# Math Help - Tangent Planes + Linear Approximation

1. ## Tangent Planes + Linear Approximation

1. Find the points on the graph of $z = x^2e^y$ at which the tangent plane is parallel to $5x - 2y + \frac{1}2z = 0$
Can someone explain this to me/start this for me?

2. Suppose that the plane tangent to the surface $z = f(x,y)$ at $(-2,3,4)$ has equations $z + 4x + 2y = 2$. Estimate $f(-2.1, 3.1)$
Same as above; I'm not really sure where I'm supposed to start. Do I integrate $-(4x + 2y) + 2$ or something?

2. So here's what I was vaguely told:
1. Tangent Plane is always z = L( x,y )
2. Knowing that "think of when they are parallel"

So if a plane is parallel that just means that it has an extra + c or something too?

$F(x,y,z) = x^2e^y-z = 0$
Let $\vec{u_1}$ be the unit vector of <10,-4,1> (I multiplied all of the components by 2) and let $\vec{u_2}$ be the unit vector of the gradient vector of F(x,y,z)
Set the components of $\vec{u_2}$ and $\vec{u_1}$ equal to each other and solve for the values. You get 3 equations and you have 3 variables.
An alternative method would be to solve both equations $(z=x^2e^y$ and $10x-4y+z=0$) for z and set them equal to each other. See where that gets you.

2. Just plug in -2.1 for x and 3.1 for y in the equation $z+4x+2y=2$ and solve for z.

4. Originally Posted by ZeroVector
So here's what I was vaguely told:
1. Tangent Plane is always z = L( x,y )
2. Knowing that "think of when they are parallel"

So if a plane is parallel that just means that it has an extra + c or something too?
No, it doesn't.

Two planes, Ax+ By+ Cz= D (which would be the same as z= (D- Ax- By)/C) and Px+ Qr+ Rz= S (which would be the same as z= (S- Px- Qy)/R) are parallel if and only if their normal vectors, <A, B, C> and <P, Q, R> are parallel- and that means that one is a multiple of the other.

A plane, Ax+ By+ Cz= D is tangent to a surface f(x, y, z)= constant, if and only if the normal vector to the plane, <A, B, C>, is parallel to the normal vector of the surface which is the same as the gradient of the function that lilaziz1 mentioned: $\left<\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right>$