# Multivariable Calculus 1 Problem

• anosh_88
In summary: You've already done the hard part. Now set the gradient of f equal to the normal for the plane and solve for x and y.
anosh_88

## 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.

## 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

Last edited by a moderator:
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
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.

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.

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.

Mod note: some text removed as too much help.

Last edited by a moderator:
Hallsoflvy, no that is the equation for P, not z. z is g(x,y).

anosh_88 said:
Hallsoflvy, no that is the equation for P, not z. z is g(x,y).
What HallsOfIvy was saying is that your equation of the plane is incorrect.

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.
This is really confusing.

To find the equation of the plane that is tangent to the graph of g, you should be working with the partial derivatives ##\frac{\partial g}{\partial x}## and ##\frac{\partial g}{\partial y}##, not "df/dx" and "df/dy" as you wrote. And in general formula should involve g and its two partial derivatives, not f.

So try again for the equation of the tangent plane. The equation should involve terms with x, y, and z, and should NOT have P in it.

Mark, I mean the partial derivative, not the actual full one. I just don't know how to type it in here. P is just a generic variable for the equation.

Regardless of the equation of P, the point at which the plane tangent to f is parallel to P will be where the x and y partial derivatives of f are equal to those of g at the point (1,2,-6), right?

Albuser, yes.

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?

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.
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:
P is just a generic variable for the equation.
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.

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?
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, I have already found the equation tangent to g. However, I'm not sure how I can make the partial derivatives equal to g and solve from there. It would be really difficult and I would not get a concrete answer.

anosh_88 said:
Albuser, I have already found the equation tangent to g.
You found an equation, but as has already been pointed out, you have an error in it.

Okay so can you guys point out the error?

anosh_88 said:
Okay so can you guys point out the error?
Actually, your equation looks OK. I'm not sure what HallsOfIvy was talking about.

The equation for P is z = 4x + 12y - 34

A normal for this plane is <4, 12, -1>
What do you get for the gradient of your other function, z = f(x, y) = 4 - x2 - y2?

The goal here is to find the point on the surface z = f(x, y) so that the gradient is equal to <4, 12, -1>.

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)

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anosh_88 said:
Hallsoflvy, no that is the equation for P, not z. z is g(x,y).
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.

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.
Halls, that is what he said. I guess I just didn't make myself clear. Sorry about that.

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)
What you wrote is incomplete. It is supposed to be an equation, but since there is no =, it's not an equation.

For a function z = f(x, y), the tangent plane at the point (x0, y0) is
z = f(x0, y0) + [∂f/∂x(x0, y0)] (x - x0) + [∂f/∂y(x0, y0)] (y - y0)

A normal to your tangent plane at (1, 2, -6) is <-4, -12, 1>, and these numbers represent the partials of g with respect to x, and y, and the partial of z with respect to z, respectively. For the tangent plane to the graph of f, you want the partials to be equal to these numbers.

Mark, are you saying that that is the answer (-4, -12, 1)?

anosh_88 said:
Mark, are you saying that that is the answer (-4, -12, 1)?
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>.

## 1. What is multivariable calculus?

Multivariable calculus is a branch of mathematics that deals with functions of more than one variable. It involves studying the rates of change and slopes of functions in multiple dimensions, and is essential in fields such as physics, engineering, and economics.

## 2. What types of problems can be solved using multivariable calculus?

Multivariable calculus can be used to solve a wide range of problems, such as optimizing functions with multiple variables, calculating volumes and areas of complex shapes, and finding rates of change and optimization in real-life scenarios.

## 3. What are some common applications of multivariable calculus?

Multivariable calculus has various applications in different fields, including physics (e.g. calculating motion of objects in multiple dimensions), economics (e.g. analyzing consumer behavior in relation to multiple variables), and engineering (e.g. optimizing designs for efficiency and performance).

## 4. What are the main concepts and techniques used in multivariable calculus?

Some of the main concepts and techniques used in multivariable calculus include partial derivatives, multiple integrals, vector calculus, and optimization methods. These tools are used to analyze functions with multiple variables and solve complex problems.

## 5. What are some helpful tips for mastering multivariable calculus?

To master multivariable calculus, it is important to have a strong understanding of single variable calculus, as well as a solid foundation in algebra and geometry. It is also helpful to practice regularly and seek out additional resources, such as textbooks and online tutorials, to supplement your learning.

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