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Question: How do you find the equations for the tangent plane to the surface \[g\left( x,y \right)={{x}^{2}}-{...

How do you find the equations for the tangent plane to the surface g(x,y)=x2−y2g\left( x,y \right)={{x}^{2}}-{{y}^{2}} through (5,4,9)\left( 5,4,9 \right)?

Explanation

Solution

To find the equations for the tangent plane to the surface g(x,y)=x2−y2g\left( x,y \right)={{x}^{2}}-{{y}^{2}} through (5,4,9)\left( 5,4,9 \right). We have to write our given equation into an equation which contains x, y, z planes and which is equivalent. Then we have to find the gradient vector field at the given point. Since the coefficients of gradient vector u.

Complete step by step solution:
To find the equations for the tangent plane to the surface g(x,y)=x2−y2g\left( x,y \right)={{x}^{2}}-{{y}^{2}} through (5,4,9)\left( 5,4,9 \right).
Let us consider the given equation as equation (1).
g(x,y)=x2−y2............(1)g\left( x,y \right)={{x}^{2}}-{{y}^{2}}............\left( 1 \right)
Now let us rewrite equation (1) and replace it with an equivalent equation (1) which contains the Z-plane.
So, the equation (1) in the equivalent form
⇒G(x,y,z)=x2−y2−z=0\Rightarrow G\left( x,y,z \right)={{x}^{2}}-{{y}^{2}}-z=0
Let us consider the above equation as equation (2)
G(x,y,z)=x2−y2−z=0........(2)G\left( x,y,z \right)={{x}^{2}}-{{y}^{2}}-z=0........\left( 2 \right)
As we know the gradient field vector for a basic equation i.e. ax+by+cz=0ax+by+cz=0 i.e. G(x,y,z)=(a,b,c)G\left( x,y,z \right)=\left( a,b,c \right).
Therefore, by applying the above concept to the equation (2) we try to find the gradient vector field by differentiating the equation with respect to z.
⇒∇G(x,y,z)=(2x,−2y,−1).........(3)\Rightarrow \nabla G\left( x,y,z \right)=\left( 2x,-2y,-1 \right).........\left( 3 \right)
Now we put the value of the point of (5,4,9)\left( 5,4,9 \right) in the gradient field to get
⇒∇G(x,y,z)=(2×5,(−2)×4,−1)=(10,−8,−1)........(4)\Rightarrow \nabla G\left( x,y,z \right)=\left( 2\times 5,\left( -2 \right)\times 4,-1 \right)=\left( 10,-8,-1 \right)........\left( 4 \right)
As we know that the gradient vector is perpendicular (normal) to the graph at this point, its components can be used as coefficients for the variables in the equation of the tangent plane.
Therefore, we can write the equation as

& \Rightarrow 10\left( x-5 \right)-8\left( y-4 \right)-\left( z-9 \right)=0 \\\ & \Rightarrow 10x-50-8y+32-z+9=0 \\\ & \Rightarrow 10x-8y-z=9 \\\ \end{aligned}$$ Let us consider the above equation as equation (5). $$10x-8y-z=9.........\left( 5 \right)$$ Therefore, the equations for the tangent plane to the surface $$g\left( x,y \right)={{x}^{2}}-{{y}^{2}}$$ through $$\left( 5,4,9 \right)$$ is $$10x-8y-z=9$$. **Here how is the value of $$G\left( -10,-8,-1 \right)$$.** **Note:** We should note a point that the gradient vector field is perpendicular to the graph at the given point but its components are used to find tangent equations. While solving the equation $$10\left( x-5 \right)-8\left( y-4 \right)-\left( z-9 \right)=0$$ students should be careful at multiplication.