УА (A) УА (C) x (B) ז x x (D) Match the following vector fields with their plots below. 1. F(x, y) = 2. F(x, y) = (0, x²) (x − y, x) - - 3. F(x, y) = (2x, 4. F(x, y) = (y, x) y)
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- Correct solution will be liked.How would I write a vector function r(t) that follows the below guidelines2. Calculate the gradient vector Vf of the function f (x, y) = x² – x + y - x²y - 2y2 at the point (2,1) and sketch it on the attached contour plot (you can save the picture, open in photo editor and use drawing tools). Explain in one paragraph (about 200-300 words) the meaning of the gradient vector Vf(2,1), negative gradient vector -Vf(2,1).
- If r(t) = u(t) x v(t), where u and v are the vector functions given below, find r'(2). u(2) = (6, 3, -6), u'(2) = (4, 0, 2), v(t) = (t, t², t³)Suppose f(x, y) (a) ▼ f(x, y) = (b) ▼ f(-1.1, 6) = = =√√tan(x) + y and u is the unit vector in the direction of (-1, 1). Then, = (c) fu (-1.1, 6) = Du f(−1.1, 6) = =#3. Consider the function if (x, y)# (0,0), if (x, y)= (0,0). f (x, y)= x2+y2 Calculate the directional derivative of f at the point (0,0) in the direction of the vector (1,3).
- (22) in the direction of (1,2). Then, Suppose f(x, y) (a) ▼ f(x, y) = = (b) ▼ ƒ(6, π) = = sin (c) ƒu (6, ñ) = Du ƒ(6, ñ) = and u is the unit vectorThe directional derivative of the function f(x, y, z) = x³y4 – 3z2 at the point (1, 1,1) in the direction of the vector v = (2, –1, 2) is Select one: 10 O a. 3 О ъ. -10 O c. None of these 10 O d. -Match the vector fields: ? curl(F)=0 ? curl(F)=2 V 2. div(F)=0, ? curl(F)=0 ? V 3. div(F)=-1, 1. div(F)=4, ? curl(F)=1 V 4. div(F)=-2, V 5. div(F)=-2, curl(F)=-1 ? curl(F)=3 6. div(F)=2, A D B E C F
- Find the derivative of the vector function r(t) = ta x (b + tc), where a = (4,-1, 5), b = (-3,-3,-1), and c = (1,2, 2). r'(t) = ( 16+16t %3D 11+14t -15-18tYou are given the derivative of a vector function r in the component form is -(e,3e",-2t , dt ,3e",-2t). You are also given that r(0) = 21-j+k. r(0) = 2i -j+k J Determine the vector function r (t) in the form r(t)= (x(t),y(t), z(t)} An efficient notation for the vector equation of a straight line in 3D(or 2D) is given by ((t)- a+ bt where t is any real number, a is the position vector from the origin to a point on the line and b b) Write the vector equation of the tangent line to the curve C generated by r(t) at the point (2, -1, 1) using the above form aSuppose f(x, y) = √/tan(x) + y and u is the unit vector in the direction of (1, 1). Then, (a) ▼ f(x, y) = (b) ▼ ƒ(−0.8, 10) = (c) fu (−0.8, 10) = Du f(−0.8, 10) =