(4) Consider the surface z = x2+4y2+1. Suppose you are walking on this surface directly above a curve C in the xy-plane, where the parameterized curve is given by C (t)cost, y(t) sin t. Find the valu...
Consider the surface z = f(x, y) = sin(x) + cos(y) and the curve C in the xy plane defined parametrically as x(t) = 2cos(t), y(t) = sin(t) a. Find z'(t). Imagine you are walking directly above the curve C in the direction of increasing t. Find the values of t for which you are walking uphill. Hint:Graph z'(t). Graph f(x, y) for -7 < x < 7 and -7 < y < 7. (You will have to find software...
Suppose S is the surface z= x2 + 4y2, lying beneath the plane z=1. Orient S by taking the inner normal n to pointing in the positive k direction. Calculate the flux integral across S in the direction n for the velocity field of a gas given by V= <y, -x,x=2>[12]
Suppose S is the surface z= x2 + 4y2, lying beneath the plane z=1. Orient S by taking the inner normal n to pointing in the positive k direction. Calculate the flux integral across S in the direction n for the velocity field of a gas given by V= <Y,-X+,x=2>[12]
Questions. Please show all work. 1. Consider the vector field F(x, y, z) (-y, x-z, 3x + z)and the surface S, which is the part of the sphere x2 + y2 + z2 = 25 above the plane z = 3. Let C be the boundary of S with counterclockwise orientation when looking down from the z-axis. Verify Stokes' Theorem as follows. (a) (i) Sketch the surface S and the curve C. Indicate the orientation of C (ii) Use the...
Find the area of the lateral surface over the curve C in 6. the xy-plane and under the surface z - f(x,y) f(x,y)-h, C:y-1 -x2 from (1,0) to (0,1) Surface: Lateral surface area - f(x, y) ds z =f(x, y) Lateral surface xy) As C: Curve in xy-plane Find the area of the lateral surface over the curve C in 6. the xy-plane and under the surface z - f(x,y) f(x,y)-h, C:y-1 -x2 from (1,0) to (0,1) Surface: Lateral surface...
(b) Let C be the closed curve formed by intersecting the cylinder x2 +y= 1 with the plane x z= 2. Let the tangent to the curve from above. point in the anti-clockwise direction when viewed Calculate the line integral (e (e sin y+ 4) dy+(e(cos z+ sin z)+ay) dz. cos x2yz) dx + (b) Let C be the closed curve formed by intersecting the cylinder x2 +y= 1 with the plane x z= 2. Let the tangent to the...
Consider the paraboloid z=x2+y2. The plane 2x−2y+z−7=0 cuts the paraboloid, its intersection being a curve. Find "the natural" parametrization of this curve. Hint: The curve which is cut lies above a circle in the xy-plane which you should parametrize as a function of the variable t so that the circle is traversed counterclockwise exactly once as t goes from 0 to 2*pi, and the paramterization starts at the point on the circle with largest x coordinate. Using that as your...
Homework 4: Problem 3 Previous Problem Problem List Next Problem (6 points) Consider the function f(x, y) - (e - x) sin(y). Suppose S is the surface z- f(x, y) (a) Find a vector which is perpendicular to the level curve of f through the point (5,5) in the direction irn which f decreases most rapidly. vector (b) Suppose u = 31 + 3/4 ak is a vector in 3-space which is tangent to the surface S at the point...
GIVEN: Ω isthe portion of the surface of the sphere centered at the origin of radius 3 above 1.2 1(xy, z) the plane, z-2: Ω: the field F = (x, x,x). a) FIND the flux of VrF through Ω in the given direction: n has positive 2-component. HINT: (radius a)on Q:(spherical coordinates) b) Parameterize the path,c-a2, (r,g,z)asin g dode with orientation to agree with the given n for Ω ANS: (a) 5 c) With positive orientation,an -e DETERMINE: F.ds ANS:...
Consider the surface given as a graph of the function g(x, y) = x∗y 2 ∗cos(y). The gradient of g represents the direction in which g increases the fastest. Notice that this is the direction in the xy plane corresponding to the steepest slope up the surface, with magnitude equal to the slope in that direction. 1. At the point (2, π), find the gradient, and explain what it means. 2. Use it to construct a vector in the tangent...