1-Find the directional derivative of f(x,y)=x/y at the point (1,-1) in the direction of the vector v=(2,5) 2-Find an equation of the tangent plane to the surface xysin(z/3)=6, at the point (3,4,1), The equation should have the form g(x,y,z)=0 and the coeffecient of x shoud be 23. 3-Find equations of the tangent plane and normal line to the surface x=4y^2+2z^2-292 at the point (2,-7,7), also find the normal line? Solution f/x (x,y) = 1/y f/y (x,y)= -x/y2 f/x (1,-1) = -1 f/y (1,-1) = -1 Therefore, the gradient is f (1,-1) = - i - j = (-1,-1) --------------(1) Letu=u1i+u2jbe a unit vector.The directional derivative at (1,-1) in the direction ofuis Du f (1,-1) = (u1i+u2j) (-i-j) = -u1 - u2 directional vector u = (2,5) /29 = (2/29,5/29) = (u1,u2) ince we are still at the point (1,-1), equation (1)is still valid. We plug in our newuto obtain Duf(1,-1) = -u1 - u2 = -2/29-5/29 = 3/29.

1. 1-How can I use my time in school to increase my personal
Solution
Through Louvain School of Management IEMBA, I developed the skills and confidence to build
leadership and do business on a global basis. The program is a valuable investment for me. It
changed my view and the perception of others on me. The lectures, the assignments and the
travels changed us into a terrific team.