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272_c13

# 272_c13 - Mat 272 Calculus III Updated on Dr Firoz Chapter...

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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 , 0 t . Thus the domain is [0, 2) 2. Determine 0 lim ( ) t r t where 3 1 sin ( ) 1 , , 1 t r t t t t = + - Solution: 0 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 =< >

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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 π . 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 0 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 ( ) 0 r t , where 0 is a zero vector. Examples: 1. Given 3 1 ( ) 1 , ,sin 1 r t t t t = + - , find ( ) r t and ( ) r t dt Solution: 2/3 2 1 1 ( ) , ,cos 3 ( 1) r t t t t - = - - and 4/3 1 2 3 3 ( ) ,ln( 1) , cos 4 r t dt t t c t c t c = + + - + - + 2. Find a unit tangent vector at (1, 0, 0) to the vector 3 ( ) 1 , ,sin 2 t r t t te t - =< + >
Mat 272 Calculus III Updated on 2/5/08 Dr. Firoz Solution: We have 2 ( ) 3 , ,2cos2 t t r t t

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272_c13 - Mat 272 Calculus III Updated on Dr Firoz Chapter...

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