Class Notes 100823
 
You might have some trouble reading the notation of these notes, so don't rack your brains over it.  The notation isn't hard to understand and doesn't take long to get used to, but we'll spend time on that Wednesday.

 

The magnitude of a vector A = a_x i + a_y j in two-dimensional space is found using the Pythagorean Theorem to be sqrt(a_x^2 + a_y)^2.

 
The magnitude of a vector A = a_1 i + a_2 j + a_3 k, in 3-dimensional space, is found by two applications of the Pythagorean Theorem to be sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^2). 
 
[ The reasons for the above: 
 
In the x-y plane we find the magnitude of a_1 i + a_2 j to be sqrt(a_1 ^ 2 + a_2 ^ 2). 
 
The sum of this vector and the vector a_3 k is our original vector A = a_1 i + a_2 j + a_3 k.  
 
The vector a_3 k being perpendicular to the entire xy plane is also perpendicular to the vector a_1 i + a_2 j. 
 
So the vectors a_1 i + a_2 j  and a_3 k are legs of a right triangle whose hypotenuse is therefore sqrt(  (sqrt(a_1 ^ 2  + a_2 ^ 2)  + a_3 ^2 ) = sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2).  ]
 
The dot product of two vectors is equal to the sum of the products of their components: 
  • If A = a_1 i + a_2 j + a_3 k andB = b_1 i + b_2 j + b_3 k, then A dot B = a_1 b_1 + a_2 b_2 + a_3 b_3.
The dot product is also equal to || A || * || B || cos(theta), where theta is the angle between the vectors.  So
 
 
So for example if we know the components of A and B, we can easily find the dot product and the magnitudes of the two vectors.  Having found the magnitudes and the dot product we can use the second relationship to get
 
cos(theta) = A dot B / || A || || B || so that
theta = arcCos( A dot B / || A || || B || ).

The cross product of two vectors A and B is a vector perpendicular to the two vectors, the direction determined by the right-hand rule, and its magnitude is || A || * || B ||  * sin(theta), where theta is the angle between the two.

The formula for the cross product is a little more complicated than the simple formula for the dot product, but it's not bad:

A `X B is the determinant of the matrix whose first row is [ i, j, k ], whose second row is [ A_1, A_2, A_3 ], the coefficients of the A vector, and whose third row is [ B_1, B_2, B_3 ], the coefficients of the B vector.

We can define curves parametrically. 
 
For example if x(t) = t^2 and y(t) = t^3, substituting t = 0, 1, 2 yields the points (0, 0), (1, 1) and (4, 8).
 
Plotting these points on an x-y coordinate system indicates a curve with positive and increasing slope.  The third point is further from the second than the second point is from the first.  If t is interpreted as time, our curve therefore represents a point moving faster and faster along the curve.  The curve is depicted below:

 
If we regard t as the clock time, it is natural to ask about the velocity of the point and how the velocity changes.  If we recall that velocity is the rate of change of position with respect to clock time, we conclude that at clock time t the object's velocity has components v_x(t) = dx/dt and v_y(t) = dy/dt.  Since dx/dt = 2 t and dy/dt = 3 t^2, we conjecture that the velocity of the particle at clock time t is given by

v(t) = v_x(t) i + v_y(t) j = dx/dt i + dy/dt j = 2 t i + 3 t^2 j.

For example, at t = 2 our velocity vector is

v(2) = 2 * 2 i + 3 * 2^2 j = 4 i + 12 j.

A unit vector is a vector whose magnitude is 1.  If you divide any vector by its magnitude you get a unit vector.  The unit vector has magnitude 1 and its direction is the same as that of the original vector.

For example the magnitude of the vector 3 i + 4 j is sqrt(3^2 + 4^2) = 5.  If you divide this vector by its magnitude you get the unit vector

u = (3 i + 4 j) / 5 = (3/5) i + (4/5) j = .6 i + .8 j.

This vector has magnitude 1, and its direction is the same as that of the original vector 3 i + 4 j.

As another example, if you divide the velocity vector v(2) = 4 i + 12 j, found just a little earlier, by its magnitude you get a unit vector. 

The magnitude of v(2) = 4 i + 12 j is || v(2) || = sqrt( 4^2 + 12^2 ) = sqrt( 160 ) = 4 sqrt(10).

We give the unit vector in the direction of v(2) the name u(2):

u(2) = v(2) / || v(2) || = (4 i + 12 j) / (4 sqrt(10) ) = sqrt(10) / 10 * i + 3 sqrt(10) / 10 * j.

u(2) is the unit vector in the direction of the velocity, when t = 2.

 

 

 

**************

some extensions for next class (optional at this time):

The rate at which the velocity changes in the x direction is the derivative d v_x(t) / dt of the x velocity function v_x(t) = 2 t.  The derivative of this function is the constant function 2.  The derivative d v_y(t) / dt of the y velocity function v_y(t) = 3 t^2 is 6 t.  So the x velocity changes at constant rate 2, while the y velocity changes at the increasing rate 6 t. 

The derivative of our velocity function v(t) is therefore

dv / dt = 2 i + 6 t j.

The x component of the velocity increases steadily, while the y component increases at an increasing rate.

The general point ( x(t), y(t) ) can be represented by a vector which originates at the origin (0, 0) and extends to the point (x(t), y(t)).  If we call this vector r(t), it is easy to see that

r(t) = x(t) i + y(t) j. 

For our current example this vector is just

r(t) = t^2 i + t^3 j.

It makes perfect sense, and will later be shown to be the case, that the derivative of this r(t) vector is found by doing the obvious::

d r(t) / dt = 2 t i + 3 t^2 j.

We have simply treated the i and j vectors as constants.

In symbols, we say that the derivative of the vector function r(t) = x(t) i + y(t) j is found by taking the derivatives of the x(t) and y(t) functions, treating the i and j vectors as constants:

d r(t) / dt = d x(t) / dt * i + d y(t) / dt * j.

As we have seen, if x(t) = t^2 and y(t) = t^3, then d x(t) / dt = 2 t and d y(t) / dt = 3 t^2, so the derivative of our r(t) vector function is

d r(t) / dt = 2 t i + 3 t^2 j.

This is the velocity vector we found earlier:

v(t) = 2 t i + 3 t^2 j,

and illustrates the general relationship

v(t) = d r(t) / dt.

The tip of the r(t) vector traces out the curve we approximated previously.

The velocity vector v(t) is at every point tangent to the r(t) curve. 

The rate of change of the velocity vector is the derivative we found earlier:

dv / dt = 2 i + 6 t j.

If we divide a vector by its magnitude, we get a vector of magnitude 1.  We call a vector of magnitude 1 a unit vector.

In the present example we can divide the velocity vector by its magnitude.  We obtain

v(t) / || v(t) || = ( 2 t i + 3 t^2 j ) / (sqrt( (2 t)^2 + (3 t^2)^2) ) = ( 2 t i + 3 t^2 j ) / (sqrt( 4 t^2 + 9 t^4 )).

This is a unit vector in the direction of the velocity.  If we evaluate ( 2 t i + 3 t^2 j ) / (sqrt( 4 t^2 + 9 t^4 )) for any value of t, except t = 0 (where the denominator is zero), we will obtain a vector of magnitude 1.