Solutions to Homework 2

11.3

#9 a)I - b)V - c)IV - d)II - e)III

#10 [3] a)IV - b)I - c)III

#13

a) The profiles look like shifted e^{−x^{2}} functions (that is like ’bumps’). For t = −1 the maximum point is to the left of the z-axis, for t = 0 on the z-axis, for t = 1 to the right of the z-axis, and for t = 2 even further to the right.

b) Increasing x (to the right).

c) Looks like a reflection of Fig.11.41 on the xz-plane. The top of the wave goes from front+left to back+right.

#18 [5]

a) (i) look like parabolas opening upwards with the vertex at (0, 1 − cos c). (ii) look like sine curves with period 2π, amplitude 1, and one of the minima at (0, c^{2}/2).

b) Fig. 11.42 has the E-axis pointing up, the x-axis to the right, and y-axis coming out of the picture. In 11.43 the E-axis is up, the y-axis to the right and the x-axis to the back.

11.4

#10 Ellipses centered at the origin, with major axis (x-axis) of length \sqrt{c}, and minor axis (y-axis) of length \sqrt{c/2}.

#14 Circles centered at the origin with radius \sqrt{− ln c}.

#27 a) π = 3, 000q_{1} + 12, 000q_{2} − 4, 000.

b) These are all decreasing lines in the first quadrant. There y- and x-intercepts are as follows. For π = 0: q_{1} = 1, 333 and q_{2} = 333. For π = 10, 000: q_{1} = 4, 666 and q_{2} = 1, 166. For π = 20, 000: q_{1} = 8, 000 and q_{2} = 2, 000. For π = 30, 000: q_{1} = 11, 333 and q_{2} = 2, 833. All slopes are −.25.

#34 [6] a)III)K) - b)V)G) - c)IV)J) - d)I)H) - e)II)F)

11.5

#4 [2] z = 2 − 2y.

#12

\begin{matrix}x\y & 3 & 4 & 5 \\ 1 & 5 & 6 & 7 \\ 2 & 8 & 9 & 10 \\ 3 & 11 & 12 & 13 \end{matrix}

#17

a) Set f(x, y) = k and solve for y to get y = (−m/n)x + (k − c)/n. These are lines with slope −m/n.

b) f(x + n, y − m) = c + m(x + n) + n(y − m) = c + mx + ny = f(x, y).

c) b) tells us that if we move n units to the right and m units down we stay on the same contour. This agrees with part a), since rise/run=−m/n.

#21 [4] The prices are $24 per CD and $16 per tape.