+x^2+y^2).

2.i) Integrate sqrt(1+x^2+y^2) on the circle 0<=x^2+y^2<=3. Changing to polar coordinates gives an easy integral which has the value 14Pi/3 as the area of the paraboloid x^2+y^2+2z=3 with z>=0.

2.ii) For the volume inside the above figure integrate z over the same circle as above to get 9Pi/4.

If you set up the integral correctly, with correct limits, but manage to get a wrong answer, then you get -1.
If the integrand is correct but the limits are wrong then you lose more points depending on how irrelevant your limits are.
Wrong integrand gets no points at all.

3) f(x,y)=x^2+xy+y^2-5x-4y where (x,y) is restricted to the triangle formed by the lines x=0, y=0, x+y=5.

f_x=2x+y-5, f_y=x+2y-4. Setting these equal to zero gives (x,y)=(2,1) which is inside the triangle.

Next check what happens on the boundary.
x=0: f=y^2-4y, critical point (0,2).
y=0: f=x^2-5x, critical point (5/2,0).
x+y=5: f=x^2-6x+5, critical point (3,2).

Adding to this list the corners as the boundaries of the boundaries gives the set of critical points as:
{(2,1), (0,2), (5/2,0), (3,2), (0,0), (0,5), (5,0)}

Evaluating f at these points gives the min and max:
f(2,1)=-7, <--- MINIMUM
f(0,2)=-4,
f(5/2,0)=-25/4,
f(3,2)=-4,
f(0,0)=0,
f(0,5)=5, <--- MAXIMUM
f(5,0)=0.

Many people solved this without a hitch. Most people however missed some of the critical points and/or got confused about where the max could be. They got between 10 and 20 depending on how close they are to reality. Some papers contained irrelevant discussions such as restricting the function to the boundaries and finding the roots; no mention of derivatives...

4) One way to do this is to observe that cos(x)^3=cos(x)cos(x)^2 =cos(x)(1/2 + (1/2)cos(2x)). Then use the expansion of cos(t) in this expression and simplify. Alternatively you can quote a `dersane' formula and write cos(x)^3= (1/4)(3cos(x)+cos(3x)) and continue from there. The answer is given in 1.iii.

There was a half page objection which was asking for a 1 point increase. No comment on this.

The integral test applies to series with nonnegative terms where the general term monotonically decreases to zero. In particular a_n=sin(n)/n is not such a term.

sin(361) is not sin(1). We are using radian measure.

To apply the alternating series test you should check first that the absolute value of the general term monotonically decreases to zero. That it is obvious from the expression should only make your answer so much easier but the answer must be on the paper.

And finally objections starting with "I only made the mistake of ..." actually confirm my opinion...

The grades given reflect only what is written on your answer sheets. They are not indications of any other value, personal or philosophical...

BELOW YOU WILL FIND THE LATEST GRADE STANDING.

Legend:
Q1, Q2, Q3 Quizzes
Q=10*{[Q1+Q2+Q3-min(Q1,Q2,Q3)]/2}
M1, M2 Midterms
F Final
D Difference between the next higher NG and yours
Math 114 1998 Spring

Q1 Q2 Q3 Q M1  M2    F   NG     LG  D
97004030 10 10 10 100 075 095 100 92.500 A
97004040 10 10 07 100 085 088 085 87.250 A   5.250
96024290 10 10 05 100 060 100 091 86.400 A   0.850
96024240 06 06 10 080 055 100 082 79.550 A   6.850
97028560 10 10 07 100 084 100 048 75.200 A-  4.350
97003920 10 10 05 100 050 100 065 73.500 A-  1.700
97003730 06 10 07 085 065 070 075 72.250 A-   1.250
96000380 06 07 07 070 060 100 063 72.200 A-   0.050
96000390 06 06 05 060 085 100 049 71.850 A-   0.350
96000350 00 00 00 000 070 090 075 70.000 B+  1.850
97003720 05 10 05 075 054 100 060 70.000 B+  0.000
96000410 07 00 05 060 065 100 056 69.650 B+  0.350
96024230 06 05 07 065 042 100 069 69.600 B+  0.050
97004070 10 10 00 100 040 100 060 69.000 B+  0.600
96024250 00 00 05 025 045 100 073 67.950 B+  1.050
96042300 10 05 07 085 045 100 057 67.550 B+  0.400
97004010 10 10 05 100 030 065 080 65.750 B    1.800
97003800 09 05 05 070 030 100 065 65.500 B    0.250
97003950 10 10 05 100 014 085 074 64.350 B    1.150
97028640 06 10 05 080 040 100 052 63.800 B    0.550
97028620 10 09 05 095 053 090 044 62.850 B    0.950
97003780 10 05 05 075 020 095 065 62.250 B    0.600
96024270 07 10 05 085 040 090 052 61.800 B    0.450
97004000 06 00 05 055 055 090 050 61.750 B    0.050
97004050 10 10 05 100 030 095 050 61.250 B    0.500
96000370 06 05 05 055 055 060 065 60.250 B    1.000
96000330 05 06 05 055 025 100 056 59.150 B-   1.100
97028580 06 09 05 075 025 085 060 59.000 B-   0.150
97004100 10 05 06 080 044 087 045 58.750 B-   0.250
96024340 05 00 05 050 023 075 072 58.300 B-   0.450
96024300 06 05 05 055 034 075 063 57.950 B-   0.350
97003820 06 05 07 065 027 090 054 57.350 B-   0.600
97028550 06 05 05 055 025 095 047 54.300 C+  3.050
97028590 05 06 05 055 035 057 063 53.700 C+  0.600
97003810 05 00 05 050 055 050 052 52.050 C+  1.650
97003600 05 10 05 075 040 065 042 50.550 C    1.500
97003970 10 09 05 095 038 070 035 50.500 C    0.050
96000100 05 06 00 055 028 095 034 49.850 C    0.650
97003750 05 00 05 050 005 095 047 48.800 C    1.050
96000120 05 00 05 050 045 060 037 46.050 C-   2.750
96000310 06 05 07 065 035 095 011 43.400 C-   2.650
97003840 06 00 00 030 039 080 025 42.750 C-   0.650
97003710 06 00 05 055 035 050 046 42.650 C-   0.100
97028660 05 00 05 050 035 065 029 41.600 C-   1.050
96000170 05 00 05 050 030 068 027 40.300 D+  1.300
97003870 07 00 05 060 035 048 031 39.150 D+  1.150
96000320 05 00 05 050 025 053 032 37.300 D+  1.850
95014830 00 00 00 000 015 085 025 35.000   2.300
97028570 05 05 00 050 033 030 034 34.350 D    0.650
95021230 05 00 00 025 070 030 017 34.300 D    0.050
97003760 05 00 05 050 005 065 024 32.100   2.200
96000290 00 00 00 000 025 063 021 30.400 D    1.700
Average                  064 041 081 052 58.338 2.61