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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 _____

[/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

[/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 $f(x,y,z)$ in the direction of the unit vector $\vec u = $ 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 ???????

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"]

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?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]

$\nabla \phi = (\phi_{x},\phi_{y},\phi_{z})$  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) .

# 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]

# 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 _____

[/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

[/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 $f(x,y,z)$ in the direction of the unit vector $\vec u = $ 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 ???????

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"]

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?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]

$\nabla \phi = (\phi_{x},\phi_{y},\phi_{z})$  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) .

# 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]

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