Line Integral : IIT JAM 2018 Question Number 20

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What are we learning ?

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/02/IIT_2018_JAM_20.png" align="center" force_fullwidth="on" _builder_version="4.2.2" min_height="388px" height="198px" max_height="207px"][/et_pb_image][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_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]If $\vec{F}(x,y)=(3x-8y)\widehat{i}+(4y-6xy)\widehat{j}$ for $(x,y) \in \mathbb{R}^2$, then $\oint \vec{F}. \mathrm d \vec{r} $, where $C$ is the boundary of triangular region bounded by the lines $x=0,y=0,\quad \textbf{and} \quad x+y=1$ oriented in the anti clock wise direction is $(A)\quad \frac{5}{2} \qquad (B)\quad 3 \qquad (C)\quad 4 \qquad (D)\quad 5 \qquad $[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.2.2" width="100%" max_width="1024px" max_width_tablet="" max_width_phone="" max_width_last_edited="on|phone"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.2.2" 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="4.2.2"]IIT JAM 2018, Question number 20[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.2.2" open="off"]Calculation of line integral of a vector point function along a closed curve.[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.2.2" open="off"]6/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.2.2" open="off"]VECTOR ANALYSIS: Schaum’s Outlines series[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.0.9" 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|0px|20px||" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs _builder_version="4.2.2" hover_enabled="0"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]Suppose we have vectors $\vec{u}=x_{1}\widehat{i}+y_{1}\widehat{j}+z_{1}\widehat{k}$ $\vec{v}=x_{2}\widehat{i}+y_{2}\widehat{j}+z_{2}\widehat{k}$ Then $\vec{u}.\vec{v}=x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}$ Now in the given context $\vec{F} (x,y)=(3x-8y)\widehat{i}+(4y-6xy)\widehat{j}$ $\mathrm d \vec{r}=\mathrm d x\widehat{i}+\mathrm d y\widehat{j}$ Then $\oint_{c} \vec{F}. \mathrm d \vec{r}=\oint_c (3x-8y) \mathrm d x+(4y-6xy)\mathrm d y$ Now we have to chose the proper path to complete the line integral. Do you want to try from here? [/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]The region is bounded by $x=0=y$ & $x+y=1$ i.e.,

Let us divide the segment into 3 parts where $I_1 : y=0 ; \quad I_2 : x+y=1; \quad I_3 : x=0 $ i.e., $\oint_c \vec{F}. \mathrm d \vec{r} = \oint_{I_1} \vec{F}. \mathrm d \vec{r} + \oint_{I_2} \vec{F}. \mathrm d \vec{r}+\oint_{I_3} \vec{F}. \mathrm d \vec{r}$ Let us calculate $\oint_{I_1} \vec{F}. \mathrm d \vec{r}$ & $\oint_{I_3} \vec{F}. \mathrm d \vec{r}$ and I'll keep $\oint_{I_2} \vec{F}. \mathrm d \vec{r}$ for your try by the end of this hint. $\oint_{I_1} \vec{F}. \mathrm d \vec{r}=\int_0^1 3x \mathrm d x$ [We have put $y=0$ so $\mathrm d y=0$ , Hence the limit will be on $x$] $=\frac{3}{2}$ $\oint_{I_3} \vec{F}. \mathrm d \vec{r}=\int_1^0 4y \mathrm d y$ [We have put $x=0$ so, $\mathrm d x=0$ therefore the limit will be on $y$; Observe the anticlockwise direction to understand the limit] $=[\frac{4y}{2}]_1^0$ $=-2$[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2" hover_enabled="0"]On $I_2$ observe that we can transform $y$ in terms of $x$ i.e., $y=1-x$ [We can also use parameterization, which is basically the same method but I prefer this one to make the calculation faster] $y=1-x $ i.e., $\mathrm d y=-\mathrm d x$ now, $\int_{I_2} \vec{F}.\mathrm d \vec{r}= \int_{I_2}(3x-8y)\mathrm d x+\int_{I_2}(4y-6xy)\mathrm d y$ $=\int_1^0(3x-8+8x)\mathrm d x+\int_1^0(4-4x-6x(1-x))(-\mathrm d x)$ [observe the anticlock wise direction limit] $=\int_1^0(11x-8)\mathrm d x +\int_1^0(10x-6x^2-4)\mathrm dx$ $=[\frac{11x^2}{2}-8x+\frac{10x^2}{2}-\frac{6x^3}{3}-4x]_0^1$ $=-\frac{11}{2}+8-5+2+4=\frac{7}{2}$ Hence our answer would be $\frac{3}{2}-2+\frac{7}{2}=3$ [/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" 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

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Volume of revolution : IIT JAM 2018 Question Number 19

Competency in Focus: Application of Calculus (Volume of Revolution)
This problem is from IIT JAM 2018 (Question number 19) and is based on the calculation of the volume of revolution.

Consider the region (D) on (yz) plane and bounded by the line (y=\frac{1}{2}) and the curve (y^{2}+z^{2}=1) where (y\geq0). If the region (D) is revolved about the (z-)axis then the volume of the resulting solid is
$ (A)\frac{\pi}{\sqrt{3}}\qquad$

$(B)\frac{2\pi}{\sqrt{3}} \qquad$

$(C)\frac{\pi\sqrt{3}}{2}\qquad$

$(D)\pi\sqrt{3} $

Do you really need a hint? Try it first!

Hint 1

Imagine that we have a portion of a curve.
$y=f(x)$ from $x=a$ to $x=b$.
In the $xy-plane$ we revolve it around a straight line - $x$-axis.
The result is called solid of revolution
Here in our next hint we will find techniques to calculate the volumes of solid of revolution.

Hint 2

Calculate the volumes of solid of revolution

Let's construct a narrow rectangle of base width $\mathrm d x$ and height $f(x)$ sitting under the curve. When this rectangle is revolved around the $x-\text{axis}$. We get a disk whose radius is $f(x)$ and height is $\mathrm d x$.
The volume of this disk $\mathrm d V=\pi[f(x)]^{2}\mathrm d x$.
So the total volume of the solid is
$V=\int_a^b \mathrm d V=\int_a^b \pi[f(x)]^{2}\mathrm d x$.
If the curve revolved around the vertical line (such as $y$-axis), then horizontal disks are used. If the curve can be solved for (x) in terms of (y). $x=g(y)$ then the formula would be.
$V=\int_a^b[g(y)]^2\mathrm d y$.
Now can you guess if the region between two curves is revolved around the axis.

HINT 3

Sometimes the region between two curves is revolved around the axis and a gap is created between the solid and the axis. A rectangle within the rotated region will become a disk with a hole in it, also known as washer. If the rectangle is vertical and extends from the curve $y=g(x)$ up to the curve $y=f(x)$, then when it is rotated around (x)- axis, it will result in a washer with volume equal to
$\mathrm d V=\pi{[f(x)]^{2}-[g(x)]^{2}}\mathrm d x$.
Which gives us
$V=\int_a^b \pi {[f(x)]^{2}-[g(x)]^{2}}\mathrm d x$.

HINT 4

Now in the given problem replace (x) by (z) , in the above discussion
$y=f(z)=\sqrt{1-z^{2}}, y=g(z)=\frac{1}{2}$
and the line $y=\frac{1}{2}$ intersects the curve at ( $\frac{-\sqrt{3}}{2},\frac{1}{2}$) , ( $\frac{\sqrt{3}}{2},\frac{1}{2}$) in terms of ($z,x)$ co-ordinate.
and hence the volume is
$V=\pi \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}[(1-z^{2})-\frac{1}{4})]\mathrm d z$.
$=\pi[\frac{3z}{4}-\frac{z^{3}}{3}]_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}$
$=\pi(\frac{3\sqrt{3}}{8}-\frac{3\sqrt{3}}{24})\times 2$
$=\pi \frac{3\sqrt{3} \times 2 \times 2}{24}$
$=\frac{\pi \sqrt{3}}{2}$

Suggested Book

Integral Calculus by Gorakh Prasad

Amc 8 Master Class

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Acute angles between surfaces: IIT JAM 2018 Qn 6

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Warm yourself up with an MCQ

[/et_pb_text][et_pb_code _builder_version="4.0.9"][/et_pb_code][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_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="4.0.9" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]In $latex \Bbb R^3$ the cosine of acute angle between the surfaces $latex x^2+y^2+z^2-9=0$ and $latex z-x^2-y^2+3=0$ at the point $latex (2,1,2)$ is

$latex \frac{8}{5\sqrt{21}}$
$latex \frac{10}{5\sqrt{21}}$
$latex \frac{8}{3\sqrt{21}}$
$latex \frac{10}{3\sqrt{21}}$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" custom_padding="|0px||||"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0.9" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.0.9" hover_enabled="0"]IIT JAM 2018 Qn no 6[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0.9" hover_enabled="0" open="off"]Multivable calculus[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0.9" hover_enabled="0" open="on"]Easy [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" hover_enabled="0" open="off"]

Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2

Tom M. Apostol

[/et_pb_accordion_item][/et_pb_accordion][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"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0.9" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][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="4.0.9"]If we are asked to give the angle between two lines then it is very easy to calculate but our forehead will get skinned whenever we will be asked to find out the acute angle between two lines and even worse if we are asked to find the angle between two surfaces. Surprisingly it is not very hard if think stepwise. Observe when you are asked to find out the angle between two lines you calculate it in terms slope. So basically you are firing putting the gun on someone else's shoulder. Here the question is to find that shoulder when it comes in finding the angle between two curves. Observe from the conception of the intersection of two curves that the tangent line of those curves also intersects and we have their corresponding slopes. Bingo! why not calculating the acute angle of the tangent lines and call them the angle between two curves. Now can you think how to calculate the acute angle between to surfaces?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0.9"]The acute angle between two surfaces would be the acute angle between their tangent plane. You can stop here and try to do the problem by your own otherwise continue... The main idea of finding tangent planes revolves around finding gradient of the corresponding surfaces. (For more info see question no 5). Can you calculate the gradient of the surfaces $latex x^2+y^2+z^2-9$ and $latex z-x^2-y^2+3$ at $latex (2,1,2)$?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0.9"]The gradient of the surfaces $latex f=x^2+y^2+z^2-9$ and $latex g=z-x^2-y^2+3$ at $latex (2,1,2)$ are $latex n_1=f_xi +f_yj+f_zk$ and $latex n_2=g_xi +g_yj+g_zk$ at $latex (2,1,2)$ which is $latex n_1=2xi+2yj+2zk=4i+2j+4k$ and $latex n_2=-2xi-2yj+k=-4i-2j+k$. Now given these two gradients, can you find out the angle between them?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0.9"]$latex n_1=2xi+2yj+2zk=4i+2j+4k$ and $latex n_2=-2xi-2yj+k=-4i-2j+k$.

This follows the cosine angles between two gradient is $latex cos \theta=|\frac{n_1.n_2}{|n_1||n_2|}|=|\frac{-16-4+4}{\sqrt{36 \times 21}}|=\frac{8}{3\sqrt{21}}$

[/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"]

Explanation of hints with graph

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Knowledge Graph

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/drawit-diagram.png" _builder_version="4.0.9"][/et_pb_image][et_pb_code _builder_version="3.26.4"]https://apis.google.com/js/platform.js

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Connected Program at Cheenta

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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://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"]

Finding Tangent plane: IIT JAM 2018 problem 5

What are we learning?

[/et_pb_text][et_pb_text _builder_version="4.0.9" text_font="Raleway||||||||" text_font_size="18px" background_color="#f4f4f4" custom_margin="50px||50px||false|false" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]We will learn to find tangent plane by solving an IIT JAM 2018 Problem. This is the Question no. 5 of the IIT JAM 2018 Solved Paper Series. Go through this link for Question no. 6. Gradient is one of the key concepts of vector calculus. We will use this problem from IIT JAM 2018 to clear our concepts.

[/et_pb_text][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_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="4.0.9" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="50px||50px||false|false" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]The tangent plane to the surface $latex z= \sqrt{x^2+3y^2}$ at (1,1,2) is given by

$x-3y+z=0$
$x+3y-2z=0$
$2x+4y-3z=0$
$3x-7y+2z=0$

[/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_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0.9" 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="4.0.9"]IIT Jam 2018[/et_pb_accordion_item][et_pb_accordion_item title="Key competency" _builder_version="4.0.9" open="off"]Gradient[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0.9" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" open="off"]

Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2

Tom M. Apostol

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Look at the knowledge graph...

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Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0.9"]Given a differentiable function $Z=f(x,y)$, Observe that when we are asked to find a tangent plane at $(x_0,y_0,z_0)$ then the picture that comes in our mind is a plane that touches the curve at a point.

When we are in dimension $2$ it is just a line, (easy to visualize), dim 3 a plane (still visible), dim 4,5,…. a surface which is hard to see, but we can plug in $x=x_0$ in the equation $z=f(x,y)$ to have $z=f(x_0,y)$ which is just a curve in 2D then we can visualize the tangent line at $y=y_0$ is a part of the tangent plane $z=f(x,y)$ isn’t it?? The same thing is true of about the tangent line at $x=x_0$ for the curve $z=f(x,y_0)$. These $f(x,y_0)$ and $f(x_0,y)$ are called sections of the curve $f(x,y)=z$ . Here $(x_0,y_0,z_0)=(1,2,3)$. So, quickly find out $f(1,y)$ and $f(x,1)$.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0.9"]You can see that $f(1,y)=\sqrt{1+3y^{2}}$ and $f(x,1)= \sqrt{x^{2}+3}$ Now observe that the tangent plane of the curve $z=f(x,y)$ is a plane right !! What will be the basic structure of a plane at $(x_0,y_0,z_0)$?

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It is a $a(x-x_0)+ b(y-y_0)+ c(z-z_0)=0$ ----------------------(1) Now see that $(x_0,y_0,z_0)=(1,1,2)$ is already given in the question. Hence the unknown is $(a,b,c)$ . Equation (1) implies $z = z_0+ \frac{a}{c}(x-x_0)+ \frac{b}{c}(y-y_0)$ Differentiating the equation by $x$ we get, $z_x= \frac{a}{c}$ Differentiating the equation by $y$ we get, $z_y= \frac{b}{c}$ Hence the equation of the tangent plane is $z=z_0+z_x|_{(x_0,y_0)}(x-x_0)+ z_y|_{(x_0,y_0)}(y-y_0)$ So calculate $z_x$ and $z_y$ at $(x_0,y_0)$

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$z_x = \frac{d}{dx}f(x,1)= \frac{2x}{2\sqrt{x^{2}+3}}|_{(1,1)} = \frac{2}{4}= \frac{1}{2}$ $z_y=\frac{d}{dy}f(1,y)=\frac{6y}{2\sqrt{1+3y^2}}|_{(1,1)}=\frac{6}{2 \times 2}=\frac{3}{2}$ So the equation of the tangent line is $z= 2+\frac{1}{2}(x-1)+\frac{3}{2}(y-1)$ $\Rightarrow 2z= 4+x-1+3y-3$

$x+3y-2z=0$ (Ans)

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Try to answer this question

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Play with graph

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Application of L'Hopital: TIFR GS 2018 Part A, Problem 2

Understand the problem

True or false: $\lim _{x \rightarrow 0} \frac{\sin x}{\log (1+\tan x)}=1$

Start with hints

Do you really need a hint? Try it first!

"Hint 1"

Observe that the following limit is of the form $\frac 00$.

Do you remember that we always solve the limits of the form $\frac 00$ and $\frac{\infty}{\infty}$ by L’Hospital’s Rule.

Hint 2

Consider $f(x)=sinx$ and $g(x)=log(1+tanx)$
Compute $f ‘(x)=cosx$ and $g ‘(x) = sec^2(x)/(1+tanx)$
Compute the limit of $f ‘(x)/g ‘(x)$

Hint 3

Observe that the $lim f ‘(x)=1$ and $lim g ‘(x) =1 $.
Hence the value of the Limit is $1$.
The statement is therefore TRUE.

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Vector Analysis

Let's discuss a beautiful problem useful for Physics Olympiad based on Vector Analysis.

Vector Analysis Problem:

Let (\vec{a}=6\vec{i}-3\vec{j}-6\vec{k}) and (\vec{d}=\vec{i}+\vec{j}+\vec{k}). Suppose that (\vec{a}=\vec{b}+\vec{c}) where (\vec{b}) is parallel to (\vec{d}) and (\vec{c}) is perpendicular to (\vec{d}). Then (\vec{c}) is
(A)(5\vec{i}-4\vec{j}-\vec{k})
(B)   ( 7\vec{i}-2\vec{j}-5\vec{k})
(C)    (4\vec{i}-5\vec{j}+\vec{k})
(D)    (3\vec{i}+6\vec{j}-9\vec{k})

Discussion:

In the given problem, (\vec{a})=(6\vec{i}-3\vec{j}-6\vec{k})
$$\vec{d}=\vec{i}+\vec{j}+\vec{k}$$ and $$\vec{a}=\vec{b}+\vec{c}...(i)$$
Now, let us consider (\vec{b}=\lambda\vec{d}) and (\vec{c}).(\vec{d})=)0.
Therefore, (b=\lambda\vec{i}+\lambda\vec{j}+\lambda\vec{j})
From (1),
$$6\vec{i}-3\vec{j}-6\vec{k}=\lambda\vec{i}+\lambda\vec{j}+\lambda\vec{k}+c$$
$$\Rightarrow\vec{c}=(6-\lambda)\vec{i}-(3+\lambda)\vec{j}-(6+\lambda)\vec{k}$$
Now,
$$\vec{c}.\vec{d}=0$$
$$\Rightarrow\lambda=-1$$
Hence,
$$\vec{c}=7\vec{i}-2\vec{j}-5\vec{k}$$

Continuous Functions and open sets

Problem: Continuous Functions and open sets

Suppose f be a continuous function from X to Y (where X and Y are domain and range). If Y is a closed set (closed interval if we working in $ R^1 $ ) then can we say that the domain is also closed?

There is a simple counter example. Suppose X = $( -3\pi,\pi)$. Then Y (the range) is [-1, 1]. Surely the range is closed and domain is open. When can we say that if the range is closed, domain is also closed? Only when the function is one-one and onto.

Continuous functions generally give a good picture of the domain set if we know the nature of range set in some detail. For example inverse image of open sets in Y (range) is always open in X (domain). This is easy to prove from the definition of continuity. After all continuity means that for every $\epsilon $ neighborhood V of f(c) we get will get a $\delta $ neighborhood U of c (c being a point in domain) such that when we pick a x from U, f(x) will lie in V. Now let V be a open set containing f(c) (that is it is some neighborhood of f(c) and V is the subset of range set Y). Then $ f^{-1} (V) $ consists of all the points in X (domain set) that maps into V. Let $ x_1 \in f^{-1} (V) $. Thus $ f(x_1) \in V $. Since V is open we will get an $epsilon$ neighborhood M of $ f(x_1) $ which is entirely inside M and such that there exists a $ \delta $ neighborhood N of $ x_1 $ such that all points in N maps to M. In other word N is in $ f^{-1} (V) $. Thus $ f^{-1} (V) $ is open (we have found a open set N containing $ x_1 $ which is entire inside $ f^{-1} (V) $ ).

As we have seen, a continuous function does not necessarily map open sets to open sets (the sin function that we discussed earlier) but preimage of open sets are open. Preimage of closed sets are however not necessarily closed (we have the previous example again to our rescue). However if the continuous function is monotone (that is it is either increasing or decreasing) a lot more can be said about the domain set by looking at the range set and vice versa (you may put some of your observations in the comment section about monotone continuous functions).

Differentiability and Uniform Continuity

Problem: Every differentiable function f: (0, 1) --> [0, 1] is uniformly continuous.

Discussion;

False

Note that every differentiable function f: [0,1] --> (0, 1) is uniformly continuous by virtue of uniform continuity theorem which says every continuous map from closed bounded interval to R is uniformly continuous. However in this case the domain is an open interval.

We can easily find counter example such as $latex f(x) = \sin ( \frac {1}{x} ) $. Intuitively speaking the function oscillates (between -1 and 1) faster and faster as we get close to x = 0. Hence we can get two arbitrarily close values of x such that their functional value's difference equals a particular number (say 1) therefore exceeding any $latex \epsilon < 1 $

An interesting discussion:

Uniform Continuity

Problem: Let f: R --> R be defined by $latex f(x) = sin (x^3) $. Then f is continuous but not uniformly continuous.

Discussion:

True

It is sufficient to show that there exists an $latex epsilon > 0 $ such that for all $latex \delta > 0 $ there exist $latex x_1 , x_2 \in R $ such that $latex | x_1 - x_2 |< \delta $ implies $latex | f(x_1) - f(x_2) | > \epsilon $ .

Assume $latex epsilon = 0.99 $ . Let $latex x_1 = (k \pi )^{\frac{1}{3}} $ and $latex x_2 = (k \pi + \frac{\pi}{2} )^{\frac{1}{3}} $.

Hence $latex |x_1 - x_2 | = (k \pi + \frac{\pi}{2})^{\frac{1}{3}} - (k \pi)^{\frac{1}{3}} < \frac {1}{k^{2/3}} $ . (This is achieved by some simple algebra like rationalization )

Now if we take $latex k > \frac {1}{\delta ^{3/2}} $ then $latex |x_1 - x_2| < \frac {1}{k^{2/3}} < \delta $ but $latex |f(x_1) - f(x_2) | = 1 > 0.99 = \epsilon $

Hence there exists an $latex \epsilon $ (= 0.99) such that for any value of $latex \delta >0 $ we will get $latex x_1 , x_2 $ such that $latex | x_1 - x_2 |< \delta $ implies $latex | f(x_1) - f(x_2) | > \epsilon $ .

Line Integral : IIT JAM 2018 Question Number 20

What are we learning ?

First look at the knowledge graph.

Next understand the problem

Start with hints

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Warm yourself up with an MCQ

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Explanation of hints with graph

Knowledge Graph

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Similar Problems

Finding Tangent plane: IIT JAM 2018 problem 5

What are we learning?

Understand the problem

Look at the knowledge graph...

Start with hints

Try to answer this question

Play with graph

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Similar Problems

Application of L'Hopital: TIFR GS 2018 Part A, Problem 2

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems

Vector Analysis

Continuous Functions and open sets

Differentiability and Uniform Continuity

Uniform Continuity