SOLUTION MTH 142 Practice Problems for Exam 2 - Spring 2000

Last changed: March 30 (a more complete answer No.15 was added since March 12)

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NOTE: The last line of answer 10 was modified. Also, please correct the following images:
  problems 4,5,6,7,8,9, the Riemann sums should begin at j = 0
  problem 12 b, change a0 = Pi/2

  1. a) Take $y=-x^2$ in $e^y = 1 + \frac{1}{1\!} y +\frac{1}{2\!} y^2 + \cdots$ to get

  2. \begin{displaymath}e^{-x^2} = 1 + \frac{1}{1\!} (-x^2) +\frac{1}{2\!} (-x^2)^2 + \cdots =1 - x^2 + \frac{1}{2} x^4 + \cdots\end{displaymath}

    This gives the answer in less time than actually calculating the coefficients one by one (which is ok too).

    b) Replace the function by the first few terms of the series to get \begin{displaymath}\lim_{x\rightarrow 0} \frac{-1+(1 - x^2 + \frac{1}{2} x^4)}{......} =\lim_{x\rightarrow 0} \frac{- 1 + \frac{1}{2} x^2}{1} =-1\end{displaymath}


  3. a) Direct calculation gives the polynomial

  4. \begin{displaymath}1 + 2 (x - \frac{\pi}{4}) +2 (x - \frac{\pi}{4})^2 + \frac{8}{3} (x - \frac{\pi}{4})^3\end{displaymath}

    b)

    From the plot above we see that the gap between f ( x ) and p ( x ) is the largest at the endpoint x = 1.2.
    Therefore the maximum error in the interval (0.7,1.2) is E = f ( 1.2 ) - P (1.2 ) = 0.20911


  5. a) The radius of convergence is given by

  6. \begin{displaymath}r = \lim_{n \rightarrow \infty} \frac{\vert a_n \vert}{\vert ......\lim_{n \rightarrow \infty} \frac{n}{(n+1) 3} =\frac{1}{3}\end{displaymath}

    b) The interval $(2 - \frac{1}{3},2+\frac{1}{3})=(\frac{5}{3},\frac{7}{3})$


  7. a) By taking sections perpendicular to the axis of rotation, we get ``washers''. At the tickmark $x_j$ the washer has inner radius $r_j = 2 x^2_j$, outer radius $R_j =1$, and thickness $\Delta x$.


  8. The Riemann sum that approximates the volumen is
    \begin{displaymath}V \approx \sum_{j=0}^n (\pi R^2_j - \pi r^2_j ) \Delta x =\sum_{j=0}^n (\pi 1^2 - \pi (2 x_j^2)^2 ) \Delta x\end{displaymath}

    The volume is obtained by taking limit as $\Delta x \rightarrow 0$. We have,
    \begin{displaymath}Vol(S) = \int_0^{\sqrt{2}/2} (\pi - \pi 4 x^4 ) dx= \frac{2 \sqrt{2} \pi}{5} \approx 1.777153\end{displaymath}


  9. a) By taking sections perpendicular to the axis of rotation, we get ``disks''. At the tickmark $y_j$ the radius $r_j = x_j = \sqrt{y_j /2}$ and thickness $\Delta y$. The Riemann sum that approximates the volumen is

  10. \begin{displaymath}\sum_{j=0}^n \pi R^2_j \Delta y =\sum_{j=0}^n \pi (\sqrt{y_j /2})^2\Delta y=\sum_{j=0}^n \pi y_j /2 \Delta y\end{displaymath}

    b) The volume is obtained by taking limit as $\Delta y \rightarrow 0$. We have,
    \begin{displaymath}Vol(S) = \int_0^{1} \frac{\pi}{2} y dy = \frac{\pi}{4}\end{displaymath}



     
     
  11. a)

  12. \begin{displaymath}\sum_{j=0}^n \frac{0.004}{1+r^2_j} 2 \pi r \Delta r \end{displaymath}
     
     

    b)
    \begin{displaymath}\int_0^{7000} \frac{(0.004) 2 \pi r}{1+r^2} \Delta r \end{displaymath}


  13. a)

  14. \begin{displaymath}\sum_{j=0}^n (2 + 0.015 x) \Delta x \end{displaymath}
     
     

    b)
    \begin{displaymath}\int_0^1 (2 + 0.015 x) dx = 2.0075 \end{displaymath}
     
     

    c)
    \begin{displaymath}\overline{x} =\frac{\int_0^1 x \, (2 + 0.015 x) dx }{\int_0^1 (2 + 0.015 x) dx }= \frac{1.0050}{2.0075} = 0.50062266\end{displaymath}


  15. A cross-section of the cone (shown in the figure) is bounded by the lines $y=\pm 2 x$ and $y=20$.


  16. Introduce tick marks in the $y$-axis. The slab $S_j$ at height $y_j$ is a disk with radius $R_j = x_j = y_j /2$ and thickness $\Delta y$, so its volume is $ \pi (y_j /2)^2 \Delta y$, and its weight is $62.4 \, \pi (y_j /2)^2 \Delta y$. The work involved in raising the slab a distance of $(20- y_j)$ to the top of the cone is
    \begin{displaymath}w_j = (20- y_j)\, 62.4 \, \pi (y_j /2)^2 \Delta y\end{displaymath}

    The total work is approximated by
    \begin{displaymath}W \approx \sum_{j=0}^n(20- y_j) \, 62.4 \, \pi (y_j /2)^2 \Delta y\end{displaymath}

    The exact work is given by
    \begin{displaymath}\int_0^{20} (20- y)\, 62.4 \, \pi (y /2)^2 \Delta y= 653451.2720\end{displaymath}


  17. a) A sketch of the dam is shown in the figure.


  18. Note that the equation of the right hand, non-horizontal side is $y = x - 30$. Introduce tick marks $y_0, y_1, \ldots, y_n$, in the y axis. At height $y_j$, the slab has area $(y_j +30) \Delta y$, and the pressure at this height is $62.4 (30-y_j )$. Therefore the force on the slab is
    \begin{displaymath}F_j = 62.4 (30-y_j ) (y_j +30) \Delta y\end{displaymath}

    The total force is approximated by
    \begin{displaymath}F \approx \sum_{j=0}^n 62.4 (y_j +30) (30 - y_j) \Delta y \end{displaymath}
    The exact value of the total force is obtained by taking the limit of the sum as $\Delta y \rightarrow 0$:
    \begin{displaymath}F = \int_0^{30} 62.4 (y_j +30) (30 - y_j) dy= 1,123,200\end{displaymath}


  19. The following table is helpful:
  20. day amount before 24hrs are up
    Jan 2 $ (0.15)\ 2$
    Jan 3 $(0.15)\ 2 + (0.15)^2\ 2$
    Jan 4 $(0.15)\ 2 + (0.15)^2 \ 2 + (0.15)^3 \ 2 $
    $\vdots$ $\vdots$
    Jan 20 $(0.15) \ 2 + (0.15)^2 \ 2 +\cdots + (0.15)^{19}\ 2$

    Then, the amount right before noon on January 20th is
    \begin{displaymath}(0.15) \ 2 + (0.15)^2 \ 2 +\cdots + (0.15)^{19}\ 2= \frac{ (0.15) 2 ( 1 - (0.15)^{20})}{1 - 0.15} \approx 0.352941176\end{displaymath}

  21. a) The Taylor polynomial of degree 2 of f(x) is

  22. $P(x) = \frac{1}{2} - \frac{1}{4} x + \frac{1}{8} x^2$
    so the absolute value of the error in approximating f(0.5) by P(0.5) is
    \begin{displaymath}\vert \ f(0.5) - P(0.5)\ \vert = \vert \ 0.4 - 0.40625 \ \vert = 0.00625 \end{displaymath}

    b) The function
    $\vert f^{(3)} (t)\vert = \vert-6/(t+2)^4 \vert$
    is decreasing on the interval (0,0.5), so it attains its maximum in that interval at t=0. The maximum is
    $M = 6/2^4 = 3/8$
    The error bound when approximating f(0.5) by P(0.5) is
    \begin{displaymath}\vert f(0.5 - p(0.5) \vert \leq M \frac{(0.5)^3}{3\! } = \frac{1}{128} \approx 0.00781\end{displaymath}


  23. a)


  24. b) we have that
    \begin{displaymath}\begin{array}{rcl}a_0 = & \frac{1}{2 \pi} \int_{-\pi}^\pi \......-\pi}^\pi \vert x \vert \sin (2 x ) dx & = 0 \\ \\\end{array}\end{displaymath}
    For details, we calculate $a_1$ step by step below.
    \begin{displaymath}\begin{array}{rl}a_1 = & \frac{1}{ \pi} \int_{-\pi}^\pi \ve......ht)\\ \\= & \frac{1}{\pi} \left(\ 2 + 2 \ \right)\end{array}\end{displaymath}
    Therefore, the Fourier polynomial we seek is $ F_2 (x ) = \frac{\pi}{2} - \frac{4}{\pi} \cos(x) $.

  25. Solution 1: Note that $\frac{n^2}{1+n^3}$ behaves like $\frac{1}{n}$ when $n$ is large, so we suspect that the series diverges. The following inequalities are clearly valid:

  26. \begin{displaymath}0 \leq \frac{n^2}{n^3 + n^3} \leq \frac{n^2}{1+n^3}, \quad n=1,2,\ldots \end{displaymath}
    The term in the center simplifies to $\frac{1}{2 n}$. Since $\sum_{n=1}^\infty \frac{1}{2n}$ diverges, so does $\sum_{n=1}^\infty \frac{n^2}{1+n^3}$. This is our original series with only the first term changed, so our original series $\sum_{n=0}^\infty \frac{n^2}{1+n^3}$ also diverges.

    Solution 2: We use the integral test. Set $f(x) = \frac{x^2}{1+x^3}$ for $x>1$. Now $\int_1^\infty f(x) dx = \lim_{b\rightarrow \infty } \int_1^b\frac{x^2}{1+x^3}......rrow \infty } \frac{1}{3} \ln \vert {1+ b^3} \vert - \frac{1}{3} \ln 2 = \infty$ Therefore, the series diverges too.


  27. Solution 1: Note that $\lim_{n\rightarrow \infty} (1+\frac{1}{2^n} = 1\not = 0$. Therefore the series diverges.

  28. Solution 2: Using comparison, we have
    \begin{displaymath}0 \leq 1 \leq 1+\frac{1}{2^n}, \quad n=0,1,\ldots\end{displaymath}

    Since $\sum_{n=0}^\infty 1$ is divergent, so is our original series.


  29. Solution 1: \begin{displaymath}\sum_{n=0}^\infty \frac{(-2)^{n+1}}{\pi^n} =
\sum_{n=0}^\inf...
...geometric series,since } \vert x\vert = \vert-2/\pi \vert < 1
\end{displaymath}

    Solution 2: Since $ \sum_{n=0}^\infty \frac{(-2)^{n+1}}{\pi^n} =
\sum_{n=0}^\infty (-1)^{n+1} \frac{(2)^{n+1}}{\pi^n} $ the series is of the alternating (sign) type. To apply Leibnitz test we must first check two things. The first is,
    \begin{displaymath}
\frac{2^1}{\pi^0} \geq \frac{2^2}{\pi^1} \geq \frac{2^3}{\pi^2} \geq
\cdots \end{displaymath}

    and this is clearly true as each new term on the right hand side is obtained from the previous one upon multiplication by $2/\pi$ which is a number less than 1. Also, we must check that
    \begin{displaymath}
\lim_{n\rightarrow \infty} \frac{(2)^{n+1}}{\pi^n} = 0
\end{displaymath}

    One can see that the above statement is true once the limit is rewritten in this form:
    \begin{displaymath}
\lim_{n\rightarrow \infty} 2 \left(\frac{2}{\pi}\right)^n = 0
\end{displaymath}

    (note that $2/\pi$ is less than 1, so raising this to a large power produces a small number). Because the two conditions to apply the test are satisfied, we have that, by Leibnitz test, the series is convergent.



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