(a) If
(b) Let
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- 1. Let F(x, y, z) be the vector field (x²(1 + y²z²), Q(x, y, z), ey cos(22)), where is a differentiable function of x, y, and z. Suppose div F = 2x(1 + y2² 2²) — 2ery sin(22) + z tan(2y+z) + 2z sec² (2y + 2); and - • curl F = (xey cos(22)-y tan(2y+z)-y sec² (2y+z), 2x²y²z-yery cos(22), -2x²yz²). Determine div(Q).arrow_forwardThe vector v = <a, 1, -1>, is tangent to the surface x2 + 2y3 - 3z2 = 3 at the point (2, 1, 1). Find a.arrow_forwardWhich of the following is the equation of the tangent plane to the surface z=x2 - x+3y² +y at the point P(1, 1,4) ? (a) x+7y =6 (b) x+7y-z=4 (c) 3x+5y=8 (d) 3x+5y-z= 4 (e) x+7y+z=12arrow_forward
- 1. Let C₁ be the directed line segment that starts at (2,0) and ends at (-3,-5) and C₂ be the parabola with parametrization x = t, y = 4 - t² for t € [−3,2]. Let F be the vector field F(x, y) = (x² – 4y) î+ (2x + 3y²) î. F.dr. (a) Evaluate Sc₁ (b) Evaluate fF.dr where C is the positively oriented path consisting of the parabola y = 4-x² starting from (2,0) to (−3,−5) and the line segment from (-3, −5) back to (2,0). (c) Use the results in (a) and (b) to evaluate ſc₂ F · dr.arrow_forward2. Let U be a vector function of position in R³ with continuous second partial deriva- tives. We write (U₁, U2, U3) for the components of U. (a) Show that V (U • U) – Ü × (▼ × Ü) = (U · ▼) Ū, (V where ((UV) U), = U₁ (b) We define the vector function of position, by setting = V × . If the condition V U = 0 holds true, show that ▼ × ((Ū · V) Ű) = (Ũ · V) Ñ - (Ñ· ▼) ū. (Hint: The representation of (UV) U from the first part might be useful.)arrow_forward6. Denote by V the vector ·(). Then we can define the operations grad (f) = f (gradient of f), div (F) = V F (divergence of F), and curl (F) = ▼ × F (curl of F). (a) Given F = (2x + 3y, 3x + z, sin x), calculate div(F) and curl(F). (b) Is the vector field F = (x + y,3y, z-x+1) conservative? What about F = (yz, xz, xy)? (c) Calculate V × (Vf) =curl(grad f) and V. (V × F) = div(curl F).arrow_forward
- Let 2 be a domain bounded by a closed smooth surface E, and f(r,y, 2) be a scalar function defined on 2 UE. Assume that f has continuous second order partial derivatives and satisfies the Laplace equation on NUE; that is, on NUE. = 0 (i) Let n be the unit normal vector of E, pointing outward. Show that I| Daf ds = 0. dS = 0. (ii) Let (ro, Y0, 20) be an interior point in 2. Show that 1 f(ro, Yo, 20) cas(r, n) |r|2 ds, 47 where r = (x – ro,y – Yo, 2 – z0) and (r, n) is the angle between the vectors r and n.arrow_forward20) For each vector field F, evaluate the circulation along the curve C consisting of the portion of y sin x from (0,0) to (7,0) followed by the line segment from (7, 0) to (0,0). = (a) F₁(x, y) = (x + ye*, ex - – 3y) (b) F2(x,y) = (2xy + cos x, x²y — sin y)arrow_forwardThe theory of complex variable comes in naturally in the study of fluid phenomena.Letus define, w = φ + iψ with φ velocity potential and ψ stream function of a two dimensional fluid flow. w is called the complex potential function.Then w turns out to be analytic because the Cauchy-Riemann conditions, ∂φ/∂x =∂ψ/∂y ; ∂φ/∂y = −∂ψ/∂x are exactly the natural flow conditions that have to be satisfied.Suppose, velocity potential for a two dimensional fluid flow is given by the functionφ(x, y) = y/x2+ y2− 2xy.Find out the stream function. Also find the complex potential function for the fluid flow.arrow_forward
- 5. Let F(x, y) = square region with vertices (0,0), (1,0), (1, 1), and (0, 1) oriented counterclockwise. Find f F · dī exactly. Simplify your answer. (tan-y, -) be a vector field and let C be the boundary of the You must show all your work in a step by step manner, clearly, by the same way you learned in this class. No calculator output or decimal approximations are accepted.arrow_forward6. Consider the vector field F(ar, y) = (8r°y+ e")î+ (Ka + e")3, where K is some constant. (a) (5 points) For which value of K is F the gradient of a function? For this value, find f such that F = Vf. (b) (5 points) For the value of K found in part (a), evaluate fe F dr, where C is the segment of the curve y = a from (0,0) to (1, 1). (c) (5 points) If instead K = 1, use Green's Theorem to evaluate f F dr, where C is the boundary of the triangle with vertices (0,0), (2,0), and (2, 1), traversed counterclockwise.arrow_forwardB onlyarrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage