Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

Sec. 18] The Vector Function of a Scalar Argument 235
In Problems 2071-2076 find the arc length
2071. x = t, y=/ 2
of the curve:
, 2-~-' from / = to t = 2.
2072. x = 2 cos /, y = 2 sin t, z = -|- 1 from / == to t = it.
2073. A: = *' cos /, y = e* sin /, z = e t
2074. y = 4~' 2 = 4~
from f = to arbitrary t .
from JC==0 to x==6 -
2075. * f = 3(/, 2jcy = 92 from the point (0, 0, 0) to M (3, 3, 2).
2076. f/ = aarcsin~, z = -|-ln ^j from the point 0(0,0,0)
to the point M(* z , j/ , ).
2077. The position of a point for any time f (f>0)
by the equations
is defined
Find the mean velocity of motion between times f = l and ^=10.
Sec. 18. The Vector Function of a Scalar Argument
1. The derivative of the vector function of a scalar argument. The vector
function a a (0 may be defined by specifying three scalar functions a x (t) 9
a y (t) and a z (t) t which are its projections on the coordinate axes:
The derivative of the vector function a-=a(t) with respect to the scalar
argument t is a new vector function defined by the equality
da a(t + M)-a(t)_dax (t) . da y (0 . da f (t)
The modulus of the derivative of the vector function is
da
dt
The end-point of the variable of the radius vector r=r(/) describes in space
the curve
r=x(t)l+y(t)J+*(t)*.
which is called the hodograph
of the vector r.
The derivative -~ is a vector, tangent to the hodograph
sponding point; here,
[
dr [_ ds
\ dt \ dt '
at the corre-
where s is the arc length of the hodograph reckoned from some initial point.
For example, Up