How to calculate parametric representation of a circle?

[geft]

Homework Statement

y^2 + 4y + z^2 = 5, x = 3

Homework Equations

The Attempt at a Solution

I know that the calculated coordinates must satisfy the above equation, but I don't know how to go about solving for those coordinates. The best I could do was to equate z = \sqrt{(-y + 5)(y + 1)}.

 

[CompuChip]

Hmm,
y^2 + 4y + z^2 = 5
doesn't really look like a circle, does it?
I mean, the general equation for a circle (in the y-z plane) is
(y - a)^2 + (z - b)^2 = r^2

Can you maybe start by rewriting it to this form?

For the next step, remember the usual parametrisation for a circle y^2 + z^2 = r^2 is
y(t) = r \sin t, z(t) = r \cos t

 

[geft]

Oh! By completing the square, I get

(y + 2)^2 + (z)^2 = 3^2

r(t) = (3, 3 \sin t, 3 \cos t)

Is this correct?

 

[CompuChip]

Almost!
The first line is right. Just think very carefully about what you call sin(t).
When in doubt, calculate
(3 \sin t)^2 + 4 (3 \sin t) + (3 \cos t)^2
and check if it gives 5.

 

[geft]

Is it r(t) = (3, 3 \sin t - 2, 3 \cos t)?

Why isn't it 3 sin t + 2? I had to change the sign to get the answer to fit.

 

[CompuChip]

Very well done.

The sign is there, because you want 3 (y + 2/3) to be 3 cos(t).

 

[geft]

Thank you very much. I have another (somewhat) related question, but I'd rather not create a new thread. I certainly hope you don't mind.

Homework Statement

Given a curve C: r(t), find a tangent vector r'(t), a unit tangent vector u'(t), and the tangent of C at P.

r(t) = (\cosh t, \sinh t), P: (\frac{5}{3}, \frac{4}{3})

Homework Equations

The Attempt at a Solution

r'(t) = (\sinh t, \cosh t)
u'(t) = (\cosh 2x)^{-1/2}(\sinh t, \cosh t)

The last part stumps me.

 

[CompuChip]

You mean about u'(t)?
It is simply a unit vector along r'(t), so you will have to divide r'(t) by its length.
To get the cosh(2x) there is just some (hyperbolic) identities magic using the formulas analagous to \cos^2 t + \sin^2 t = 1 and \cos^2 t - \sin^2 t = \sin 2t.