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# What are we learning?

Gradient is one of the key concepts of vector calculus. We will use this problem from IIT JAM 2018 will use these ideas

# 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

# 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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