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Let \phi (x,y,z) = 3y^2+3yz for (x,y,z) \in \mathbb{R}^{3} Then the absolute value of the directional derivative of \phi in the direction of the line \frac{x-1}{2} = \frac{y-2}{-1}= \frac{z}{-2} , at the point (1,-2,1) is _____

Source of the problem
IIT JAM 2018
Topic
Directional Derivative 
Difficulty Level
Easy
Suggested Book

Generalized Directional Derivatives  By Pastor Karel

 

Start with hints

Do you really need a hint? Try it first!

The rate of change of f(x,y,z) in the direction of the unit vector \vec u = <a,b,c> is called the directional derivative and is denoted by D_{\vec u}f(x,y,z) . The defination of directional derivative is  D_{\vec u}f(x,y,z) = \lim h \to 0 \frac{f(x+ah,y+bh,z+ch)-f(x,y)}{h} . Here if you consider u to be (1,0,0), (0,1,0) and (0,0,1) then we will get D_{\vec u}f(x,y,z) to be f_x, f_y and f_z respectively. If you observe closely, this is nothing but differentiating the function f w.r.t x,y and z keeping all the other thing constant.

Now \nabla (f)=(f_x,f_y,f_z) by the above definition can you try to solve the problem ???????

 The points on the straight line \frac{x-1}{2} = \frac{y-2}{-1}= \frac{z}{-2} = k \implies x=1+2k , y=2-k , z=-2k So any point on the straight line would be of the form (x,y,z)= (1,2,0) + (2,-1,-2)k Here the direction of the straight line would be (2,-1,-2)

Now we have to consider the unit vector in that direction and so the unit vector would be u= \frac{(2,-1,-2)}{\sqrt{4+1+4}} = (\frac{-2}{3},\frac{-1}{3},\frac{-2}{3}) Here the absolute value of the directional derivative would be \nabla \phi . u . Would you like to calculate this dot product?

\nabla \phi = (\phi_{x},\phi_{y},\phi_{z})   i.e the co-ordinates involving partial derivative of \phi . So, \nabla \phi = (0, 6y+3z , 3y)|_{(1,-2,1)} = (0,-9,-6)

Then \nabla \phi . u= \frac{9}{3} + \frac{6 \times 2}{3} = 3+4 = 7 (Ans) .

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