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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 \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 _____

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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 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 ???????

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 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?

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\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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A graphical view point

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Take A Look At This Knowledge Graph 

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Similar Problems

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