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Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents 13.1 Curves in Space and their Tangents Vector-Valued Function ● vector function for short, sometimes called space curves ● takes one or more real variables and returns a vector ● we’ll focus on single variable ( t ) vector functions that give three-dimensional vectors represents time in most applications r ( t ) = f ( t ) , g (t ) , h (t ) = f (t ) i + g (t ) j + h ( t ) k real valued functions called component functions of r We want to look at the calculus of these vector functions. Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents Interesting vector valued functions: r ( t ) = cos ( t ) ,sin ( t ) , t called the circular helix For a nice animation go to : http://math.bu.edu/people/paul/225/circular_helix.html r (t ) = ( 2 + cos ( )) cos t , ( 2 + cos ( ) ) sin ( t ) ,sin ( ) 3t 2 3t 2 called the trefoil knot For a nice animation go to : http://math.bu.edu/people/paul/225/trefoil_knot.html 3t 2 Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents Limit of Vector Functions r ( t ) = f ( t ) , g (t ) , h (t ) Limit of a vector function - taken component-wise lim r ( t ) = lim f ( t ) , lim g ( t ) , lim h ( t ) t →a provided the component function limits exist t →a t →a t →a Let r ( t ) = sin ( t ) 2t , e , ln (1 − t ) . Find lim r ( t ) t →0 t sin ( t ) lim r ( t ) = lim , lim e2t , lim ln (1 − t ) t →0 t →0 t →0 t →0 t 1 1 0 lim r ( t ) = 1,1, 0 = i + j t →0 Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents Differentiation of Vector Functions r ( t ) = f ( t ) , g (t ) , h (t ) Derivative of a vector function - taken component-wise r′ ( t ) = f ′ ( t ) , g ′ ( t ) , h′ ( t ) provided the component function are differentiable Let r ( t ) = sin −1 t , 1 − t 2 , ln (1 + 3t ) . Find r′ ( t ) r′ ( t ) = r′ ( t ) = 1 , 1 1− t 2 1− t 2 ( 2 1 1− t2 , −t 1− t 2 , −1/2 1 ) ( −2t ) , 1 + 3t (3) 3 1 + 3t Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents r ( t ) is the position vector r′ ( t ) is the tangent vector r′ ( t ) will now be called the velocity vector, v ( t ) r′ ( t ) is called the speed of the particle r′′ ( t ) is called the acceleration vector, a ( t ) ( 0, 2 ) t =0 r ( t ) = sin t , 2 cos t v ( t ) = r ′ ( t ) = cos t , −2 sin t a ( π6 ) a ( t ) = r′′ ( t ) = − sin t , −2 cos t (1, 0) r ′ ( t ) = cos 2 t + 4sin 2 t π r = 6 1 2 , 3 π v = 6 t= 3 2 , −1 2. 3. 4. 5. 6. π 2 π a = − 1 ,− 3 2 6 Differentiation Rules: 1. v ( π6 ) d u ( t ) + v ( t ) = u′ ( t ) + v′ ( t ) dt d cu ( t ) = cu′ ( t ) dt d f ( t ) u ( t ) = f ′ ( t ) u ( t ) + f ( t ) u′ ( t ) dt d u ( t ) ⋅ v ( t ) = u′ ( t ) ⋅ v ( t ) + u ( t ) ⋅ v′ ( t ) dt d u ( t ) × v ( t ) = u′ ( t ) × v ( t ) + u ( t ) × v′ ( t ) dt d u ( f ( t ) ) = f ′ ( t ) u′ ( f ( t ) ) dt Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents Show that if r ( t ) = c (constant), then r′ ( t ) ⊥ r ( t ) for all t. Assume r ( t ) = c. 2 The fact that r ( t ) = r ( t ) ⋅ r ( t ) , gives us r ( t ) ⋅ r ( t ) = c 2 . Taking the derivative of both sides gives d r ( t ) ⋅ r ( t ) = 0. dt Using the properties of differentiation, we get r ′ ( t ) ⋅ r ( t ) + r ( t ) ⋅ r ′ ( t ) = 0. The dot product is commutative, so r ′ ( t ) ⋅ r ( t ) + r ′ ( t ) ⋅ r ( t ) = 0 or 2 r′ ( t ) ⋅ r ( t ) = 0. So, r′ ( t ) ⋅ r ( t ) = 0 and thus r ′ ( t ) ⊥ r ( t ) ( for all t ) Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents Most vector functions do not have constant magnitude. The helix example from earlier: r ( t ) = cos ( t ) ,sin ( t ) , t r ( t ) = cos 2 ( t ) + sin 2 ( t ) + t 2 r (t ) = 1 + t 2 It’s magnitude isn’t constant (it depends on t). Consider its derivative however: r′ ( t ) = − sin ( t ) , cos ( t ) ,1 r′ ( t ) = sin 2 ( t ) + cos 2 ( t ) + 1 = 2 r′ ( t ) has constant magnitude, so r′ ( t ) ⊥ r ′′ ( t ) So for the helix, the velocity vector will always be orthogonal to the acceleraton vector ( don't think that this always happens ) Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents Show that if r ( t ) is a vector function such that r′′ ( t ) exists, then d r ( t ) × r′ ( t ) = r ( t ) × r′′ ( t ) dt Using the properties of differentiation, we get d r ( t ) × r ′ ( t ) = r ′ ( t ) × r ′ ( t ) + r ( t ) × r′′ ( t ) dt anytime you cross a vector with itself, you get the zero vector So, d r ( t ) × r ′ ( t ) = r ( t ) × r ′′ ( t ) . dt When taking the derivative of the cross product of a vector function and its velocity, you can just take the cross product of the vector function and its acceleration. Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents Geometry of the derivative secant vector vector : z r (t + h ) − r (t ) P Q r (t ) r (t + h ) y Tangent vector z x P r (t ) r (t + h ) − r (t ) h →0 h r′ ( t ) = lim tangent vector Q r (t + h ) − r (t ) h r (t + h ) Unit tangent vector y T (t ) = x r′ ( t ) r′ ( t ) ( it's the velocity vector with magnitude 1) Math 114 – Rimmer 13.1 Vector Functions (Space Curves) and Their Tangents The tangent line to a smooth curve r ( t ) = f ( t ) , g ( t ) , h ( t ) = f ( t ) i + g ( t ) j + h ( t ) k at t = t0 passes through the point ( f ( t0 ) , g ( t0 ) , h ( t0 ) ) , and has the same direction as the velocity vector at t0 f ′ ( t0 ) , g ′ ( t0 ) , h ′ ( t0 ) . So the parametric equations of the tangent line are: x = f ( t0 ) + y = z = g ( t0 ) + h ( t0 ) + f ′ ( t0 ) t g ′ ( t0 ) t h′ ( t0 ) t