Practice Exercises 8-23. Sketching vector fields Sketch the following vector fields. 8. F = (1,0) 9. F = (−1, 1) 10. F = (1,y) 11. F = (x, 0) 12. F = (x,y) 13. F = (x, −y) 14. F = (2x, 3y) 15. F = (y, -x) 16. F = (x + y, y) 17. F = (x,y - x)
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- The position vector r describes the path of an object moving in the xy-plane. Position Vector Point r(t) = (2e-t, 6e) (2,6) (a) Find the velocity vector v(t), speed s(t), and acceleration vector a(t) of the object. v(t) s(t) = a(t) (b) Evaluate the velocity vector and acceleration vector of the object at the given point. v(0) = a(0) =A particle moves at constant speed 1445 units along the curve of intersection of the two the two surfaces y = x and z = 2 in the direction of increasing x. Find its velocity when it is at the point (- 12, 144, - 1152). A the point (-12, 144, - 1152), the velocity vector is V = i + + k.A net is dipped in a river. Determine the flow rate of water across the net if the velocity vector field for the river is given by v=(x-y,z+y+7,z2) and the net is decribed by the equation y=1-x2-z2, y20, and oriented in the positive y- direction. (Use symbolic notation and fractions where needed.)
- 2. 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).The position vector r describes the path of an object moving in the xy-plane. Position Vector Point r(t) = ti + (-t2 + 8)j (1, 7) (a) Find the velocity vector v(t), speed s(t), and acceleration vector a(t) of the object. v(t) s(t) a(t) (b) Evaluate the velocity vector and acceleration vector of the object at the given point. v(1) = a(1)a C d b Find a vector-valued function whose graph is the plane x+y+z=3₁ a. r(u,v) = ui-vj-(3-u-v)k b. r(u,v) = ui+vj+(3-u-v)k c. r(u,v) = ui+vj+(3+u+v)k d. r(u,v) = ui+vj-(3+u+v)k e. r(u,v)=-ui-vj+(3-u-v)k
- (5) Let ß be the vector-valued function 3u ß: (-2,2) × (0, 2π) → R³, B(U₁₂ v) = { 3u² 4 B (0,7), 0₁B (0,7), 0₂B (0,7) u cos(v) VI+ u², sin(v), (a) Sketch the image of ß (i.e. plot all values ß(u, v), for (u, v) in the domain of ß). (b) On the sketch in part (a), indicate (i) the path obtained by holding v = π/2 and varying u, and (ii) the path obtained by holding u = O and varying v. (c) Compute the following quantities: (d) Draw the following tangent vectors on your sketch in part (a): X₁ = 0₁B (0₂7) B(0)¹ X₂ = 0₂ß (0,7) p(0.4)* ' cos(v) √1+u² +Find the directional derivative of f (x, y, z) = 2z²x + y³ at the point (2, 1, 1) in the direction of the vector √5 (Use symbolic notation and fractions where needed.) directional derivative: + 2 √5A force field is given by the equation. A particle is moved from (1,0,0) to (1,1,1) along a straight line. Calculate the work done by the force field on the particle.
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