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Unformatted text preview: 5 / 2) s 2 . Furthermore, ( t ) = 1 / (5 t ). Problem 5. Find the equation of the normal plane of the curve x = t,y = t 2 ,z = t 3 at the point (1 , 1 , 1 , ). Solution: We have r (1) = h 1 , 2 , 3 i is the normal vector to the normal plane. So, the equation of the normal plane is: ( x1)+2( y 1 )+3( z1) = 0, or x + 2 y + 3 z = 6. Problem 6. Find the tangential and normal components of the acceleration vector r ( t ) = t i + t 2 j + 3 t k . 1 Solution: We have a T =  r ( t ) r 00 ( t )   r ( t )] = 4 t 4 t 2 + 10 . Furthermore, a N =  r ( t ) r 00 ( t )   r ( t )] = 2 10 4 t 2 + 10 ....
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This note was uploaded on 10/20/2011 for the course MAC 2313 taught by Professor Keeran during the Fall '08 term at University of Florida.
 Fall '08
 Keeran
 Calculus, Geometry

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