Find the equation of the tangent plane to $xy+yz+zx=11$ when $x=1$ and $y=2$ giving the answer in the form $f(x,y,z)=k$, where $k$ is a constant and $k\in \Bbb{Z}$.
So I know that the tangent plane at a given point $(a,b,f(a,b))$ is given by: $$z=f(a,b) +(x-a)f_x(a,b)+(y-b)f_y(a,b)$$
But would someone be able to explain to me how I would go about solving this as I understand the basics of partial differentiation but not implicit partial differentiation. Any help would be appreciated.
On
Hint:
If $xy+yz+zx=11$, then we can think of $z$ as a function of $x$ and $y$, namely $$z=f(x,y)=\frac{11-xy}{x+y}$$ Now, use what you know about partial differentiation and the above formula.
On
The normal to a surface specified by an equation of the form $$f(x,y,z)=\text{constant}$$ is given by $\nabla f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)$.
In this case, $\nabla f=(y+z, z+x, x+y)$. And when $x=1$ and $y=2$, $z$ is equal to $3$, so $\nabla f=(5,4,3)$.
So all you need to do is find the equation of a plane through $(1,2,3)$ and normal to $(5,4,3)$.
On
$f(x,y,z):=xy+yz+zx-11=0.$
At $x=1,y=2 :$
$2+2z+z=11; z=3;$
Tangent plane passes through $(1,2,3).$
Normal to $f(x,y,z)=0:$
$\vec n= \nabla f(x,y,z)= (f_x,f_y,f_z)=$
$(y+z,x+z,x+y).$
At $(1,2,3):$ $\vec n= (5,4,3).$
Equation of plane:
$\vec n (\vec r -\vec r_0)=0;$
$(5,4,3)((x,y,z)-(1,2,3))=0;$
$5x+4y+3z -5-8-9=0;$
$5x+4y+3z=22.$
The equation of the tangent plane to the surface $F(x,y,z) = 0$ at the point $(x_0,y_0,z_0)$ is given by $$\nabla F(x_0,y_0,z_0)\cdot\big((x,y,z) - (x_0,y_0,z_0)\big)=0$$
In your case $F(x,y,z) = xy+yz+zx-11$ so
$$\nabla F(x,y,z) = (y+z,x+z,x+y)$$
If $x_0=1$ and $y_0=2$, we have $$11 = F(x_0,y_0,z_0)=2 + 2z_0 + z_0 \implies z_0=3$$ so our point is $(x_0,y_0,z_0) = (1,2,3)$.
Hence the equation of the tangent plane is $$0 = \nabla F(1,2,3)\cdot\big((x,y,z) - (1,2,3)\big)=(5,4,3)\cdot(x-1,y-2,z-3) = 5(x-1)+4(y-2)+3(z-3)$$ We can rewrite this as
$$5x+4y+3z=22$$