Parametrized Curve on the Function f(x, y) = x^2 - y^2 + 4?

[icosane]

Homework Statement

6. Show that the parametrized curve r=<t+(1/t),t-(1/t),8> lies on the curve f(x, y) = x^2 - y^2 + 4? Show your calculations.


2. The attempt at a solution

I don't know where to start. I've just been plugging in random x and y and t values and haven't had any luck or insight on how to solve this problem. Any help appreciated.

 

[LCKurtz]

Are you sure the problem is stated correctly? You have given a curve in 3-space and your equation f(x,y) defines a surface in 3-space, not a curve. You might write its equation as:

z = x² - y² + 4

Check whether r(t) lies on that surface (plug in the values).

 

[Galadirith]

Hi icosane,

Well first ill assume that f(x,y) = z. Perhaps this is something that you forgot to include, or perhaps it inst specified in the question, which might then be where all the confusion lies. So considering the two equations:

$$\textbf{r}\ = \ <t+(1/t),t-(1/t),8> \ \ (i)$$

$$f(x, y) = x^2 - y^2 + 4$$

or

$$z = x^2 - y^2 + 4 \ \ (ii)$$

now consider equation (ii), we have three variables, do we have any expression we can substitute for them, do I hear (i) calling. From hear you should be able to show using equation (i) that the components of (i) satisfy (ii). Have a go :D

 

[icosane]

The exact wording of the question is,

6. Which of the following parametrized curves lie on the graph of the function f(x, y) =
x^2 - y2 + 4? Show your calculations.
r1(t) = (t + 1/t)i + (t - 1/t )j + 8k.
r2(t) = sin(t)i + cos(t)j + 4k

This question was written by the professor and it likely was the wording that confused me. I see that if I let f(x,y) = z, then let x = t+1/t, y = t-1/t, and z = 8 * plug this into the original equation x^2-y^2+4 = z everything seems to work out. Thanks guys.