Consider the following surface parametrization. x-5 cos(8) sin(φ), y-3 sin(θ) sin(p), z-cos(p) Find an expression for a unit vector, n, normal to the surface at the image of a point (u, v) for θ in [...
16, Let x: U R2-, R, where x(8, φ) (sin θ cos φ, sin θ sin φ, cos θ), be a parametrization of the unit sphere S2. Let and show that a new parametrization of the coordinate neighborhood x(U) = V can be given by y(u, (sech u cos e, sech u sin e, tanh u Prove that in the parametrization y the coefficients of the first fundamental form are Thus, y-1: V : S2 → R2 is a conformal...
(5) The image of the parametrization Φ(u, u) = (a . sin(u) . cos(u), b . sin(u) . sin(e), c . cos(u)) sin(u sin() cosu with b < a, 0 r, 0 2π parametrizes an ellipsoid. u u a) Show that all the points in the image of Φ satisfy the Cartesian equation of an ellipsoid E b) Show that the image surface is regular at all points. c Write out the integral for its surface area A(E). (Do not...
1a throught 1d please < G- (20 points total) A surface is parametrized as: x(φ, θ)-25in(n) * Cos(0), y (φ, θ)- 25in(p) * Sin(θ), z-4Cos(p) for 0 φ π and 0 f 2m (a) Does this surface pass the point (1,2, 3)? If ye reason. (b) What is the normal vector of the surface at any given (p, 0) within the parameter domain? (e) What is the normal vector at ) (d) What is the equation for the tangent plane...
Problem 3 (12 points) The curve with parametric equations (1 + 2 sin(9) cos(9), y-(1 + 2 sin(θ)) sin(0) is called a limacon and is shown in the figure below. -1 1. Find the point (x,y 2. Find the slope of the line that is tangent to the graph at θ-π/2. 3. Find the slope of the line that is tangent to the graph at (,y)-(1,0) ) that corresponds to θ-π/2. Problem 3 (12 points) The curve with parametric equations...
true or false is zero. F 9. The plane tangent to the surface za the point (0,0, 3) is given by the equation 2x - 12y -z+3-0. 10. If f is a differentiable function and zf(x -y), then z +. T 11. If a unit vector u makes the angle of π/4 with the gradient ▽f(P), the directional derivative Duf(P) is equal to |Vf(P)I/2. F 12. There is a point on the hyperboloid 2 -y is parallel to the plane...
NO.25 in 16.7 and NO.12 in 16.9 please. For the vector fied than the vecto and outgoing arrows. Her can use the formula for F to confirm t n rigtppors that the veciors that end near P, are shorter rs that start near p, İhus the net aow is outward near Pi, so div F(P) > 0 Pi is a source. Near Pa, on the other hand, the incoming arrows are longer than the e the net flow is inward,...