Vector Calculus

Date: 2/8/96 at 15:3:17
From: Anonymous
Subject: Vector Calculus

1)  Find the total distance travelled by a particle along the path.  
    Sketch the path.
	
	x(t) = (3t^2,t^3), -1<=t>=1


2)  For each of the following curves, find the equation of the tangent
    line at t1.  Sketch the curve and tangent at t1.

	a)  x(t) = ((e^-t) cost, (e^-t) sint), t1=pi
	b)  x(t) = (3t^2, t^3),  t1=1/2

Date: 8/3/96 at 9:37:45
From: Doctor Jerry
Subject: Re: Vector Calculus

For question 1, if a curve C is described parametrically by 
   x=x(t)
   y=y(t),
for t <= a <= b,

then the length of C is the integral of the length of the tangent 
vector.  

The  tangent vector is {x'(t),y'(t)} and its length is 
sqrt(x'(t)^2+y'(t)^2).  So the arc length of C is integral 
from a to b of sqrt(x'(t)^2+y'(t)^2) dt.

So, just integrate sqrt((6t)^2+(3t^2)) on the interval -1 to 1.

For question 2, I'll just given the equation of the tangent line.

Use the standard equation for a line

r(t) = p + q*t,   -infinity < t < infinity,

where p is a point through which the line passes and q is a vector in 
the direction of the line.  For p, use x(pi) or x(1/2).  For q, 
differentiate x(t) and then use x'(pi) and x'(1/2).  

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/