Problem 2. Suppose f(x, y) = 10e – 7xe", P =(-2,1), and u is the unit vector in the direction of (5, 2). A. Find the gradient of f(x, y) and evaluate at the point P = (-2,1). Vf = (-7(x*e^x+e^x),20*y*e^(y^2)) (Vf) (P) (-7*(2*e^2+e^2),20*e) B. Find the tangent plane to f(x, y) at the point (-2, 1). z = -7*(2*e^2+e^2)*(x+2)+20*e*(y-1)+10*e+14e^(-2) C. Find the differential df and evaluate at the point (-2, 1) df = -7(x*e^x+e^x)*dx+20*y*e^(y^2)*dy %3D df(-2, 1) = -7*(2*e^2+e^2)*dx+20*e*dy Use the differential at the point (-2, 1) to estimate the change in the values of f(x, y) if x changes by -0.5 and y changes by 0.2. df -7*(2*e^2+e^2)*0.5+20*e*0.2 D. Find the directional derivative of f(x,y) at P = (-2, 1) in the direction of u. D,f = -7*(2*e^2+e^2)*5/sqrt(29)+20*e*2/sqrt(29) E. At the point P = (-2,1), find the direction of maximum ascent. Your answer should be a unit vector. u <-7*(2*e^2+e^2)/sqrt(49*(2*e^2+e^2)^2,400*e^2).: Find the maximum rate of change of f(x, y) at P = (-2, 1). <-7*(2*e^2+e^2)/sqrt(49*(2*e^2+e^2)^2+400°e^2),20*e/sqrt(4 F. Find a vector v such that Dy f(-2,1) = 0. V =

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