# What are we learning?

**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

- \(x-3y+z=0\)
- \(x+3y-2z=0\)
- \(2x+4y-3z=0\)
- \(3x-7y+2z=0\)

##### Source of the problem

##### Key competency

##### Difficulty Level

# Look at the knowledge graph…

# Start with hints

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 \(f(x,1)\).

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)

# Try to answer this question

# Play with graph

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