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Lisa [10]
3 years ago
9

Find the directional derivative of f(x,y,z)=z3−x2yf(x,y,z)=z3−x2y at the point (−5,5,2)(−5,5,2) in the direction of the vector v

=⟨−3,2,−4⟩v=⟨−3,2,−4⟩.
Mathematics
2 answers:
olga_2 [115]3 years ago
7 0

We are given

f=z^3 -x^2y

Firstly, we can find gradient

so, we will find partial derivatives

f_x=0 -2xy

f_x=-2xy

f_y=0 -x^2

f_y=-x^2

f_z=3z^2

now, we can plug point (-5,5,2)

f_x=-2*-5*5=50

f_y=-(-5)^2=-25

f_z=3(2)^2=12

so, gradient will be

gradf=(50,-25,12)

now, we are given that

it is in direction of v=⟨−3,2,−4⟩

so, we will find it's unit vector

|v|=\sqrt{(-3)^2+(2)^2+(-4)^2}

|v|=\sqrt{29}

now, we can find unit vector

v'=(\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })

now, we can find dot product to find direction of the vector

dir=(gradf) \cdot (v')

now, we can plug values

dir=(50,-25,12) \cdot (\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })

dir=(-\frac{150}{\sqrt{29} } - \frac{50}{\sqrt{29} } - \frac{48}{\sqrt{29} })

dir=-\frac{248\sqrt{29}}{29}.............Answer



Diano4ka-milaya [45]3 years ago
5 0

Answer:

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