[/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"]
| Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2 | Tom M. Apostol |
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 \((x_0,y_0,z_0)\)?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0.9"]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 \((x_0,y_0)\)
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0.9"]\(z_x = \frac{d}{dx}f(x,1)= \frac{2x}{2\sqrt{x^{2}+3}}|_{(1,1)} = \frac{2}{4}= \frac{1}{2}\)\(z_y=\frac{d}{dy}f(1,y)=\frac{6y}{2\sqrt{1+3y^2}}|_{(1,1)}=\frac{6}{2 \times 2}=\frac{3}{2}\)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)
[/et_pb_tab][/et_pb_tabs][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="20px||20px||false|false" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]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"]

In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]

Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.

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