Check that the point (−1,−1,3)(−1,−1,3) lies on the given surface. Then, viewing the surface as a level surface for a function f(x, y,z)f(x, y,z), find a vector normal to the surface and an equation for the tangent plane to the surface at (−1,−1,3)(−1,−1,3).

Answavelength; amplitude;

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.

F= (mass)(acceleration) = mam = 0.25 kgvi = 16 m/st = 2 svf = 0 m/s (since it was put to stop)a=(vf-vi)/ta=(0-16)/2a = 8 m/s^2 (decelerating)f = ma = (0.25 kg)(8 m/s^2)f = 2 nhope this !