CALCULUS III MATH. 2934, FALL 2021
                                   MW1:30-2:45 PM, F 1:30-2:20 PM, 115 PHSC
                                         HOMEWORK, EXAMS, SOLUTIONS
                                                       TOMASZPRZEBINDA
                                                            Contents
                   1.   Some Linear Algebra                                                                        1
                   2.   Motion along a curve                                                                       3
                   3.   Functions of several variables                                                             7
                   4.   Multiple integrals.                                                                       13
                   5.   Exam 1 (10/13/2021 in class), study guide                                                 16
                   6.   Exam 1 (10/13/2021 in class)                                                              17
                   7.   Vector calculus.                                                                          19
                   8.   Exam 2 (12/01/2021 in class), study guide                                                 29
                   9.   Exam 2 (12/01/2021 in class)                                                              30
                   10.   Exam 2’ (12/06/2021 in class)                                                            32
                   11.   Final Exam study guide, (Friday, 12/17/2021, 8:00-10:00 AM, in the usual
                         classroom 115 PHSC)                                                                      34
                                                   1. Some Linear Algebra
                Here we follow sections 11.1, 11.2 in the text. The following homework is due on 9/1/201.
                Problem 1. Exercise 6, page 405 in the text.
                There are many such examples. Here is one:
                                                      (0,0,1),     (1,−1,0).
                Both are perpendicular to (1,1,0) and to each other.
                Problem 2. Exercise 11, page 405 in the text.
                Statement 1 is false. Vectors
                                        u=(cos(π),sin(π)),         v = (cos(π),−sin(π))
                                                   3       3                  3          3
                                                                  1
            2                            TOMASZPRZEBINDA
            make the angle 30 degrees with the vector w = (1,0), but the sum of them is equal to
                                        u+v=(2cos(π),0),
                                                     3
            which makes angle zero with w = (1,0).
              Statement 2 is true. Two nonzero vectors make an angle of ninety degrees means that
            their dot product is zero. Thus the assumption is that u · w = 0 and v · w = 0. Hence
                                    (u+v)·w=u·w+v·w=0.
              Statement 3 is false. Let u = (1,0,0), v = (1,1,0) and w = (0,0,1). Then u · w = 0,
            v · w = 0 but u · v = 1 6= 0.
            Problem 3. Exercise 43, page 406 in the text.
            Statement (a) is false. Let W = aV for some number a Then
                                         2          2        2  2
                                  |V +W| =|V +aV| =(1+a) |V|
            and
                                        2      2       2   2
                                      |V | +|W| = (1+a )|V| .
            Since, for a > 0,
                                          2           2      2
                                    (1 +a) = 1+2a+a >1+a ,
            we see that in our example for V 6= 0,
                                              2     2     2
                                       |V +W| >|V| +|W| .
              Statement (b) is true. Since we are talking about a real triangle, we have V 6= 0 and
            W6=0. The equality
                                        |V +W|=|V|+|W|
            is equivalent to
                                              2            2
                                      |V +W| =(|V|+|W|) ,
            which means that
                                                     2                2
                           V · V +2V·W +W ·W =|V| +2|V|·|W|+|W| .
            This is equivalent to
                                          V·W =|V|·|W|,
            which means that the cosine of the angle between V and W is 1. Hence W is a multiple
            of V . Therefore they can’t form a real triangle.
              Statement (c) is obviously true.
              Statement (d) is false. The vectors perpendicular to (1,1,1) form a plane, hence they
            can’t be on one line.
                              MATH. 2934, FALL 2021. HOMEWORK, EXAMS, SOLUTIONS.               3
              Problem 4. Exercise 10, page 414 in the text.
              The equation is
                                      0 = (1,1,1)·(x−x ,y −y ,z −z ).
                                                        0     0      0
              Equivalently
                                           x+y+z=x +y +z .
                                                        0   0    0
              Problem 5. Exercise 36, page 415 in the text.
              The vectors orthogonal to these two panes are (1,1,5) and (3,2,1). They are not parallel
              to each other. Hence the planes have a non-empty intersection. Therefore the distance
              between them is zero. In fact the point
                                                  (0,−2, 13)
                                                      9 9
              is in the intersection of the two planes.
              Problem 6. Get familiar with wolframalpha.com in order to be able to check your answers
              and do complicated computations.
                                          2. Motion along a curve
              Here we follow sections 12.1, 12.2 and 12.4 in the text. The following homework is due
              on 9/8/201.
              Problem 7. Exercise 16, page 457 in the text.
              The position of the projectile is
                                                       1 2
                                   R(t) = (v0cos(α)t,−2gt +v0sin(α)t,0).
              Hence, the velocity
                                 v(t) = R′(t) = (−v0cos(α),−gt+v0sin(α),0).
              and the speed              p
                                  |v(t)| = (−v cos(α))2 +(−gt+v sin(α))2.
                                               0                  0
              Clearly,
                                              |v(t)| ≥ v0 cos(α).
              Hence the minimum is v cos(α).
                                    0
                The t varies between 0 and the time when the projectile hits the ground, i.e. when
                                              1 2
                                            − gt +v sin(α)t = 0,
                                              2      0
              which happens for
                                                t = 2v0sin(α) .
                                                       g
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                Wecompute                          p
                                         |v(0)| =    (v0 cos(α))2 + (−v0sin(α))2 = v0
                and
                                                      |v(2v0sin(α))| = v .
                                                              g            0
                As we checked in class, the trajectory is a parabola. Hence the maximum is v0.
                Problem 8. Exercise 20, page 452 in the text.
                By assumption
                                                 0 = (x(t),y(t)) · (x′(t),y′(t)).
                Explicitly,
                                                          ′           ′
                                                    x(t)x (t) + y(t)y (t) = 0.
                Hence, by the chain rule,
                                                      d      2        2
                                                      dt(x(t) +y(t) ) = 0.
                Therefore the function
                                                           x(t)2 + y(t)2
                is constant, i.e.
                                                           2        2     2
                                                       x(t) +y(t) = R ,
                where R2 does not depend on t.
                Problem 9. Exercise 24, page 452 in the text.
                x(t) = 1 +2cos(t), y(t) = 3+2sin(t).
                                   2                     2
                Problem 10. Exercise 26, page 452 in the text.
                Since                                         Z
                                                                 t
                                                       s(t) = t |v(u)|du,
                                                                0
                the Fundamental Theorem of Calculus implies that
                                                           ds = |v(t)|.
                                                           dt
                Hence
                                                          2
                                                         d s = d |v(t)|.
                                                            2
                                                         dt      dt
                On the other hand, the acceleration
                                                           a(t) = v′(t).
                Thus we are interested in a possible equality
                                                        |v′(t)| = d |v(t)|.                                    (1)
                                                                  dt