MATH 251 Lecture 1

« previous | Wednesday, January 18, 2012 | next »

Multivariable Calculus

James Vargo
math.tamu.edur/~vargo/courses/m251.html
BLOC 620c

(No TA)

No calculators on exams.

Homework

• WebAssign homework
• Written homework (due each Wednesday)

The Celestial Sphere

Vectors and Trigonometry

Spherical Coordinates

Define a point in space using

• $\rho$ (distance to origin; radius; nonnegative integer)
• $\varphi$ (colatitude = $\pi /2$ − latitude; angle made with $z$-axis; 0 to $\pi$)
• $\theta$ (longitude; angle made from $x$; 0 to $2\pi$)

Converting to Cartesian Coordinates

• $x=\rho \sin \varphi \cos \theta$
• $y=\rho \sin \varphi \sin \theta$
• $z=\rho \cos \varphi$

What are the latitude/colatitude of College Station?

lat. = 30.6°

colat = 90° − lat = 59.4°

Assuming Earth is Fixed, what direction does the celestial sphere rotate? =

(left-hand rule)

In a day, how long is Capella above the horizon

For a star (Capella) in celestial sphere,

• Declimation = latitude from horizon (46°, so colatitude is 90-46 = 44°)
• Right-Ascension = longitude (we don't care about this)

The star makes a complete revolution once every 24 hours. If we slice the sphere along Capella's celestial path, we get a circle. If we find the angle $\theta$ that represents half of the angle at which Capella is above the horizon (from its rising to when it's directly overhead)

$24\times \left(\frac{\theta }{\pi }\right)$

For the observational vector (straight up; $\overrightarrow{v}$ and the vector from Earth to the declination on the horizon $\overrightarrow{w}$:

$\overrightarrow{v}\cdot \overrightarrow{w}=0$

We find that $\overrightarrow{v}=(\rho =1,\theta =0,\varphi =59.4^{\circ })=⟨\sin 59.4^{\circ },0,\cos 59.4^{\circ }⟩$ and $\overrightarrow{w}=(\rho =1,\theta ,\varphi =44^{\circ })=⟨\sin 44^{\circ }\cos \theta ,\sin 44^{\circ }\sin \theta ,\cos 44^{\circ }⟩$

For a different problem (e.g. the sun), just change the declination ($\varphi$ in $\overrightarrow{w}$)