3.1: 2) it's the graph of y = 1/x
4.) The graph looks like a bowtie
8.) x(t) = (5cos(t), 3sin(t)), v(t) = x'(t) = (-5sin(t), 3cos(t), speed = (9 + 16sin^2 t)^{1/2}, a(t) = x''(t) =
(-5cos(t),
-3sin(t))
= -x(t)
10.) x(t) = (e^t, e^{2t}, 2e^t), v(t) = x'(t) = (e^t, 2e^{2t}, 2e^t), speed = 5e^{2t} + 4e^{4t})^{1/2}
a(t) = x''(t) = ( e^t, 4e^{2t}, 2e^t)
18.) f(t) = (cos(e), 2, 1) + (-e * sin(e), -2, 1)(t-1) = (cos(e) + e * sin(e) - (e * sin(e))t, 4 -2t, t
)
22.) v_0 = 32[55(3)^{1/2}]^{1/2} ~ 312.329
Recommended HW:
p. 72 true/false:
14.) F
18.) T
20.) T
p. 170 true/false:
2.) F, domainof f = {(x, y) | x does not equal to -y, y not equal to 0}
4.) F
6.) F
12.) F
16.) F, Df is a 4 x 3 matrix.
18.) F
2.8:
2a.) domain = R^3, range = {(3a, -2a, a) | a any real numberber}
2b.) [(3x - 2y + z)/14](3, -2, 1)
20.) (t, 2t - 1, -10t + 192/24)
22.)
The surface S_1 is equivalent to the level surface of $F_1(x, y, z) = xy - z.
The surface S_2 is equivalent to the level surface of $F_1(x, y, z) = s + y^2 - (3/4)x^2.
The gradients are normal to the tangent planes
grad F_1(2, 1, 2) = (1, 2, -1), grad F_2(2, 1, 2) = (-3, 2, 1).
There dot product is 0, so the two surfaces intersect perpendicularly at (2, 1, 2).
34.) T(x, y, z) = 10(xe^{-y^2} + ze^{-x^2}), grad T = 10(e^{-y^2} + -2xze^{-x^2}, -2xy e^{-y^2}, e^{-x^2})
a.) v = (2, 3, 1) - (0, 0, 1) = (2, 3, 0)
D_v(T)(2, 3, 1) = 10(e^{-9} + -4e^{-4}, -12 e^{-9}, e^{-4}) * (2, 3, 0)
= 10(2e^{-9} -8e^{-4} - 36 e^{-9})
= 10( -8e^{-4} - 34 e^{-9})
= -80e^{-4} - 340 e^{-9} deg/cm (assuming the point (x, y) is x cm north and y cm east of the origin)
b.) grad T = 10(e^{-y^2} + -2xze^{-x^2}, -2xy e^{-y^2}, e^{-x^2})
at (2, 3, 1): (10e^{-9} -40e^{-4}, -120 e^{-9}, 10e^{-4})
rate of change = 10[145e^{-18} - 8e^{-13} + 17 e^-8}]^{1/2} deg/cm
c.)
rate of change of T in max direction = 10[145e^{-18} - 8e^{-13} + 17 e^-8}]^{1/2} deg/cm ( 3cm/sec)
= 30[145e^{-18} - 8e^{-13} + 17 e^-8}]^{1/2} deg/sec