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 $$\cos^2 t - \sin^2 t = \sin 2t$$.