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MATH 263C CCE — SAMPLE EXAMINATION
You will gain the most benefit from this sample exam if you take it as if it were the supervised examination. Set yourself a time limit of 3 hours and do not use your textbook or notes. Practice writing out all the steps in your solutions to the problems, since you will have to show your work on the problems for the supervised examination.
- Determine if the following sequences converge or diverge, and if a sequence converges, find.
- Determine which of the following series are convergent.
(a) ;
(b).
- Determine if the series
is absolutely convergent, conditionally convergent, or divergent.
- (a) Prove that the series
converges absolutely.
(b) The series
is a geometric series. Find the interval of convergence and the sum of the series.
- Find the interval of convergence of the series
- Use the fact that has the power series representation
to find a power series representation of:
(a)
(b)
- Use the first two terms of an appropriate infinite series and approximate the integral.
Hint: Note the power series in powers of x for is
.
- (a) Write the first three non-zero terms in the Taylor series for.
(b) Write the first three terms in the Maclaurin series for.
- (a) Find the rectangular coordinate equation of the conic having parametric equations where the parameter t satisfies. (b) Find the equation of the parabola whose vertex is at the origin (0,0), and whose axis is the y - axis, if the parabola passes through the point. Then determine the coordinates of the focus of the parabola, and the equation of the directrix.
- For the hyperbola with equation
,
(a) Find the coordinates of the foci, and the equation of the asymptotes. (b) Sketch a rough graph of the hyperbola, including the asymptotes.
ANSWER KEY TO SAMPLE EXAMINATION
- (a).
In , note the terms are:
Clearly, the terms oscillate back and forth between positive and negative values; so the sequence diverges. Alternatively, note there are two sub-sequences which do converge. The sub-sequence converges to , and the sub-sequence
converges to 1. Since there are two infinite subsequences converging to different values, the given sequence diverges.
To handle , where n is a positive integer, define.
Note: letting , that. Therefore,.
Alternatively, you can use L’Hôpital’s Rule. Thus,
Hence,.
- (a) First, observe that the series
is a positive-term series. It is easy to see, intuitively, that this series converges, because for
large n , , and converges
( p -series, p =. To be more rigorous, which is the idea here, you can prove the given series converges using either the ordinary comparison test, or the limit comparison test.
Using the Ordinary Comparison test, observe that
for every positive integer n. Since converges (because does, as it is a p -
series with ), then the given series converges, by the Ordinary Comparison test.
To apply the Limit Comparison test in testing the given series ,
note that for large n , the n -th term is essentially like. Further,
Therefore, since converges, so does the given series , by the Limit Comparison test.
- (b) The given series
diverges, because.
- Applying the Absolute Ratio test to
, we get
Hence, the given series converges absolutely (and therefore converges) for .
Checking the endpoints , the given series is
which converges by the Alternating Series test. For , the given series is
which diverges (harmonic series). Therefore, the interval of convergence of the given series is the interval.
- (a)
Hence,
(b)
- First, in the power series
replace. This gives
and
with error from the true sum.
- Here,
Now,
The Taylor series for in powers of is
(b)
- (a) (b)
In (b), to change the equation to rectangular coordinates, observe
or. This is the rectangular coordinate equation of a circle with center at , and radius 1.
- (a) The graph of the lemniscate is roughly as sketched below.
Note: Since the equation isn’t changed when r is replaced by , the graph is symmetric with respect to the pole (the origin here). The area is therefore twice the area of the loop in the first quadrant, and is
- (b) The graphs of the circles are depicted below.
The circles intersect where
(i) The area inside the graphs of both circles is
(ii) The area outside the graph of and inside the graph of is
.
- Let denote the position vector from the origin to a general point on the curve. Here,
Clearly, the point corresponds to ; so you want to compute first.
Note ; so.
The desired unit tangent vector is.
- (a)
(b) Since the plane whose equation we seek is parallel to the plane containing the vectors , normal vectors to these two planes have the same direction. Now recall that is normal to the plane containing the vectors. Thus, as a normal vector to the plane containing the point ,we can take the vector computed in 16(a); equally well, we can use as the normal vector to this plane the vector
. The equation of the plane through is thus , or equivalently,.
