Describe the level surfaces of the function. f(x,y,z)=x+3y+5z

wurmiana6d

wurmiana6d

Answered question

2021-11-13

Describe the level surfaces of the function.
f(x,y,z)=x+3y+5z

Answer & Explanation

mylouscrapza

mylouscrapza

Beginner2021-11-14Added 22 answers

f(x,y,z)=x+3y+5z 
k is constant k
x+3y+5z=k 
Equation of a plane is ax+by+cz=d, with a surface normal of <a,b,c>

Varying l wi;; result in different parallel planes. So f describes a familly of parallel planes. 

Eliza Beth13

Eliza Beth13

Skilled2023-05-14Added 130 answers

Answer:
x+3y+5zc=0
Explanation:
To begin, we can set up the equation x+3y+5z=c. This equation represents a plane in three-dimensional space. The level surfaces of the function are precisely these planes for different values of c.
Now, let's rewrite the equation in standard form to get a clearer picture of the level surfaces. We rearrange the terms to obtain x+3y+5zc=0. This form allows us to identify the coefficients of x, y, and z as 1, 3, and 5, respectively.
We can represent the equation as follows:
x+3y+5zc=0
Thus, the level surfaces of the function f(x,y,z)=x+3y+5z are planes in three-dimensional space, given by the equation x+3y+5zc=0, where c represents the constant value.
Don Sumner

Don Sumner

Skilled2023-05-14Added 184 answers

Step 1:
To describe the level surfaces of the function f(x,y,z)=x+3y+5z, we need to find the values of x, y, and z that satisfy the equation f(x,y,z)=k, where k is a constant.
Let's start by setting up the equation:
x+3y+5z=k
Step 2:
To visualize the level surfaces, we can rewrite the equation in terms of z as the dependent variable:
z=kx3y5
This equation represents a plane with normal vector (1,3,5) passing through the point (x,y,kx3y5).
The level surfaces of the function are therefore planes with varying values of k. Each plane is parallel to the plane defined by the equation x+3y+5z=0, which passes through the origin.
By varying the value of k, we can obtain different parallel planes. When k=0, the plane passes through the origin and represents the level surface corresponding to the value k=0. For other values of k, the plane will be shifted parallel to the original plane.
In summary, the level surfaces of the function f(x,y,z)=x+3y+5z are a family of parallel planes with varying values of k, all of which are parallel to the plane defined by the equation x+3y+5z=0.
madeleinejames20

madeleinejames20

Skilled2023-05-14Added 165 answers

By setting f(x,y,z) equal to c, we can find the equations of the level surfaces.
x+3y+5z=c
This equation represents a plane in three-dimensional space. For each value of c, there is a corresponding level surface that is a plane parallel to the original plane. The value of c determines the distance of the plane from the origin.
By rearranging the equation, we can express it in the general form of a plane equation:
x+3y+5zc=0
In this form, the coefficients of x, y, and z represent the normal vector to the plane.
Thus, the level surfaces of the function f(x,y,z)=x+3y+5z are planes with equations of the form x+3y+5zc=0, where c is a constant.

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