We've seen how to get the first and second derivatives of a vector function. If a vector function is locating a moving particle through a plane or through space( the particle is moving on the space curve or plane curve that is the graph of the vector function ) it is referred to as a displacement function. The first and second derivative functions for a displacement function are referred to as the velocity and acceleration functions. They are, of course, vector functions and, as such, result in vectors. We expect the velocity vector to give us direction and speed for the moving particle. And, we expect the acceleration function to give us information concerning the forces acting on the particle in motion.That the velocity and acceleration functions do as we expect is the purpose of the following text.
Suppose the we have a vector function that is a displacement function and it is designated by r(t). Graphically, the vector [r(t + h) - r(t)] is across the tips of the vectors r(t + h) and r(t) and this vector has the zero vector as its limit as h approached 0. (I'm using h instead of delta t in this discussion; you'll "see" delta t when we go back to the text book) When we divide r(t + h) - r(t) by the scalar h, we are actually multiplying the vector r(t + h) - r(t) by the reciprocal of h and h is not allowed to be 0. You will recall that such a multiplication just changes the length but not the direction of the vector so that [r(t + h) - r(t)]divided by h has the same direction as [r(t + h) - r(t)]. From the Calculus I course we know that the limit as h approaches 0 of the quotient may exist and need not be 0.( this came under the "heading" of 0 divided by 0 limit problems which were referrred to as indeterminant forms ) This is the first derivative and when it exists it is our velocity vector by definition. From a picture of r(t + h) - r(t), across the tips of the vectors r(t +h) and r(t), we are able to "see" that a non zero velocity vector would be in the direction of the motion of the particle moving on the space or plane curve graph of the displacement function.
Look at the picture of this in the text:See 12.3 Velocity and Acceleration, Pg. 848
The fact that the magnitude or norm of the velocity vector "gives" us the speed ( remember we expected this ) can be derived from the formula for arc length that was studied in previous Calculus courses. If the integral is thought of in terms of "summing-up" and the delta t in the integrand as an increment of time, then the sum of the speed multiplied by time should give distance.
Look at the formula for arc length in the text:See 12.5 Arc Length , Pg. 867The integrand, the square root of the sum of the squares of the first derivative functions in the parametric representation for the curve, is just the norm of the velocity vector function. Hopefully, the arc length formula makes more sense to us now.
The fact that the vector function called "acceleration" gives information concerning forces acting on the moving particle can be seen by studying this next example of motion in a circle at a constant speed. Our expectation is of a force called "centripetal force" that is "pointing" at the center.
The next example is interesting because the path is similar to that for comets in our univeerse.
See 12.3 Example 2 Figure 12.12
The next two examples illustrate that at times we are differentiating to get vector function results but at other times we would need to integrate to get the desired result.
While time does not permit inclusion, you can notice an excellent explanation for Projectile Motion (11.3 Projectile Motion) if you were to need one in your studies of Physics.
Look at the following solved problems
in the Exercises at the end of
12.3 : 1,3,5,9,11,13,15,41,43 Then
do the following unsolved
problems as homework: 2,12,14,16,42(a),44