 # What are we learning?

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.

# Understand the problem

The tangent plane to the surface $z= \sqrt{x^2+3y^2}$ at (1,1,2) is given by
1. $x-3y+z=0$
2. $x+3y-2z=0$
3. $2x+4y-3z=0$
4. $3x-7y+2z=0$
IIT Jam 2018
Easy
##### Suggested Book
 Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2 Tom M. Apostol

# Look at the knowledge graph… Do you really need a hint? Try it first!

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 $f(x,1)$.

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 $(x_0,y_0,z_0)$?

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 $(x_0,y_0)$

$z_x = \frac{d}{dx}f(x,1)= \frac{2x}{2\sqrt{x^{2}+3}}|_{(1,1)} = \frac{2}{4}= \frac{1}{2}$ $z_y=\frac{d}{dy}f(1,y)=\frac{6y}{2\sqrt{1+3y^2}}|_{(1,1)}=\frac{6}{2 \times 2}=\frac{3}{2}$ 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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