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Line Integral : IIT JAM 2018 Question Number 20

[et_pb_section fb_built="1" _builder_version="4.2.2" width="99.8%" custom_margin="||||false|false" custom_padding="||||false|false"][et_pb_row _builder_version="4.2.2" width="100%" custom_margin="||||false|false" custom_padding="||||false|false"][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_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

What are we learning ?

[/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"]Competency in Focus: Application of Calculus This is problem from IIT JAM 2018 is based on calculation of Line integration of a vector point function along a closed curve.[/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"]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://www.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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Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/amc-8-american-mathematics-competition/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="4.0.9" 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_nav in_same_term="off" _builder_version="4.0.9"][/et_pb_post_nav][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]
[et_pb_section fb_built="1" _builder_version="4.2.2" width="99.8%" custom_margin="||||false|false" custom_padding="||||false|false"][et_pb_row _builder_version="4.2.2" width="100%" custom_margin="||||false|false" custom_padding="||||false|false"][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_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

What are we learning ?

[/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"]Competency in Focus: Application of Calculus This is problem from IIT JAM 2018 is based on calculation of Line integration of a vector point function along a closed curve.[/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"]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://www.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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Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/amc-8-american-mathematics-competition/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="4.0.9" 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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