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MIDTERM EXAM II–Solutions Math 21D Temple-F
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Problem 1. (20pts) (a)Calculate the gradient ∇f (x, y, z) :
f (x, y, z) = sin(xy^2 z^3 )
Soln: ∇f = y^2 z^3 sin(xy^2 z^3 )i+2xyz^3 sin(xy^2 z^3 )j+3xy^2 z^2 sin(xy^2 z^3 )k.
(b) Calculate Curl F(x, y, z) =
i j k ∂ ∂x
∂ ∂y
∂ ∂z M N P
when
F(x, y, z) = xyz^3 i + xzj + z^10 k.
Soln:
CurlF =
i j k ∂ ∂x
∂ ∂y
∂ ∂z xyz^3 xz z^10
= i(0 − x) − j(0 − 3 xyz^2 ) + k(z − xz^3 )
(c) Calculate the divergence div F(x, y, z) = Mx + Ny + Pz :
F(x, y, z) = xyz^3 i + xzj + z^10 k
Soln: div F = yz^3 + 0 + 10z^9
(d) Label the arrows with div, ∇ and Curl and the (?)’s with correct dimensions of the space on which these operators act, ordered so that two in a row make zero.
R?^ → R?^ → R?^ → R?
Soln: R^1 → ∇ → R^3 → Curl → R^3 → div → R^1
Problem 2. (20pts) Evaluate the line integral ∫ C F·T^ ds,^ where C is the straight line from P = (− 1 , − 1 , 0) to Q = (1, 0 , 1), and F = xi + 3j − yk.
Soln: x(t) = P + t(Q − P ) = (−1 + 2t, −1 + t, t) so dx = 2 dt, dy = dt, dz = dt. Thus ∫
C F^ ·^ T^ ds^ =
∫ C xdx^ + 3dy^ −^ ydz^ =
∫ (^1) 0 (−1 + 2t)2dt^ + 3dt^ −^ (−1 +^ t)dt =
∫ (^1) 0 (2 + 3t)dt^ = 2t^ +
t^2 ]^10 = 2 +
Problem 3. (20pts) Let R denote the box 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 in the xy-plane, and let C denote the boundary of R oriented counterclockwise. Assume that a fluid of density δ = y kg/m^2 and velocity v = xi + yj m/s is flowing through the box R.
(a) Find the mass flux vector F = δv, and determine its dimen- sional units.
Soln: F = xyi + y^2 j has dimensions of (^) m skg.
(b) The line integral ∫ C F^ ·^ n^ ds^ gives the rate at which mass passes outward through C. Write this as the integral of a 1-form using the components of F. (Do not evaluate it.)
Soln: ∫ C F·n^ ds^ =^
∫ C (M, N^ )·(Ty,^ −Tx)^ ds^ =^
∫ ∫^ C^ (M, N^ )·(dy,^ −dx) = C M dy^ −^ N dx^ =^
∫ C xydy^ −^ y^2 dx
Problem 5. (20pts) Prove that if F = M (x, y)i + N (x, y)j = ∇f (x, y), then ∫ C F^ ·^ T^ ds^ =^ f^ (Q)^ −^ f^ (P^ ) for any smooth curve taking P to Q. (You may assume F is smooth everywhere.)
Soln: Let x(t) = (x(t), y(t)), a ≤ t ≤ b be any smooth pa- rameterization of C so that x(a) = P and x(b) = Q. Then we
can write: ∫ C F^ ·^ T^ ds^ =^
∫ (^) b a F^ ·^ v(t)^ dt^ =^
∫ (^) b a
{ ∂f ∂x
dx dt +^
∂f ∂y
dy dt
} dt = ∫ (^) b a d dt f^ (x(t), y(t))dt^ =^ f^ (x(b), y(b))−f^ (x(a), y(a)) =^ f^ (Q)−f^ (P^ ).
