Midterm 2 Review Solutions

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Math 2300 – Calculus II – University of Colorado

Fall 2010 – Review for Midterm II

1. Rotating the ellipse \displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 about the x-axis generates an ellipsoid. Compute its volume.

Solution: In this situation, the a and b would be numbers. But, in general, we have that

\displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \displaystyle=1
\displaystyle\frac{y^{2}}{b^{2}} \displaystyle=1-\frac{x^{2}}{a^{2}}
\displaystyle y^{2} \displaystyle=b^{2}-\frac{b^{2}}{a^{2}}x^{2}
\displaystyle y \displaystyle=\sqrt{b^{2}-\frac{b^{2}}{a^{2}}x^{2}}

This object is symmetric with respect to the x-axis and y-axis, meaning that we only need to find one of the limits of integration. Setting y=0 in the above equation implies

\sqrt{b^{2}-\frac{b^{2}}{a^{2}}x^{2}}=0\qquad\Leftrightarrow\qquad b^{2}-\frac{b^{2}}{a^{2}}x^{2}=0\qquad\Leftrightarrow\qquad b^{2}\left(1-\frac{x^{2}}{a^{2}}\right)=0.

When we create such objects, we never have b=0 and so we must have 1=x^{2}/a^{2} and a^{2}=x^{2}. So, the x-value where the ellipsoid has a zero y-value must be a. (This makes perfect sense from the definition, if you should know it or look it up.)

So, we consider slices in the ellipsoid revolved around the x-axis, where the radius is given as


and the width of a Riemann disk is dx. So, we find that the volume is

\displaystyle\int _{{-a}}^{a}\pi r^{2}\, dx \displaystyle=\int _{{-a}}^{a}\pi\left(b^{2}-\frac{b^{2}}{a^{2}}x^{2}\right)\, dx
\displaystyle=\pi\int _{0}^{a}\left(b^{2}-\frac{b^{2}}{a^{2}}x^{2}\right)\, dx
\displaystyle=\frac{2\pi ab^{2}}{3}

2. Consider the region \mathcal{R} enclosed between the curves y=a (with a>0) and y=x^{2}, situated to the right of the y-axis. How high does a have to be so that the volume of the solid obtained by revolving \mathcal{R} about the x-axis is V=16/3?

Solution: The volume in question is given by

\displaystyle\int _{0}^{{\sqrt{a}}}\pi a^{2}\, dx-\int _{0}^{{\sqrt{a}}}\pi x^{4}\, dx \displaystyle=\pi a^{2}\left[x\right]_{{x=0}}^{{x=\sqrt{a}}}-\pi\left[\frac{1}{5}x^{5}\right]_{{x=0}}^{{x=\sqrt{a}}}
\displaystyle=\pi a^{2}\sqrt{a}-\pi\frac{1}{5}a^{{5/2}}
\displaystyle=\pi a^{{5/2}}-\pi\frac{1}{5}a^{{5/2}}
\displaystyle=\frac{4}{5}\pi a^{{5/2}}

Solving a for when this quantity is 16/3 gives


3. Given a continuous function f, we say that a function g is the “arclength function” for f if, for all values of x, g(x) represents the arc length of the graph of f between 0 and x:

g(x)=\int _{0}^{x}\sqrt{1+\left[f^{{\prime}}(t)\right]^{2}}\, dt
  1. (a)

    Show that g has to satisfy the following properties:

    1. (i)


      Solution: This is so that the distance traveled along the function (i.e., the arc length) at the instant you start at x=0 is zero.

    2. (ii)

      g is increasing.

      Solution: As you travel along the function, the arc length must increase.

    3. (iii)

      g^{{\prime}}(x)\le 1 (Hint: Show that f(x)=\displaystyle\int _{0}^{x}\sqrt{\left[g^{{\prime}}(x)\right]^{2}-1}\, dt

      Solution: Using the fundamental theorem (twice) with the hint for f(x), we get

      \displaystyle\int _{0}^{x}\sqrt{1+\left[\frac{d}{dx}\int _{0}^{x}\sqrt{\left[g^{{\prime}}(x)\right]^{2}-1}\, dt\right]^{2}}\, dt \displaystyle=\int _{0}^{x}\sqrt{1+\sqrt{\left[g^{{\prime}}(x)\right]^{2}-1}^{2}}\, dt
      \displaystyle=\int _{0}^{x}\sqrt{1+\left[g^{{\prime}}(x)\right]^{2}-1}\, dt
      \displaystyle=\int _{0}^{x}\sqrt{\left[g^{{\prime}}(x)\right]^{2}}\, dt
      \displaystyle=\int _{0}^{x}g^{{\prime}}(x)\, dt

      and so f(x) does satisfy the given equality. Since the square root is only capable of being calculated when g^{{\prime}}(x)\ge 1, we see that part (iii) should have been worded better. Oops!

  2. (b)

    Find a function f whose arc length from 0 to x is 3x.

    Solution: This can be accomplished with f(x)=3x.

  3. (c)

    Find a function f whose arc length from 0 to x is 2x+1.

    Solution: This cannot work since the arc length at 0 is 1.

  4. (d)

    Find a function f whose arc length from 0 to x is x/2.

    Solution: This cannot work since the arc length is less than the distance traveled along the x-axis.

4. With x and b in meters, a chain hangs in the shape of the catenary \cosh x=\displaystyle\frac{1}{2}\left(e^{x}+e^{{-x}}\right) for -b\le x\le b. If the chain is 10m long, how far apart are the ends?

Solution: Here, the arc length is 10m. So, we have

\displaystyle 10 \displaystyle=\int _{{-b}}^{b}\sqrt{1+\left[\frac{d}{dx}\left(\frac{1}{2}\left(e^{x}+e^{{-x}}\right)\right)\right]^{2}}\, dx
\displaystyle=\int _{{-b}}^{b}\sqrt{1+\left[\frac{1}{2}\left(e^{x}-e^{{-x}}\right)\right]^{2}}\, dx
\displaystyle=\int _{{-b}}^{b}\sqrt{1+\sinh^{2}x}\, dx
\displaystyle=\int _{{-b}}^{b}\sqrt{\cosh^{2}(x)}\, dx
\displaystyle=\int _{{-b}}^{b}\cosh x\, dx

Thus, b\approx 2.31244 and the ends are twice as far as that apart.

5. and 6. These are nice problems, but we'll review something a bit more down-to-earth on Wednesday before the exam. Make sure to review area and arc length in polar coordinates.

7. I'm not entirely sure what this is asking. We'll also mention centers of mass during our review in class.

8. Show that the sequence defined by s_{n}=2n^{2}-n has s_{1}=1 and satisfies the recurrence relation s_{n}=s_{{n-1}}+4n-3 for n>1.

Solution: To verify the first question, just plug in n=1 and it works. The second question, showing the recurrence relation, isn't quite as easy. Notice that s_{n}=s_{{n-1}}+4n-3 is equivalent to s_{n}-s_{{n-1}}=4n-3. So, we'll use the first definition of s_{n} to calculate s_{n} and s_{{n-1}} and show that subtracting them actually gives us 4n-3.

\displaystyle s_{n}-s_{{n-1}} \displaystyle=(2n^{2}-n)-(2(n-1)^{2}-(n-1))

9. In each case below, define the discrete sequence corresponding to a function f:s_{n}=f(n) for n\ge 1, and for each decide whether this sequence converges. Justify your answer.

  1. (a)


    Solution:s_{n}=\sqrt{n}. This quantity keeps increasing and never settles down at a single value, and so the sequence does not converge.

  2. (b)

    f(x)=\sin(\pi x)

    Solution:s_{n}=\sin(\pi n)=0 and so the sequence converges to zero as all of the terms are zero.

  3. (c)


    Solution:s_{n}=\displaystyle\sin\left(\frac{\pi}{2}n\right). This one is a bit more tricky, since the sequence looks like


    In any case, it doesn't converge as it keeps cycling through these values.

  4. (d)


    Solution: We have s_{n}=\displaystyle\sin\left(\frac{\pi}{n}\right) and one should notice that as n gets larger, then the fraction inside the sin function will become smaller and smaller, approaching zero. In fact, this causes the sequence itself to converge to \sin(0)=0.

10. and 11. We'll consider these in class on Wednesday.

© 2011 Jason B. Hill. All Rights Reserved.