Can You Solve These Advanced Calculus Problems?

[jherasjr]

Calculus Questions - HELP!

1. Consider the cube determined by the planes x=-1, x=3, y=5, y=9,
z=0, and z=4.

a) Give the coordinates of the eight vertices and center of
the cube.

b) Determine an equation of the largest sphere contained in
the cube.

c) Determine an equation of the largest sphere that would fit
between the sphere found in (b) and the cube.

2. Determine an equation of a plane that intersects the plane
x+y+z=3 at an angle of 60 degrees.

3. a) Determine an equation of the tangent plane to the surface
given by x^2*y+y^2*z+z^2*x=1 at the point (1,0,1).

b) Determine an equation of the line that is normal to the
surface given by x^2*y+y^2*z+z^2*x=1 at the point (1,0,1).

4. Prove that u+v+w=0, then u*v=v*w, and u*w=w*v. What is the
geometric interpretation of these relationships?

5. Suppose f(x,y)=A*X^3+B*X*Y+C*Y^2, where A, B, and C are
constants. For what values of A, B, and C does f have a
critical value at (-2,1)? Determine what type of critical
point it is.

6. Determine the maximum value of f(x,y,z)=(x*y*z)^2 subject to
the constraint x^2+y^2+z^2=c^2, where c is not equal to zero.

 

[HallsofIvy]

Why is this not in the homework section?

Presuming they are homework, show us what you have done on these problems.

 

[M98Ranger]

1. a) The eight vertices of the cube are (-1,5,0), (-1,5,4), (-1,9,0), (-1,9,4), (3,5,0), (3,5,4), (3,9,0), (3,9,4). The center of the cube is (1,7,2).

b) The largest sphere contained in the cube has a radius of 2 and is centered at (1,7,2). Therefore, its equation is (x-1)^2 + (y-7)^2 + (z-2)^2 = 4.

c) The largest sphere that would fit between the sphere found in (b) and the cube would have a radius of 1 and be centered at (1,7,2). Therefore, its equation is (x-1)^2 + (y-7)^2 + (z-2)^2 = 1.

2. To find the equation of a plane that intersects x+y+z=3 at an angle of 60 degrees, we need to find a vector that is perpendicular to the plane. We can use the cross product of two vectors in the plane, for example (1,0,1) and (1,-1,0) to find this perpendicular vector. This gives us the vector (1,-1,-1). The equation of the plane can then be written as x-y-z=d, where d is a constant. To find d, we can plug in the coordinates of the point of intersection (1,1,1) into the equation, giving us d=3. Therefore, the equation of the plane is x-y-z=3.

3. a) To find the equation of the tangent plane at the point (1,0,1), we can use the gradient of the surface given by x^2*y+y^2*z+z^2*x=1. The gradient is ∇f = (2xy+z^2, x^2+2yz, 2zx+y^2). Plugging in the coordinates of the point (1,0,1), we get ∇f(1,0,1) = (1,1,1). Therefore, the equation of the tangent plane is x+y+z=3.

b) The normal vector to the surface at the point (1,0,1