Calculus 3 :

Section 11.6: Problem 1 [1 point} Find the directional derivative of x, 3:) = 12y3+ 2543,- at the point [3, 3} in the direction 3 = inf-1. The gradient off is: mm) = DE]: va, 3) = DE]: The directional derivative is: C] Section 11.6: Problem 10 (1 point) Consider a function f(x, y) at the point (1, 3). At that point the function has directional derivatives: 5 in the direction (parallel to) (2, 3), and 13 in the direction (parallel to) (1, 4). /17 The gradient of f at the point (1, 3) isSection 11.6: Problem 11 [1 point} Consider the surface a: 2 55:2 + 332 1?8. Find an equation of the tangent plane to the surface at the point {11.1, 4, ). Find a 1"rector equation of the normal line to the surface at {1,4, } rft] = E]. Section 11.6: Problem 12 [1 point} Consider the surface 3 ID = rev-sous 2:. Find an equation of the tangent plane to this surface at ( Hi, i], 1}]. E] Find a 1"rector equation for the normal line to the surface at [10, i], D}. rft] = E] Section 11.6: Problem 18 (1 point) 140 The temperature at any point in the plane is given by T(x, y) =- x2+ +1 (a) What shape are the level curves of T? O A. hyperbolas OB. circles O C. parabolas OD. lines O E. ellipses OF. none of the above (b) At what point on the plane is it hottest? What is the maximum temperature? (c) Find the direction of the greatest increase in temperature at the point (-1, -3). What is the value of this maximum rate of change, that is, the maximum value of the directional derivative at (-1, -3)? (d) Find the direction of the greatest decrease in temperature at the point (-1, -3). What is the value of this most negative rate of change, that is, the minimum value of the directional derivative at (-1, -3)?Section 11.6: Problem 2 (1 point) Consider the function f(x, y, 2) = my + y2 + 12 Find the gradient of f: Find the gradient of f at the point (1, -4, -5). C Find the rate of change of the function f at the point (1, -4, -5) in the direction u = (5/v38, -2/v38, 3/V38).Section 11.6: Problem 3 [1 point] Find the directional derivative of x, 5:, z) = :3 $23: at the point (4, 3, 2] in the direction of the vector v = (1, 2, 1). C] Section 11.6: Problem 4 (1 point) Find the directional derivative of f(x, y, 2) = x + 2 - 52 at the point P = (2, 4, -5) in the direction of the origin.Section 11.6: Problem 5 [1 point]: Find the maximum rate of change of f{z,y, z] = z + E at the point {2, 2, 5) and the direction in which it occurs. 2".- Maximum rate of change: E] Direction in which it occurs: E] Section 11.6: Problem 6 (1 point) Find the maximum rate of change of f(x, y) = In(x + y") at the point (1, -3) and the direction in which it occurs. Maximum rate of change: Direction in which it occurs:Section 11.6: Problem 7 (1 point) Consider the function f(x, y) = y - x2. Find the the directional derivative of f at the point (-1, 4) in the direction given by the angle 0 = = Find the vector which describes the direction in which f is increasing most rapidly at (-1, 4).Section 11.6: Problem 8 [1 point]: Suppose f{z,y} = E ,P = (4,4) and v = 3i+ 2} y 1. Find the gradient of 1-". var, y) = D i+ Dr 2. Find the gradient of f at the point P. w = E] a: 3. Find the directional derivative of _f at P in the direction of v. Duf{P} = E] 4. Find the maximum rate of change of f at P. C] 5. Find the (unit) direction vector in which the maximum rate of change occurs at P. DHDJ Section 11.6: Problem 9 {1 point} Suppose that you are climbing a hill whose shape is given by , and that you are at the point In which direction should you proceed initially in orderto reach the top of the hill fastest? If you climb in that direction, at what angle above the horizontal will you be climbing initially [radian measure}