Quiz 4: Curves and surfaces in 3-dimensional space

This is an ellipse with centre (1,2) and with axes of length 4 in the x–direction and 6 in the y–direction.

The curve is an ellipse centre (1,2) with axes length 6 in the x direction and 4 in the y direction.

This is an ellipse with centre (-1,-2) and with axes of length 6 in the x–direction and 4 in the y–direction.

This is an ellipse with centre (-1,-2) and with axes of length 4 in the x–direction and 6 in the y–direction.

This curve has the following graph.

In this curve the z values increase as the square of t, making the spiral much ‘steeper’ than the curve shown in the question. The parametric equations given in this option correspond to the following graph.

This curve has the following graph.

The curve circles around the z-axis and the distance of the curve from the z-axis increases linearly with t (and hence z).

As t increases from large negative values through zero to large positive values, y starts out negative then becomes positive, then negative and finally positive again. This fits the curve shown. It is actually quite hard to “see” where in space this curve lies, from the diagram provided, so don’t worry if you didn’t get this one correct.

The cost and sint terms make this curve wind around the z–axis, and the z = t³ term makes the height of the curve above the xy–plane increase rapidly.

This is a “half” straight line because z = y = x + 1, for all t. (It is only half a line because x ≥-1, y ≥ 0 and z ≥ 0 since t² ≥ 0 for all t.) (The box is there to help visualize the line).

This cannot be the graph of f(x,y). Look again at the formula for f(x,y)! It tells us that f(x,y) ≥-1 for all x,y. In fact, this option is the graph of z = 1 - (x² + y²).

This is the graph of a regular cone (it has “straight sides”), whereas f(x,y) is a paraboloid. In fact, this is the graph of z = .

Even though this is the graph of a paraboloid it cannot be the graph of z = f(x,y) since f(0,0) = -1 whereas this graph goes through (0,0,0). In fact, this is the graph of z = x² + y².

This is the only graph listed which is paraboloid passing through (0,0,-1).

The function f(x,y) = 4 - (x² + y²) is an inverted paraboloid. It has the following graph.

This function does not depend on x. Its graph is:

This surface lies entirely below the xy–plane so it cannot be the function pictured. Its graph is

This graph has a parabolic cross-section z = -x² which does not depend on the value of y.

This surface looks like an egg carton, as shown below. The surface in the question can’t be given by the equation z = cosxcosy because in the plane x = y, for example, we have z = cos²x and so z ≥ 0, which is not the case.

As sinxsiny = cos(x + )cos(y + ) this surface is just a shifted version of the surface z = cosxcosy.

As cosxsiny = cos(x)cos(y + ) this surface is just another shifted version of the surface z = cosxcosy.

This function is a surface of revolution, so it could be the graph in the question. However, the ripples in this function get closer together as gets large, so it cannot be the right answer.

This is a surface of revolution rotated about the z-axis and its equation is given by either z = cos or z = sin. When x = y = 0 we have z = 1 (from the graph) so it is z = cos.

This is very similar to the graph in the question; however, this cannot be the right function because z = 0 when (x,y) = (0,0) which does not agree with the graph in the question.

This cannot be the right graph because as . In fact, the function has the graph:

This function looks like a very “steep” paraboloid of revolution. It cannot be the function in the question because z = 0 when (x,y) = (0,0). In fact, has the following graph.

This function looks like a very “steep” inverted paraboloid of revolution. It cannot be the function in this option because here, z is always less than 1 and becomes very large and negative as x² + y² increases. In fact, has the following graph.

This is a surface of rotation about the z-axis. As x² + y² →∞ as (x,y) → (0,0) it is the only function listed which is consistent with the graph in the question.

These level curves are ellipses with equations of the form + = 1, which are not the same equations as the level curves for the function with equation z = x(x + y).

The level curve at height z = c satisfies the equation c = x(x + y); that is, .

These level curves appear to have an asymptote at x = 2 whereas the level curves of z = f(x,y) have an asymptote at x = 0.

These level curves are ellipses with equations of the form + = 1, which are not the same equations as the level curves for the function with equation z = x(x + y).

The level curves of this function are circles centred at the origin.

The level curves of this function are ellipses centered at (-1,-2).

The level curves of this function are ellipses centered at (1,2). The semi-major axis of each ellipse is vertical and the semi-minor axis is horizontal. That is, the ellipses are taller than they are wide, and so this option doesn’t match the given set of curves.

The level curves of this function are ellipses centered at (1,2). The semi-major axis of each ellipse is horizontal and the semi-minor axis is vertical. The ratio of these two axes is 3 : 2, matching that shown on the set of curves.

For example, these could be some of the level curves of the surface given by .

For example, these could be some of the level curves of the surface given by .

For example, these could be some of the level curves of the surface given by .