[/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"]
[/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"]
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"]
[/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"]
[/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"]
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"]
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"]
