S E C T I O N15.5The Gradient and Directional Derivatives(ET Section 14.5)689∇T(3,9,4)=e9−41,3,−3=e51,3,−3The rate of change of the bug’s temperature at the starting pointPis the directional derivativeDuf(P)= ∇T(3,9,4)·u=e51,3,−3·132,−2,−1= −e53≈ −49.47The answer is−49.47 degrees Celsius per meter.34.Suppose that∇fP=2,−4,4 . Isfincreasing or decreasing atPin the directionv=2,1,3 ?SOLUTIONWe compute the derivative offatPwith respect tov:Dvf(P)= ∇fP·v=2,−4,4·2,1,3=4−4+12=12>0Since the derivative is positive,fis increasing atPin the direction ofv.35.Letf(x,y)=xex2−yandP=(1,1).(a)Calculate∇fP.(b)Find the rate of change offin the direction∇fP.(c)Find the rate of change offin the direction of a vector making an angle of 45◦with∇fP.SOLUTION(a)We compute the gradient off(x,y)=xex2−y. The partial derivatives are∂f∂x=1·ex2−y+xex2−y·2x=ex2−y1+2x2∂
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