Lateral Surface Area In Exercises 65-72, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface
Trending nowThis is a popular solution!
Chapter 15 Solutions
Multivariable Calculus
- What dilation maps triangle ABC onto triangle A'B'C' below? B (x, y) → (2x, 2y) B. (x, y) → (0.5x, 0.5y) 24) C. (x, y) (3x, 3y) D. (x, y) → (-0.5x, -0.5y)arrow_forwardCheck that the point (−1,−1,1) lies on the given surface. Then, viewing the surface as a level surface for a function f(x,y,z) find a vector normal to the surface and an equation for the tangent plane to the surface at (−1,−1,1) x^2−3y^2+z^2=−1arrow_forwardConsider the function f(x,y)=|x|+|y| Sketch the surface z = f(x,y) Draw the level curves of the functionarrow_forward
- Dteermine the equation of the tangent plane to the surface given equation G(u, v) = (2u + v, u - 4v, 3u) at the point where: u = 1 and v = 4. %3Darrow_forwardTangent of x?/3 + y2/3 + z2/3 = a²/3 surface at any point ( xo , Yo ,Zo ) Show that the sum of the squares of the intersecting axes of the plane is constant.arrow_forwardConsider the surface What is the equation of the tangent plane to the surface at the point P(3,3,-2)? O 32³ +3zy+zyz² = 0 O 34z + 7y - 12z = 147 O 3x+3y-2z = 0 O 34z + 7y-12z = 0 O 3x+3y-2z=147 What is the equation of the line normal to the surface at the point P(3,3,-2) ? 3 -0-0)-C) 3 +t 21 -2 O O O NC O =t No wo wo -2 102 -0-0-0) 21 +t 3 -36 102 102 =1 21 -36 -36 3 2 3 -O-C-(C) <=8 3 +t 102 32³ +3zy+zyz² = 144. 21 -36arrow_forward
- Check that the point (-2, 2, 4) lies on the surface cos(x + y) = exz+8 (a) View this surface as a level surface for a function f(x, y, z). Find a vector normal to the surface at the point (-2, 2, 4). -4i + 4k (b) Find an implicit equation for the tangent plane to the surface at (-2, 2, 4). X-Z +6=0arrow_forwardParameterize the line of intersection of the surfaces: x2 - y2 = z - 1 and x2 + y2 = 4arrow_forwardThe tangent plane at a point Po (f(u0.Vo) 9(40.vo).h(4o.vo)) u(uo.vo) xry (uo.vo), the cross product of the tangent vectors ru (uo.vo) and r, (uo.vo) at Po. Find an equation for the plane tangent to the surface at Po. Then find a Cartesian on a parametrized surface r(u,v) = f(u,v) i+ g(u,v) j+ h(u,v) k is the plane through P, normal to the vector equation for the surface and sketch the surface and tangent plane together. The circular cylinder r(0,z) = (4 sin (20)) i+ (8 sin 0) j+z k at the point Po (2/3,2,2) corresponding to (0,z) =arrow_forward
- Check that the point (-2, 2, 4) lies on the surface cos(x + y) = exz+8 (a) View this surface as a level surface for a function f(x, y, z). Find a vector normal to the surface at the point (-2, 2, 4). (b) Find an implicit equation for the tangent plane to the surface at (-2, 2, 4).arrow_forwardSketch the surface x = 2y2 +3z2arrow_forwardClassify and sketch the surface x − y2 − 4z2 = 0.arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage