317 review mt2 - MATH 317 REVIEW 1 VECTOR FUNCTIONS 1.1 Vector Functions and Space Curves r(t = hx(t y(t z(t)i = x(t)i y(t)j z(t)k or r(t = hf(t g(t

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MATH 317 REVIEW 1. VECTOR FUNCTIONS 1.1. Vector Functions and Space Curves. r ( t ) = x ( t ) , y ( t ) , z ( t ) = x ( t ) i + y ( t ) j + z ( t ) k or r ( t ) = f ( t ) , g ( t ) , h ( t ) = f ( t ) i + g ( t ) j + h ( t ) k . (1) Space curves: parametric form (2) Domain and range of r ( t ) . (3) Limits: lim t a r ( t ) = lim t a f ( t ) , lim t a g ( t ) , lim t a h ( t ) . (4) Continuity of r ( t ) at t = a. EXAMPLES: (1) Straight lines are space curves: r ( t ) = x 0 + at, y 0 + bt, z 0 + ct , where at least one of a, b, c is non-zero. (2) r ( t ) = cos t, sin t, t is a helix. (3) Find the vector equation of the space curve that is the intersection of the cylinder y 2 + z 2 = 2 and the plane ax + by + cz = d, where a = 0 . (4) Computer sketches (5) A twisted cubic: r ( t ) = t, t 2 , t 3 . It lies on the surface y = x 2 and also on the surface z = x 3 . In fact it is the intersection of these surfaces. (6) Can you extend the definition of r ( t ) = t sin(1 /t ) , t 2 , e t so it is continuous at t = 0? What about r ( t ) = sin(1 /t ) , t 2 , e t ? 1.2. Derivatives and Integrals of Vector Functions. The definition of the derivative of a vector function is r ( t ) = lim h 0 r ( t + h ) - r ( t ) h . If r ( t ) = x ( t ) , y ( t ) , z ( t ) = x ( t ) i + y ( t ) j + z ( t ) k or r ( t ) = f ( t ) , g ( t ) , h ( t ) = f ( t ) i + g ( t ) j + h ( t ) k , then r ( t ) = x ( t ) , y ( t ) , z ( t ) = x ( t ) i + y ( t ) j + z ( t ) k or r ( t ) = f ( t ) , g ( t ) , h ( t ) = f ( t ) i + g ( t ) j + h ( t ) k . (1) Interpretation of r ( t ) as a tangent vector or velocity vector. (2) Interpretation of r ( t ) as an acceleration vector. (3) The unit tangent vector T ( t ) := r ( t ) | r ( t ) | . (4) Suppose r ( t ) = a cos t, b sin t, ct . Then r ( t ) = - a sin t, b cos t, c and the unit tangent vector is T ( t ) = r ( t ) | r ( t ) | = 1 a 2 sin 2 t + b 2 cos 2 t + c 2 - a sin t, b cos t, c . (5) Equation of the tangent line to the space curve r ( t ) = x ( t ) , y ( t ) , z ( t ) at t = t 0 . 1
2 Rules of differentiation: (1) d dt [ λ u ( t ) + μ v ( t )] = λ u ( t ) + μ v ( t ) . (2) d dt [ f ( t ) u ( t )] = f ( t ) u ( t ) + f ( t ) u ( t ) . (3) d dt [ u ( t ) v ( t )] = u ( t ) v ( t ) + u ( t ) v ( t ) . (4) d dt [ u ( t ) × v ( t )] = u ( t ) × v ( t ) + u ( t ) × v ( t ) . (5) d dt [ u ( f ( t ))] = f ( t ) u ( f ( t )) . Example: If | r ( t ) | = c then r ( t ) is perpendicular to r ( t ) . Let r ( t ) = f ( t ) , g ( t ) , h ( t ) . Then b a r ( t ) dt is defined as a limit of Riemann sums and therefore b a r ( t ) dt = b a f ( t ) dt, b a g ( t ) dt, b a h ( t ) dt 1.3. Arc Length and Curvature. (1) The length of a plane curve x = f ( t ) , y = g ( t ) , a t b, is L = b a f ( t ) 2 + g ( t ) 2 dt. The corresponding formula for a space curve x = f ( t ) , y = g ( t ) , z = h ( t ) , a t b is L = b a f ( t ) 2 + g ( t ) 2 + h ( t ) 2 dt = b a | r ( t ) | dt. (2) Example: find the length of the circular helix x = t, y = cos t, z = sin t, 0 t 2 π. (3) The arc length function s ( t ) = t a f ( u ) 2 + g ( u ) 2 + h ( u ) 2 du, a t b ds dt = f ( t ) 2 + g ( t ) 2 + h ( t ) 2 = | r ( t ) | (4) Compute the length of the space curve r ( t ) = t cos t, t sin t, t , 0 t 2 π. (5) Parametrizing a curve by arc length: x = x ( s ) , y = y ( s ) , z = h ( s ) , 0 s L, where L is the total length of the curve from a to b. (6) The curvature of a space curve is κ := d T ds , where T ( s ) is the unit tangent vector parametrized by arc length. Using the chain rule d T dt = d T ds ds dt we have κ = d T /dt ds/dt = | T ( t ) | | r ( t ) | .
3 (7) Example: the curvature of a circle of radius r is 1 /r. To see this parametrize the circle by x = r cos t, y = r sin t, z = 0 . Then r ( t ) = < - r sin t, r cos t, 0 > and T ( t ) = < - sin t, cos t, 0 > and therefore κ = 1 /r.