Question

(1 point) A stone is thrown from a rooftop at time t 0 seconds. Its position at time t (the components are measured in meters) is given by r()-бі-50+ (24.5-49:2) k. The origin is at the base of the bulding, which is standing on flat ground. Distance is measured in meters. The vector i points east,j points north, and k points up. (a) How high is the rooftop? meters. (b) When does the stone hit the ground? seconds (c) Where does the stone hit the ground? (Your answer should be the position vector of that point.) (in meters) (d) How fast is the stone moving when it hits the ground? meters per second) (in (1 point) Find the equations of the tangent plane and the normal line to the surface x 6y + 3z2 - 517 at the point (-1, -6, 10). Equation of the tangent plane (make the coefficient of x equal to 1): Vector equation of the normal line: lt)-(- +1 HINT. Represent the given surface as a level surface of a certain function f(r, y, z), that is, re-write the equation of the surface in the form f(x, y, z)c, where f(x, y, z) is a function of three variables and c is a constant. Then find the gradient vector of the function f at the given point. The tangent plane to the surface at that point is orthogonal to the gradient vector, and the normal line is parallel to it. This will allow you to write the equations of both tangent plane and normal line. (1 point) Find the differential of f(x, y) - Vx3 + y2 at the point (1, 3). The answer should be an expression of the form a * dx + b * dy, where a and b are numbers. df Then use the differential to estimate the change in f(x, y) as (x, y) changes from (1,3) to (1.06,2.93) f(1.06, 2.93) -f(1,3) ~

Answer 1

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Anu:

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