Vector-Valued Functions

Exercise. Sketch the curve defined by the vector function

r( t ) = ( 2 - 4t )i + ( −1 + 3t )j + ( 3 + 2t )k , where 0≤t≤1

Expert Solution

Solution

The parametric equations for the curve are

x = 2 - 4t , y = -1 + 3t , z = 3 + 2t

which are parametric equations of the line passing through the point

( 2 , -1 , 3 ) with direction numbers -4 , 3 , and 2. Because the parameter interval is the closed interval [ 0 , 1 ] , we see that the curve is a straight line segment : Its initial point ( 2 , -1 , 3 ) is the terminal point of the vector

r(0)2i-j +3k , and its terminal point ( -2 , 2 , 5 ) is the terminal point of the vector r(1) = -2i + 2j + 5k . ( See Figure 4. )

SUN BUNRA answered 2 years ago

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