How do I determine a constant (call it 'c') so that the plane $$2x+2y+z=c$$ is a tangent to the surface $$x+y^{2}+z^{4}=1$$ ? This is a problem in _multiple variable calculus_

Question

abb0t · Answer

I think you start by finding the gradient first and some points, since the derivative IS the tangent. So partial derivatives.

anonymous · Answer

The gradient for the surface gives me
$$\nabla f=(1,2y,4z^{3})$$

But the plane equation gives me
$$\frac{df}{dx}(x-j)+\frac{df}{dy}(y-k)+\frac{df}{fz}(z-l)=0$$ for some arbitrary points $$j,k,l \in R$$

Which will give me the tangent plane of the surface:
$$(x-j)+2y(y-k)+4z^3(z-l)=0$$

From now on I'm lost: 
But according to the plane given:
$$1*(x-j)=2x (\rightarrow x=-j)$$
$$2y(y-k)=2y \rightarrow y-k=1 (\rightarrow y=k+1)$$
$$4z^3(z-l)=z \rightarrow 4z^ 3=4z^ 2l$$

phi · Answer

I like to think of the equation of a plane as the dot product of its normal with point P= (x,y,z)

$$ \overrightarrow{N}\cdot \overrightarrow{P} = c$$
your equation
$$ 2x+2y+z=c $$
has a normal  (2,2,1)

the gradient of a surface is normal to the surface.  so you want to find the (x,y,z) point on the surface where the gradient points in the same direction as the normal to the plane.

we want 
$$ (1,2y,4z^3) = a(2,2,1) $$
where $a$ is a scalar.  to get 1 = 2a, we set a =1/2
we want 2y = 1  and  4 z^3 = 1/2
we get y=1/2  and z= 1/2
use the original equation
$$x+y^2+z^4=1 $$
to find x

once you have x,y,z coordinates of the point on the surface that has the correct gradient,  plug those values into the equation of the plane to find c

If I did the arithmetic correctly (no guarantee)  x= 11/16

ant the point is (11,8,8)/16
which gives  c = 23/8

anonymous · Answer

Amazing. Just amazing.