the straight line in the xy plane from the point (0, 4) to the point (2,0)

Transcription

1 Math 238 Review Problems for Final Exam For problems #1 to #6, we define the following paths vector fields: b(t) = the straight line in the xy plane from the point (0, 4) to the point (2,0) c(t) = the path along the circle x 2 + y 2 = 25 in the xy plane from the point ( 5,0) to the point (0, 5) d(t) = (3cos(2t),3sin(2t),t) for 0 t 4 F = G = V = 3x 2 y 3 i + 3x 3 y 2 j (z 3 y 2x)i + (x + y + z)j + xyzk 3x 3 y 2 i 3x 2 y 3 j 1. Do each of the following for the path d(t): Find the velocity vector and the acceleration vector. Evaluate each of the velocity vector and the acceleration vector at t= /6. Find the arc length of the path between t=7 /6 and t=3 /2. 2. Consider the line integral of G over the path d(t). Write the integral in the form integral., but do not attempt to evaluate this

2 3. Do each of the following for the vector field F: Decide whether or not this vector field could be the curl of another vector field, and say why or why not. If possible, find a function f(x,y) where the vector field is f; if this is not possible, give a reason why. of the vector field over a path from the point (4,2) to the point ( 7,8); if this is not possible, give a reason why. of the vector field over a path which begins and ends at the point (4,2). 4. Do each of the following for the vector field G: Decide whether or not this vector field could be the curl of another vector field, and say why or why not. If possible, find a function f(x,y,z) where the vector field is f; if this is not possible, give a reason why. of the vector field over a path from the point (4,2,1) to the point ( 7,8,3); if this is not possible, give a reason why. of the vector field over a path which begins and ends at the point (4,2,1).

3 5. Do each of the following for the vector field V: Decide whether or not this vector field could be the curl of another vector field, and say why or why not. If possible, find a function f(x,y) where the vector field is f; if this is not possible, give a reason why. of the vector field over a path from the point (4,2) to the point ( 7,8); if this is not possible, give a reason why. of the vector field over a path which begins and ends at the point (4,2). 6. Parametrize each of the following paths by writing each path as (x(t),y(t),z(t)) and stating the bounds a t b): b(t) c(t) 7. Parametrize the surface consisting of the part of the cone 3z 2 = x 2 + y 2 which lies above the xy plane and inside the sphere x 2 + y 2 + z 2 = 400, by writing the surface as (x(u,v),y(u,v),z(u,v)) and stating the bounds for u and v.

4 8. Consider the region in the xy plane determined by x 4 + y Use polar coordinates to write a double integral which will give us the area of this region, but do not attempt to evaluate this integral. 9. Consider the surface area of the portion of the plane z x 2y = 0 which lies above the triangle in the xy plane determined by x = 0, y = 8, and y = 2x. Write a double integral which will give us this surface area, but do not attempt to evaluate this integral. 10. Consider the surface area of the portion of the sphere x 2 + y 2 + z 2 = 100 which lies inside the cone z 2 = 3x 2 + 3y 2 in the part of R 3 where y 0 and z 0. Recall that the sphere x 2 + y 2 + z 2 = 100 can be parametrized by (10 cosu sinv, 10 sinu sinv, 10 cosv) for 0 u < 2 and 0 v. Find limits on u and v which describe desired surface area. (Hint: it will be helpful to consider the angle the cone makes with the z axis in either the xz plane or the yz plane.) Write a double integral of an algebraically simplified function f(u,v) which will give us this surface area, but do not attempt to evaluate this integral.

5 11. Consider the volume of the region inside both the sphere x 2 + y 2 + z 2 = 25 and the cone z 2 = (x 2 + y 2 )/3 in the part of R 3 where x 0, y 0, and z 0. Use spherical coordinates to write a triple integral which will give us this volume, but do not attempt to evaluate this integral. 12. Consider the volume of the region inside both the sphere x 2 + y 2 + z 2 = 25 and the cylinder 9(x 2 + y 2 ) = 64 in the part of R 3 where z 0. Use cylindrical coordinates to write a triple integral which will give us this volume, but do not attempt to evaluate this integral. 13. Consider the area in the first quadrant of R 2 between the two circles x 2 + y 2 = 9 and (x 2.5) 2 + y 2 = Use polar coordinates to write the sum of two double integrals which will give us this area, but do not attempt to evaluate these integrals. 14. Find and classify all critical points for each of the following functions: f(x,y) = x 3 + y 2 + 6xy f(x,y) = x 3 + y 3 + 3xy Find the extreme values of f(x,y) = xy - 7x on the circle x 2 + y 2 = 15.

6 16. Consider the following integral: Sketch the region over which the integral is taken. Rewrite the integral by reversing the order of integration. 17. Consider the function f(x,y) = x 2 ln(y). Find the second degree Taylor polynomial about (2,e) (but you do not have to algebraically simplify it). Find the directional derivative at the point (2,e) in the direction of the vector (3,4). 18. Find the equation of the plane tangent to the surface defined by 5x + z 2 y z + 1 = 0 at the point (-3,4,2).

7 Answers 1. d (t) = ( 6sin(2t),6cos(2t),1) d (t) = ( 12cos(2t), 12sin(2t),0) d ( /6) = ( 3 3, 3, 1) d ( /6) = ( 6, 6 3, 0) 2. The integral is 3. Since div(f) = 6xy 3 + 6x 3 y =/ 0, then F could not be the curl of another vector field. Since curl(f) = 0, then F is a gradient vector field, and F = f where f(x,y) = x 3 y 3. Since F = f, the line integral over any path from the point (4,2) to the point ( 7,8) must be f( 7,8) f(4,2) = = Since F = f, the line integral over any path which begins and ends at the point (4,2) must be zero (0).

8 4. Since div(g) = 1 + xy =/ 0, then G could not be the curl of another vector field. Since curl(g) = (xz 1)i + (3z 2 yz)j + 2k =/ 0, then G cannot be a gradient vector field. Since G is not a gradient vector field, the line integral over a path from the point (4,2,1) to the point ( 7,8,3) cannot be found, unless it is specified which path is chosen. Since G is not be a gradient vector field, the line integral over a path which begins and ends at the point (4,2,1) cannot be found, unless it is specified which path is chosen. 5. Since div(v) = 0, then V could be the curl of another vector field. Since curl(v) = ( 6xy 3 6x 3 y)k =/ 0, then V cannot be a gradient vector field. Since V is not a gradient vector field, the line integral over a path from the point (4,2) to the point ( 7,8) cannot be found, unless it is specified which path is chosen. Since V is not a gradient vector field, the line integral over a path which begins and ends at the point (4,2) cannot be found, unless it is specified which path is chosen.

9 6. One possible answer is b(t) = (2t, 4t 4, 0) for 0 t 1 One possible answer is c(t) = (5cost, 5sint, 0) for t 3 /2 7. The sphere and the cone intersect when x 2 + y 2 = 300 and z = 10. One possible answer is (u cosv, u sinv, u/ 3) for 0 u 10 3 and 0 v First, we recognize that the plane z x 2y = 0 is just the function f(x,y) = x + 2y. The triangle in the xy plane determined by x = 0, y = 8, and y = 2x can be described by either 4 x 0 or by 0 y 8 2x y 8 y/2 x 0 The desired surface area is or.

10 10. 0 u < and 0 v /6 u (u,v) = ( 10 sinu sinv, 10 cosu sinv, 0) v (u,v) = (10 cosu cosv, 10 sinu cosv, 10 sinv) u (u,v) v (u,v) = ( 100 cosu sin 2 v, 100 sinu sin 2 v, 100 sinv cosv) u (u,v) v (u,v) = 100 sinv The desired surface area is

11 14. (0,0) is a saddle point (6, 18) is a local minimum (0,0) is the only critical point, but the Second Derivative Test fails to classify this point. However, once we realize that f(0,0)=0 and f(x,y) can be negative or positive, we then see that (0,0) must be a saddle point. 15. The extrema are at ( 12.75, 1.5) which gives the minimum and ( 12.75, 1.5) which gives the maximum (x 2) + (4/e)(y e) + (x 2) 2 (2/e 2 )(y e) 2 + (4/e)(x 2)(y e) 12/5 + 16/(5e) 18. 5x + 4y +15z = 31