Question

Need some step by step helps with these 3 questions.

(a) [4 marks] Given the scalar function g(x, y, 3)(1-2) sin y+ e cos z calculate v-Vo and find ▽·ν. (b) [4 marks] In what direction from the point P(-1,0,7) and at what rate does function g increase in the quickest way? (c) [4 marks] What is the rate of change of the above function in the direc- tion of vector a = (3,4,0) at point P?

Thanks

Answer 1

Given that g (x, y, z) = (1 - x^2)siny + e^y cosz nablag = i partial differentialg/partial differentialx + j partial differentialg/partial differentialy + partial differentialg/partial differentialz = (- 2xsiny)i + (1 - x^2)cosy + e^y cosz) j + (-e^y sinz)k v = nablag = (-2xsiny)i + ((1 - x^2) cosy + e^y cosz) j + (- e^y sinz) k therefore nabla v = (i partial differential/partial differentialx + j partial differential/partial differential y + k partial differential/partial differentialz) ((- 2xsiny)i + ((1 - x^2)cosy + e^y cosz) j + (- e^y sinz) k) = partial differential/partial differentialx (- 2xsiny) + partial differential/partial differentialy ((1 - x^2)cosy + e^y cosz) + partial differential/partial differentialz (-e^y sinz) = -2 siny- (1 - x^2) siny + e^y cosz - e^y cosz = -2 siny - (1 - x^2) siny given that nabla g = i partial differentialg/partial differentialx + j partial differentialg/partial differentialy + partial differentialg/partial differentialz = (- 2xsiny)i + (1 - x^2)cosy + e^y cosz) j + (- e6y sinz) k at the point p = (-1, 0, phi) nablag = -j since nabla g at p is negative so in opposite direction from the point p the function g has quick increasing given that nabla g = i partial differentialg/partial differentialx + j partial differentialg/partial differentialy + partial differentialg/partial differentialz