Vector CalculusDate: 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/ |
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