VECTOR-VALUED FUNCTIONS MATH 200 MAIN QUESTIONS FOR TODAY Whats a - PowerPoint PPT Presentation

MATH 200 WEEK 3 - WEDNESDAY VECTOR-VALUED FUNCTIONS

MATH 200 MAIN QUESTIONS FOR TODAY ▸ What’s a vector valued function? ▸ How do we find the domain of a vector valued function? ▸ How do we compute the derivative of a vector-valued function? ▸ Do the derivative rules from calc 1 (e.g. the product rule) still apply? ▸ How do we integrate vector-valued functions?

MATH 200 VECTOR-VALUED FUNCTIONS ▸ A vector-valued functions has scalars as input and vectors as output . ▸ Domain: Scalars ▸ Range: Vectors ▸ E.g. ▸ r (t) = <3t, t 2 , t 3 +1> ▸ r (t) = <cos(t), sin(t), t> ▸ r (t) = <ln(t), 1, t 2 >

MATH 200 IT’S PERFECTLY FINE TO JUST THINK OF THIS AS A DIFFERENT WAY OF WRITING PARAMETRIC EQUATIONS  x = 2 t 2  r ( t ) = ⟨ 3 t 2 , t 3 − 1 , ln( t ) ⟩ =  y = t 3 − 1 ⇒ ⃗  z = ln( t )  …AND YOU CAN VISUALIZE THE CURVE AS BEING DRAWN OUT BY THE VECTORS YOU GET WHEN YOU PLUG IN VARIOUS T VALUES

MATH 200 DOMAIN ▸ Domain: the set of allowable inputs for a given function. ▸ Natural vs. Given Domain ▸ Sometimes, we’re given the domain for a function. ▸ e.g. r (t) = <cos(t),sin(t),1>, 0<t< π /4 ▸ Other times, we aren’t given the domain and we have to figure out what domain makes sense (or is a natural choice) ▸ e.g. r (t) = <ln(t), t + 1, e t > ▸ Looking at the components one at a time, we can say that the domain is t>0

MATH 200 DOMAIN EXAMPLES ▸ Determine the (natural) domain for each of the following vector valued functions. The easiest way to do so may be to figure out what needs to be omitted for each component. � 1 � √ 1. � r ( t ) = t, 2 t − 1 t , � � � 3 2. � r ( t ) = ln( t ) , arcsin( t ) , ( t ) � 1 � t 2 − 4 , 4 , t 3 3. � r ( t ) =

MATH 200 DERIVATIVES ▸ Differentiating vector-valued functions is actually pretty straightforward: we just differentiate each component ▸ E.g. r (t) = <3t 2 , ln(t), 4> ▸ r ’(t) = <6t, 1/t, 0> ▸ For practice’s sake, compute the derivatives of the following vector-valued functions: ▸ r (t) = <sin(t), t 3 , 1 - t> r ’(t) = <cos(t), 3t 2 , -t> ▸ r (t) = <2tcos(t), t 2 , t-1> r ’(t) = <2cos(t) - 2tsin(t), 2t, 1>

MATH 200 TANGENT VECTORS ▸ In calc 1, we use the derivative to find, among other things, the slope of the tangent line to the original function . ▸ E.g. If I want the slope of the line tangent to f(x) = x 2 at x=-1, I first find f’ and then plug in x=-1. ▸ f’(x) = 2x ▸ f’(-1) = -2 ▸ So, the slope of the tangent line to f at x = -1 is 2 ▸ Now, we have a vector-valued function as the derivative, so what do we do with that.

MATH 200 TANGENT VECTORS CONT. ▸ Consider the function r (t) = <cos(t), sin(t), t>. ▸ Taking the derivative we get… ▸ r ’(t) = <-sin(t), cos(t), 1> ▸ Let’s plug in a value for t, say π /2: ▸ r ’( π /2) = <-1, 0, 1> ▸ What does this vector tell us? ▸ At t = π /2, we’re at r ( π /2) = <0, 1, π /2> ▸ Let’s plot r ’( π /2) = <-1, 0, 1> at (0, 1, π /2)

MATH 200 ▸ The vector r ’( π /2) is tangent to the curve at r ( π /2) ▸ Now, find parametric equations for the line tangent to r at t= π /2 ▸ Point: (0, 1, π /2) ▸ Vector: <-1, 0, 1>  x = − t   L : y = 1  z = π 2 + t 

MATH 200 EXAMPLE ▸ Consider the vector-valued function r (t) = <3t-1,t 2 ,ln(t-1)> ▸ Find parametric equations for the line tangent to r at the point (5,4,0) ▸ Express the line as a vector-valued function

MATH 200 INTEGRATION ▸ Since differentiation works with vector-valued functions, it stands to reason that anti-differentiation also works, and in the same way… � b �� b � b � b � r ( t ) dt = x ( t ) dt, y ( t ) , dt z ( t ) dt ⃗ a a a a WE JUST INTEGRATE TERM BY TERM ▸ The interpretation isn’t quite the same as in calc 2… ▸ It’s not area under the curve anymore ▸ The interpretation is context-dependent

MATH 200 INTEGRATION EXAMPLE ▸ Suppose v (t)=<4t,e t > is the velocity vector for some particle moving along the plane. ▸ If the particle’s initial position (t=0) is (2,3), find its position function r (t) � r (0) = ⟨ 2 , 3 ⟩ ⃗ r ( t ) = v ( t ) dt � � � 2(0) 2 + C 1 , e 0 + C 2 � = � 2 , 3 � �� � � e t dt = 4 y dt, � C 1 , 1 + C 2 � = � 2 , 3 � 2 t 2 + C 1 , e t + C 2 � � = C 1 = 2 , C 2 = 2 r = ⟨ 2 t 2 + 2 , e t + 2 ⟩ ⃗