Download Vector-Valued Function: Differentiating and Integrating and more Study notes Calculus in PDF only on Docsity! Vector-Valued Function Differentiation and Integration Page 1 of 8 Differentiation and Integration of Vector-Valued Functions A graph of the vector-valued function ( ) ( ) ( )3 32 cos cos , 2 cos sin 2 2 r t t t t t⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ is shown in Figure 1. Find the values of ( )r t and ( )r t′ at 0,t = , 2 t π= ,t π= and 3 2 t π= . Sketch each of the derivative vectors with their tails at the corresponding point on ( )r t . x y Figure 1: ( )r t Table 1: ( )r t and ( )r t′ t ( )r t ( )r t′ 0 2 π π 3 2 π Three properties of ( )0r t′ when ( )0 0r t′ ≠ . Vector-Valued Function Differentiation and Integration Page 2 of 8 Figure 2 shows the function ( ) ( ) ( )sin , cosr t t t= along with the velocity vector ( ( ) ( )v t r t′= ) and acceleration vector ( ) ( )( )a t r t′′= at three different values of t. Establish that the path described by ( )r t is truly circular. What appears to be the relationship between ( )v t and ( )a t at every value of t? Confirm this relationship. Figure 2 Vector-Valued Function Differentiation and Integration Page 5 of 8 Show that the speed functions for both ( ) ( ) ( )1 sin , cosr t t t= and ( ) ( ) ( ) 3 / 22 / 3 2 / 3 2 3 1 1 3 1 1 , 3 2 t t r t ⎡ ⎤+ − + −⎣ ⎦= are constant and that the speed function for ( ) ( ) ( )2 23 sin , cosr t t t= is not constant. Vector-Valued Function Differentiation and Integration Page 6 of 8 Show that if ( )s t is constant, then ( ) ( ) 0s t s t t′⋅ = ∀ . If ( )s t is constant, then at any value of t what are the possible geometric relationships between ( )s t and ( )s t′ ? Vector-Valued Function Differentiation and Integration Page 7 of 9 Net displacement vs. distance travelled In MTH 252 we talked about motion that takes place along a line. In that context, a position function is a scalar function. If ( )s t is a scalar position function, then ( ) ( )v t s t′= is that function’s velocity function. Since the motion is linear, the velocity value can describe the direction of motion with its sign; one direction corresponding to positive velocity and the other to negative velocity. In this context, we define the net displacement over the time interval [ ],a b to be ( ) ( )s b s a− . ( ) ( )s b s a− is the distance between the starting and ending points of the motion and the sign on ( ) ( )s b s a− indicates the direction of the net movement (for example, a positive sign generally means that the net movement was either rightward or upward). From the Total Change Theorem (a weak statement of The Fundamental Theorem of Calculus), we know that ( ) ( ) ( ) ( ) b b a a s b s a s t dt v t dt′− = =∫ ∫ ; in other words, integrating the velocity function over [ ],a b gives us the net displacement over [ ],a b . We can create a contrived situation to come up with the actual distance travelled over [ ],a b ; if all of the motion is in the positive direction, then the distance traveled is equal to the net displacement. We can force all of the motion into the positive direction by forcing the velocity to always be positive, i.e., by taking the absolute value of the velocity. Thus, the distance traveled is given by ( ) b a v t dt∫ . Since ( )v t is the speed function, we conclude that integrating the speed function over [ ],a b results in the total distance traveled during the time interval [ ],a b . In Figure 5, the net displacement and total distance traveled over [ ]0,3 are, respectively: ( ) 3 0 3s t dt′ =∫ and ( ) 3 0 5s t dt′ =∫ . Net displacement over [ ]0,2 Net displacement over [ ]2,3 Net displacement over [ ]0,3 Figure 5: ( ) 24s t t t= −
