272_c13

272_c13 - Mat 272 Calculus III Updated on 2/5/08 Dr. Firoz...

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Unformatted text preview: Mat 272 Calculus III Updated on 2/5/08 Dr. Firoz Chapter 13 Vector Functions Section 13.1 Vector Functions and Space Curves A vector valued function or simply a vector function is a function whose domain is a set of real numbers and whose range is a set of vectors. Representation of a vector function: The vector ( ) r t along with components ( ), ( ), ( ) f t g t h t , which are simple functions of t, represented as ( ) ( ), ( ), ( ) ( ) ( ) ( ) r t f t g t h t f t i g t j h t k =< >= + + Important properties : 1. lim ( ) lim ( ),lim ( ),lim ( ) t a t a t a t a r t f t g t h t =< > , provided each limit exists 2. The parametric representations ( ), ( ), ( ) x f t y g t z h t = = = is a space curve on I, t is a parameter. Examples: 1. Find the domain of 3 ( ) ,ln(2 ), r t t t t =<- > Solution: For, 3 t , t is in I, set of real numbers. For ln(2 ) t- , t < 2 and for t , t . Thus the domain is [0, 2) 2. Determine lim ( ) t r t where 3 1 sin ( ) 1 , , 1 t r t t t t = +- Solution: lim ( ) 1, 1,1 t r t =< - > 3. Describe the curve defined by ( ) 1 ,2 5 , 1 6 r t t t t =< + +- + > Solution: The parametric form of the curve is 1 , 2 5 , 1 6 x t y t z t = + = + = - + , which represents a straight line thru (1, 2, -1) in the direction of the vector 1,5,6 < > 4. Sketch the curve ( ) cos ,sin , r t t t t =< > Mat 272 Calculus III Updated on 2/5/08 Dr. Firoz Solution: The parametric form of the curve is cos , sin , x t y t z t = = = , which represents a curve that spirals around a circular cylinder with level curves 2 2 1 x y + = . The curve is known as circular helix. See the figure at page # 851 in you text. 5. Find the vector function that represents the curve of intersection of the cylinder 2 2 1 x y + = and the plane 2 y z + = Solution: Consider cos , sin x t y t = = , then 2 sin z t = - , 0 2 t . Now ( ) cos ,sin ,2 sin cos sin (2 sin ) r t t t t i t j t k t =<- >= + +- 6. Find the vector equation of a line thru the points P(1, 0, 1) and Q(2, 3, 1) Solution: Use the formula thru two given points as 1 ( ) (1 ) , 0 1 r t t r tr t =- + We have ( ) (1 ) 1,0,1 2,3,1 , 0 1 r t t t t =- < > + < > 7. Identify the curve ( ) cos10 , sin10 , t t t r t e t e t e--- =< > The curve is a spiral around a cone whose level curves are circles 2 2 t x y e- + = . Observe that 2 2 2 x y z + = Section 13.2 Derivatives and Integrals of Vector Functions For given vector function ( ) ( ), ( ), ( ) ( ) ( ) ( ) r t f t g t h t f t i g t j h t k =< >= + + The derivative is defined as ( ) ( ), ( ), ( ) ( ) ( ) ( ) r t f t g t h t f t i g t j h t k =< >= + + and the integral is defined as ( ) ( ) , ( ) , ( ) r t dt f t dt g t dt h t dt =< > Smooth curve : A curve given by a vector function ( ) r t on an interval I is called smooth if ( ) r t is continuous and ( ) r t , where 0 is a zero vector....
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This note was uploaded on 05/09/2008 for the course MAT 272 taught by Professor Firoz during the Spring '08 term at ASU.

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272_c13 - Mat 272 Calculus III Updated on 2/5/08 Dr. Firoz...

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