Question

Which of the following surfaces matches the level curves below? ​

Answer 1

Observe that
`y=\pm x` form a level curve for the surface in question. This means that if we substitute
`y=x` or
`y=-x` in to the equation for the surface, then
`z=f(x,y)=c` is a constant.

This is only true for the surface in option A.

• 
  `z = x^2 - y^2 + 6`

`y = x \implies z = x^2 - x^2 + 6 \implies z = 6`

`y = -x \implies z = x^2 - (-x)^2 + 6 \implies z = 6`

`z` is constant, so both of
`y=\pm x` are level curves.

• 
  `z^2 = x + y + 6`

`y=x \implies z^2 = x + x + 6 \implies z = \pm√(2x + 6)`

`y=-x \implies z^2 = x + (-x) + 6 \implies z = \pm\sqrt6`

`y=x` is not a level curve.

• 
  `z = √(x^2 + y^2 + 6)`

`y = x \implies z = √(x^2 + x^2 + 6) = √(2x^2 + 6)`

`y = -x \implies z = √(x^2 + (-x)^2 + 6) = √(2x^2 + 6)`

Neither are level curves.

• 
  `z = √(x^2 + y^2)`

`y = x \implies z = √(x^2 + x^2) = \sqrt2 |x|`

`y=-x \implies z = √(x^2 + (-x)^2) = \sqrt2 |x|`

Again, neither are level curves.

Answer 2

(a) z = x² -y² +6

Explanation:

Each of the equations can be interpreted as describing a family of curves based on the value of z.

(a)

The difference of squares will give rise to hyperbolic curves, matching the given diagram

(b)

The sum of linear terms will give rise to straight lines.

(c)

Squaring both sides gives ...

z² -6 = x² +y²

which gives rise to a family of circles.

(d)

This is the same formula as (c), but with circles of a different radius.

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Additional comment

The standard form equation for a hyperbola is ...

(x/a)² -(y/b)² = 1 . . . . . centered at the origin, with semi-axes 'a' and 'b'.

Here, we have a=b=√(z-6).