Multivaraible Calculus. surfaces in R2 andR3

[ScullyX51]

Homework Statement

In class, we studied functions ⃗r : I → R^3, where I ⊂ R is some interval.
Let us now consider a function
⃗r : U → R^3, U ⊂ R^2
That is, we have a function, ⃗r, which sends a point (u, v) in the plane to a
point (vector) in R3 . You may call it a “vector function of two arguments”
or a “vector function of a vector argument”. To be specific:
Let
U = {(u, v), v > 0} ⊂ R^2
⃗r(u, v) = (√v/2 cosh u, √v sinh u, v)
or, equivalently,
x = √v/2 cosh u
y = √v sinh u
z = v

1.Verify that the points ⃗r(u, v) satisfy the equation z = 4x^2− y^2
.
2.Identify the surface S given by the equation z = 4x^2-y^2
Is it an ellipsoid, paraboloid, hyperboloid (and which one), cone, cyllinder?

Homework Equations

The Attempt at a Solution

I know that the surface given by the equation is a hyperbolic parabloid, but I have no idea how to approach showing that the points satify the equation. When I tried just plugging the values of x,y, and z into the equation I end up with something very messy, and I'm not quite sure what I am supposed to be solving for. If anyone could explain this to me it would be great!

 

[mighty2000]

Hello! Thank you for your post. Let's take a look at the first part of the problem, verifying that the points ⃗r(u, v) satisfy the equation z = 4x^2− y^2.

To do this, we can simply substitute the values of x, y, and z from the given vector function ⃗r(u, v) into the equation z = 4x^2− y^2. So, we have:

z = v = 4(√v/2 cosh u)^2 - (√v sinh u)^2
z = v = 4v/4 cosh^2 u - v sinh^2 u
z = v = v cosh^2 u - v sinh^2 u
z = v(cosh^2 u - sinh^2 u)

Now, we can use a trigonometric identity to simplify cosh^2 u - sinh^2 u. The identity is:

cosh^2 x - sinh^2 x = 1

So, we have:

z = v(1)
z = v

And we can see that this is indeed equal to z, which is what we wanted to show. So, the points ⃗r(u, v) do satisfy the equation z = 4x^2− y^2.

Now, for the second part of the problem, we need to identify the surface S given by the equation z = 4x^2− y^2. This is a hyperbolic paraboloid. We can see this by rearranging the equation to get:

z + y^2 = 4x^2

This can be rewritten as:

(z/4) + (y/2)^2 = x^2

This is the standard form of a hyperbolic paraboloid. So, we can conclude that the surface S is a hyperbolic paraboloid. I hope this helps! Let me know if you have any further questions or need clarification.