The discussion centers on finding the point on the graph of the function f(x,y) = 4 - x² - y² that has a tangent plane parallel to the tangent plane P of the function g(x,y) = 8 - 2x² - 3y² at the point (1, 2, -6). The tangent plane P is correctly identified as z = 4x + 12y - 34. Participants clarify that the equation for P should not include the variable P itself and emphasize the importance of using the correct partial derivatives of g to find the tangent plane of f. The goal is to equate the gradients of both functions to find the desired point on f.
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How did you find the plane P? Note that P is just a name for the plane - P should not be in the equation of the plane.anosh_88 said:Homework Statement
Let P be the tangent to the graph of g(x,y) = 8-2x^2-3y^2 at the point (1, 2, -6). Let f(x,y) = 4-x^2-y^2. Find the point on the graph of f which has tangent plane parallel to P. [/B]Homework Equations
g(x,y) = 8-2x^2-3y^2 at (1, 2, -6)
f(x,y) = 4-x^2-y^2
The Attempt at a Solution
I found P, which is P = 4x+12y-34
You mean, I presume, z= 4x+ 12y- 34. But that is incorrect. The first, g, graph is z= 8- 2x^2- 3y^2 or 2x^2- 3y^2+ z= 8.anosh_88 said:Homework Statement
Let P be the tangent to the graph of g(x,y) = 8-2x^2-3y^2 at the point (1, 2, -6). Let f(x,y) = 4-x^2-y^2. Find the point on the graph of f which has tangent plane parallel to P. [/B]Homework Equations
g(x,y) = 8-2x^2-3y^2 at (1, 2, -6)
f(x,y) = 4-x^2-y^2
The Attempt at a Solution
I found P, which is P = 4x+12y-34
What HallsOfIvy was saying is that your equation of the plane is incorrect.anosh_88 said:Hallsoflvy, no that is the equation for P, not z. z is g(x,y).
This is really confusing.anosh_88 said:Took df/dx, df/dy, and I know that g(x,y) = 8-2x^2-3y^2. I also followed the general formula, which is: f(x,y)+[df/dx(x,y)](x-x0)+[df/dy(x,y)](y-y0). In the derivatives, I plugged in the point.
And those are what I wrote, partial derivatives of g, ##\frac{\partial g}{\partial x}## and ##\frac{\partial g}{\partial y}##. If you're working with g, you shouldn't write the partials of the general formula in terms of f. That was my point.anosh_88 said:Mark, I mean the partial derivative, not the actual full one. I just don't know how to type it in here.
P is NOT a variable. It is only a symbolic name for the plane. You could call the plane Fred instead of P, but the equation of the plane will not involve Fred or P.anosh_88 said:P is just a generic variable for the equation.
The OP has already found the partial derivatives, and has attempted to find the equation of the tangent plane, but has an error in his equation of the plane.albuser said:So why not simply take the partial derivatives of g at that point, find the general partial derivatives of f for any x and y and set the two equal to each other?
You found an equation, but as has already been pointed out, you have an error in it.anosh_88 said:Albuser, I have already found the equation tangent to g.
Actually, your equation looks OK. I'm not sure what HallsOfIvy was talking about.anosh_88 said:Okay so can you guys point out the error?
No, you are the one who is wrong. You wrote P= 4x+ 12y- 34. That is impossible. P, on the left, is a plane while 4x+ 23y- 34, on the right, is a number. The cannot be equal, they are different kinds of things.anosh_88 said:Hallsoflvy, no that is the equation for P, not z. z is g(x,y).
Halls, that is what he said. I guess I just didn't make myself clear. Sorry about that.No, you are the one who is wrong. You wrote P= 4x+ 12y- 34. That is impossible. P, on the left, is a plane while 4x+ 23y- 34, on the right, is a number. The cannot be equal, they are different kinds of things.
What is true is that the plane, P, is given by the equation z= 4x+ 12y- 34.
What you wrote is incomplete. It is supposed to be an equation, but since there is no =, it's not an equation.anosh_88 said:Mark, this is exactly where I got stuck: on where to go after I found the gradient of g(x,y); and how I can connect that to f(x,y) in order to find what the point f is. The gradient of f(x,y) is:
∂f/∂x = -2x
∂f/∂y = -2y
f(x,y) = z = 4 - 2x2 - 2y2
That is as far as I got and then I didn't know how to proceed.
All I know is that the general form is
f(x0, y0) + [∂f/∂x(x0, y0)] (x - x0) + [∂f/∂y(x0, y0)] (y - y0)
No, that's not what I said. I said that a normal to the tangent plane was <-4, -12, 1>. You need to find the point on the graph of z = f(x, y) at which the tangent plane to this surface is also perpendicular to <-4, -12, 1>.anosh_88 said:Mark, are you saying that that is the answer (-4, -12, 1)?