Eliminating the parameter for a vector valued function: R(t) = cosh(t)i + sinh(t)j

[MathRookie]

Hello! I am a math newbie and this is my first post. I am asked to eliminate the parameter of the vector valued function: R(t) = cosh(t)i + sinh(t)j

Here is what I have done so far:
t = arccosh(x)
y = sinh(arcosh(x))

Seems fairly straightforward, but I would like someone to confirm that I have done this correctly.

Next, I need to graph y = sinh(arcosh(x)). Also pretty straightforward; it is a sinusoidal, vertical curve that begins at (1, 0) and extends in the positive y direction.

The next part is what has me confused. I am asked to graph the vector valued function using my calculator set in parametric mode, from -3 <= t <= 3. The result is something that looks like half of a hyperbola and doesn't at all resemble my first graph. Shouldn't they look the same? Or have I done something wrong?

 

[Deleted member 4993]

Hello! I am a math newbie and this is my first post. I am asked to eliminate the parameter of the vector valued function: R(t) = cosh(t)i + sinh(t)j

Here is what I have done so far:
t = arccosh(x)
y = sinh(arcosh(x))

Seems fairly straightforward, but I would like someone to confirm that I have done this correctly.

Next, I need to graph y = sinh(arcosh(x)). Also pretty straightforward; it is a sinusoidal, vertical curve that begins at (1, 0) and extends in the positive y direction.

The next part is what has me confused. I am asked to graph the vector valued function using my calculator set in parametric mode, from -3 <= t <= 3. The result is something that looks like half of a hyperbola and doesn't at all resemble my first graph. Shouldn't they look the same? Or have I done something wrong?

What would you have done:

If it was:

R(t) = r_o*cos(t)i + r_o*sin(t)j

How would you eliminate 't'. Hint use sin²(t) + cos²(t) = 1

What is the similar relationship in hyperbolic function?

For your problem use, cosh2(t) - sinh2(t) = 1

 

[MathRookie]

Thank you for replying! I realized some time after my post what I was missing. In hyperbolic trig function, sinh^2(t) - cosh^2(t) = 1. Letting x = cosh(t) and y = sinh(t), I have x^2 - y^2 = 1, which is a hyperbola, so it fits with the result in my calculator.

However, the result in my calculator only displays the right half of the hyperbola. I don't understand why the left half is missing.

I also don't understand why my original strategy of solving for t did not work!