Directional Derivative : IIT JAM 2018 Problem 42

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Please give a warm-up quiz

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let $latex \phi (x,y,z) = 3y^2+3yz $ for $latex (x,y,z) \in \mathbb{R}^{3}$ Then the absolute value of the directional derivative of $latex \phi $ in the direction of the line $latex \frac{x-1}{2} = \frac{y-2}{-1}= \frac{z}{-2} $ , at the point $latex (1,-2,1)$ is _____

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[/et_pb_code][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1"]IIT JAM 2018[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Directional Derivative [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.1" open="off"]

Generalized Directional Derivatives  By Pastor Karel

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Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1" hover_enabled="0"]The rate of change of $latex f(x,y,z) $ in the direction of the unit vector $latex \vec u = <a,b,c>$ is called the directional derivative and is denoted by $latex D_{\vec u}f(x,y,z) $. The defination of directional derivative is $latex 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 $latex u$ to be $latex (1,0,0), (0,1,0)$ and $latex (0,0,1)$ then we will get $latex D_{\vec u}f(x,y,z)$ to be $latex f_x$, $latex f_y$ and $latex f_z$ respectively. If you observe closely, this is nothing but differentiating the function $latex f$ w.r.t $latex x,y$ and $latex z$ keeping all the other thing constant.

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

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

Now we have to consider the unit vector in that direction and so the unit vector would be $latex 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 $latex \nabla \phi . u $ . Would you like to calculate this dot product?

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

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

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A graphical view point

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Take A Look At This Knowledge Graph 

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Similar Problems

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