This is a solution of the second exam in MA114 Sections 001 to 006. Question statements may not be verbatim. Each problem is 10 points.

1. Complete the following statements. You do not need to show work for this problem.
   1. By definition `sum_(i=1)^oo a_i = lim_(n->oo)`

      `sum_(i=1)^oo a_i = lim_(n->oo) sum_(k=1)^n a_k`, i.e. the infinite sum is the limit of the partial sums.

   2. For which values of `r` does the series `sum_(i=0)^oo r^i` converge absolutely?

      It converges absolutes when `|r| < 1`.

   3. For which values of `p` does the series `sum_(i=0)^oo 1/(i^p)` converge absolutely?

      It converges absolutes when `p > 1`.

   4. The alternating series `sum_(n=1)^oo a_n` converges if `lim_(i->oo) |a_i| = 0` and

      The additional condition is that `|a_i| >= |a_(i+1)|` for all sufficiently large `i`.

   5. According to the ratio test, `sum_(n=1)^oo a_n` is absolutely convergent if

      The condition is that `lim_(n->oo) |(a_(n+1))/(a_n)| < 1`.

2. 1. The partial sum `S_8 = sum_(j=1)^8 a_j` of the series `sum_(j=1)^oo (-1)^j j is

      The sum partial sum is `-1+2-3+4-5+6-7+8 = 4`.

   2. Use the alternating series test to find an `n` such that `S_n = sum_(i=1)^n ((-1)^i)/i` is within 0.05 of `S = sum_(i=1)^oo ((-1)^i)/i`. Justify your answer.

      The alternating series test is applicable because the series is alternating, the absolute value of thee terms is decreasing and approaches 0 as `i` approaches `oo`. The test includes the truncation error estimate of `|a_(n+1)| = 1/(n+1)`. This is at most 0.05 when `n+1 > 1/0.05 = 20` and this is true when `n` is 19 or larger.

3. SHOW YOUR WORK!
   1. Find the radius of convergence of the power series `sum_(n=1)^oo (3x^n)/(n^4+9n^2)` and its interval of convergence.

      Use the ratio test:

      `1 > lim_(n->oo) |a_(n+1)/a_n| = lim_(n->oo) (3|x|^(n+1) (n^4 + 9n^2))/(3|x|^n ((n+1)^4 + 9(n+1))) = lim_(n->oo) (|x|(1+9/(n^2))/((1+1/n)^4 +9(1+1/n)^2)(1/n^2)) = |x|` and so the radius of convergence is 1 and the interval of convergence is `-1 <= x <= 1`. (The justification for the endpoints is given below.)

   2. Analyze the behavior at the left endpoint of the interval of convergence.

      The left endpoint is at `x = -1` and the series there is `S = sum_(n=0)^oo (3(-1)^n)/(n^4+9n^2)`. We will show that this is absolutely convergent by replacing the terms with their absolute values. One has for the new terms: `0 <= 3/(n^4+9n^2) = 3/(n^2)(1/(n^2+9n^2)) <= 3 (1/n^2)`. By the comparison test applied to 3 times the convergent p-series `sum_(n=1)^3 1/(n^2)`, it follows that the series is absolutely convergent at `x=-1`.

   3. Analyze the behavior at the right endpoint of the interval of convergence.

      The right endpoint is at `x=1`. In the last part, we already saw that the series converges.

4. SHOW YOUR WORK. A car travels counterclockwise around a circular track at a constant speed, circling the track six (6) times per hour. The track has a radius of 2 miles. If the track is assumed to be the graph of `x^2+y^2 = 4` and at time `t = 0` the car is at the point (0, 2).
   1. Find the position of the car at time t hours expressing your answer parametrically.

      The parametric equations for the circle are `x = 2 cos(theta), y = 2 sin(theta)`. The angle `theta` can be calculated as initial value plus angular velocity times `t`, i.e. `theta = pi/2 + (2pi)6t`. So the parametric equations are:

      `x = 2cos(pi/2 + 12pi t)` and `y = 2 sin(pi/2 + 12 pi t)`

   2. Find the position of the car at time `t = 1/4`.

      Substitute into the parametric equations to get `x = 2 cos(pi/2 + 12 pi (1/4)) = 2 cos((3pi)/2) = 0` and `y = 2 sin(pi/2 + 12 pi (1/4)) = 2 sin((3pi)/2) = -2`.

5. SHOW YOUR WORK!
   1. Find the radius of convergence of the power series `sum_(n=1)^oo (3(7-5x)^n)/n`.

      Using the ratio test, one has

      `1 > lim_(n->oo) |a_(n+1)/a_n| = lim_(n->oo) (3|7-5x|^(n+1) n)/(3|7-5x|^n (n+1)) = lim_(n->oo) 5|7/5-x|/(1+1/n) = 5|7/5-x|` and so the radius of convergence is 1/5 and the interval of convergence is at least `|7/5-x| < 1/5` or `6/5 < x < 8/5`. The interval contains only the right endpoint as is justified below. So the interval is actually `(6/5,8/5]`.

   2. Analyze the behavior at the left endpoint of the interval of convergence.

      The left endpoint is at `x = 6/5` where `7 - 5x = 1` The series is `sum_(n=1)^oo 3/n` which is three times the harmonic series which we know diverges.

   3. Analyze the behavior at the right endpoint of the interval of convergence.

      The right endpoint is at `x = 8/5` where `7-5x = -1`. The series is `sum_(n=1)^oo 3((-1)^n)/n`. This is an alternating series where the absolute values of the terms are clearly decreasing and have limit 0 as `n-> oo`. So the series converges.

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6. Use the integral test to determine whether the series `sum_(n=1)^oo (2n)/(n^2+1)` is convergent or divergent. You must use the integral test and must show that it applies.

   The terms are `a_n = f(n)` where `f(x) = (2n)/(n^2+1)`. The function f(x) is continuous and positive for `x >=1`. Furthermore,

   `f'(x) = ((x^2+1)2 - 2x(2x))/((x^2+1)^2) = (2-2x^2)/((x^2+1)^2) < 0` when `x > 1` and so `f(x)` is decreasing for `x >= 1`. So, we can apply the integral test. One has `int_1^oo f(x)dx = int_1^oo (2x)/(x^2+1)\ dx = {:ln(x^2+1) |_1^oo = oo` and so the series diverges.

7. Calculate the exact values of the following. SHOW YOUR WORK.
   1. `T = sum_(i=0)^oo (1/3)^i`

      This is a geometric series of the form `sum_(i=0)^oo r^i = 1/(1-r)` provided that `|r| < 1`. So `T = 1/(1-1/3) = 3/2`.

   2. `P = sum_(n=1)^oo 1/(n(n+1)) = sum_(n=1)^oo (1/n - 1/(n+1))`

      The partial sum of the first n terms is

      `S_n = (1-1/2)+(1/2-1/3)+(1/3-1/4)+ ... + (1/n - 1/(n+1)) = 1 - 1/(n+1)` because of sum telescopes and the sum of all the intermediate terms is zero. So the sum of the series is `lim_(n->oo) S_n = lim_(n->oo) 1 - 1/(n+1) = 1`.

8. SHOW YOUR WORK. A 10 foot ladder leans against the right wall of a narrow passageway with the bottom of the ladder touching the left wall as shown in the figure. The right wall is moving to the right (i.e. the passageway is getting wider) at a constant speed of 3 feet per minute. At time `t=0`, the passageway is 4 feet wide. Give parametric equations describing `P(t)` the point of contact of the ladder with the right wall for times between `t=0` and `t=3` minutes. The diagram represents the time `t=0`.

   Since the wall is moving at 3 feet per minute from an initial value of 4 feet, the abscissa of `P(t)` is `x = 4 + 3t` feet at time t minutes. The Pythagorean Theorem gives `x^2 + y^2 = 10^2` where `y >= 0` is the ordinate of `P(t)`. So, `y = sqrt(100 - (4+3t)^2)` feet at time t minutes.

9. SHOW YOUR WORK. The curve C is described parametrically by `x(t) = 7 sin(t)`, `y(t) = 3 + 4t+t^2`.
   1. Calculate a parametric form for the tangent line to C at the point corresponding to `t=1`.

      One has `x'(t) = 7 cos(t)` and `y'(t) = 4 + 2t`. So, the tangent line goes through the point `(x(1), y(1)) = (7sin(1), 8)` and its slope is `(y'(1))/(x'(1)) = 6/(7cos(1))`. The line can therefore be parameterized as `x(t) = x(1) + t x'(1) = 7 sin(1) + 7t cos(1)`, `y(t) = y(1) + t y'(1) = 8 + 6t`.

   2. What is `(dy)/(dx)` as a function of t?

      `(dy)/(dx) = (y'(t))/(x'(t)) = (4+2t)/(7 cos(t))`.

      What is the value of `(dy)/(dx)` at the point where `t=0`.

      Substitute `t=0` into our formula to get `4/(7 cos(0)) = 4/7`.

   3. Calculate `(d^2y)/(dx^2)` as a function of `t`. DO NOT SIMPLIFY YOUR ANSWER.

      Using the chain rule, one has

      `(d^2y)/(dx^2) = ((d((dy/dx)))/(dt))/((dx)/(dt)) = ((d((4+2t)/(7cos(t))))/(dt))/(7 cos(t)) = (7 cos(t) 2 - (4+2t)(-7sin(t)))/((7 cos(t))^3)`