[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]Please give a warm-up quiz
[/et_pb_text][et_pb_code _builder_version="4.1"][/et_pb_code][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]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 _____[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_code _builder_version="3.26.4"][/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
Start with hints
[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.1" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first![/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 ???????
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"] 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?
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]$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) .
[/et_pb_tab][/et_pb_tabs][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="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]A graphical view point
[/et_pb_text][et_pb_video src_webm="https://cheenta.com/wp-content/uploads/2020/01/2020-01-21-04-29-14.mp4" _builder_version="4.1" custom_margin="77px||77px||true|"][/et_pb_video][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="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]Take A Look At This Knowledge Graph
[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/Untitled-Diagram-6.png" align="center" _builder_version="4.1"][/et_pb_image][et_pb_video _builder_version="4.1"][/et_pb_video][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" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]Connected Program at Cheenta
[/et_pb_text][et_pb_blurb title="College Mathematics Program" url="https://cheenta.com/collegeprogram/" image="https://cheenta.com/wp-content/uploads/2018/03/College-1.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" link_option_url="https://cheenta.com/collegeprogram/" border_color_all="#e02b20"]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://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
[/et_pb_text][et_pb_post_slider include_categories="12" image_placement="left" _builder_version="4.1"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]
