Calc 3 Lecture Notes Section 11.2 Page 1 of 7
Section 11.2: The Calculus of Vector-Valued Functions Big idea: Our basic foundational notions of the calculus of scalar functions can be extended to vector-valued functions by applying limits to all three component functions simultaneously.
Big skill: You should be able to take the limit, derivative, and integral of a vector-valued function, as well as determining the smoothness of and tangent vectors to a curve defined by a vector-valued function.
First Calculus Notions: Limits and Continuity Definition 2.1: Definition of the Limit of a Vector-Valued Function Given a vector-valued function r (t) = f( t), g( t) , h( t ) , we define the limit of r (t) as t approaches a as follows:
limr (t) = lim f( t) , g( t) , h( t) = lim f( t) , lim g( t) , lim ( t) , t a t a t a t a t a
lim r (t) provided all three of the above limits exist. If any one of the limits does not exist, then t a does not exist.
Thus, finding the limit of a vector function involves finding the limits of three scalar functions, if possible, and the resulting vector (assuming the limits exist) is just a constant vector.
Practice: sin t 4 1. Find lim e 3t , , t0 t t 2
2 2. Find limt+ 2,cosh( t) , tan ( t) t p / 2 Calc 3 Lecture Notes Section 11.2 Page 2 of 7
Definition 2.2: Continuity of a Vector-Valued Function We say the vector-valued function r (t) = f( t), g( t) , h( t) is continuous at t = a if
lim r(t) = r( a) t a .
So, r (t) is continuous at t = a if the limit and the value of the function at t = a are equal. Similar to our study of continuity of scalar functions in Calculus 1, this roughly corresponds to being able to draw the curve traced out by r (t) through t = a without lifting your pencil from the paper. Using the above definition, one can show that r (t) is continuous at t = a if and only if all three of the scalar functions f, g, and h are continuous at t = a, which leads to the following theorem:
Theorem 2.1: Condition for the Continuity of a Vector-Valued Function A vector-valued function r (t) = f( t), g( t) , h( t) is continuous at t = a if and only if all of f, g, and h are continuous at t = a.
Practice: 1. Determine where r (t) =sinh( t) , ln( t - 1) , tan ( t) is continuous.
Second Calculus Notion: Derivatives Definition 2.3: Definition of the Derivative of a Vector-Valued Function The derivative r (t) of the vector-valued function r (t) is defined by
r(t+ D t) - r ( t) r (t )= lim Dt 0 Dt at any value of t for which this limit exists. If this limit exists at some value t = a, we say r is differentiable at a.
The above definition leads to the following theorem: Calc 3 Lecture Notes Section 11.2 Page 3 of 7
Theorem 2.2: Derivative of a Vector-Valued Function in Terms of its Components If r (t) = f( t), g( t) , h( t) , and the components f, g, and h are all differentiable for some value of t, then r is also differentiable at that value of t and its derivative is given by rⅱ(t) = f( t), g ⅱ( t) , h( t ) .
Practice: d 1. Compute sinh(t2 ) ,( t- 1) ln( t - 1) , tan ( et ) . dt
Theorem 2.3: Derivatives of Combinations of Vector-Valued Functions and Scalars Suppose r (t) and s(t) are differentiable vector-valued functions, f (t) is a differentiable scalar function, and c is a constant. Then
d (i) 轾r(t) + s( t) = rⅱ( t) + s ( t) dt 臌
d (ii) 轾cr( t) = c r ( t) dt 臌
d (iii) 轾f( t)r( t) = f'( t) r( t) + f( t) r ( t ) dt 臌
d (iv) 轾r(t)� s( t) � rⅱ( t) s( t) r( t) s ( t) dt 臌
d (v) 轾r(t)� s( t) � rⅱ( t) s( t) r( t) s ( t) dt 臌
Some selected proofs: Calc 3 Lecture Notes Section 11.2 Page 4 of 7
Implication of a zero derivative: Recall that for a scalar function f(x), that any extrema of f would only occur at a critical value, which is defined as any x = a such that f (a) = 0 or is undefined. For a vector-valued function r (t) = f( t), g( t) , h( t ) , values of t that result in r (t) = 0 can result in cusps in the curve traced out by the function. Thus, we say that the curve traced out by r(t) is smooth on an interval I if r is continuous on I, and r (t) 0 , except possibly at the endpoints.
Examples: The curves traced out by r (t) = t3, t 2 and r (t) = t3, t 2 , t
Graphical interpretation of r (t) : It is a vector tangent to the curve at r(t)
See VectorValuedDerivative.wp2 …
Practice: Draw the curve traced out by r (t) = 2cos( t) ,2sin ( t) and its derivative for various t values. Calc 3 Lecture Notes Section 11.2 Page 5 of 7
The answer we got above is a specific case of a more general result:
Theorem 2.4: Orthogonality of a Constant-Magnitude Vector-Valued Function and Its Derivative r (t) is constant throughout some interval if and only if r(t) and r(t) are orthogonal for all t in that interval. Proof: 2 If r (t) = c , then r(t)� r( t) = r ( t) c2 . Differentiating: d d 轾r(t)� r ( t) c2 dt臌 dt rⅱ(t)� r( t) � r( t) r ( t) 0 2r(t)� r ( t) 0 r(t)� r ( t) 0 Thus, r(t) and r(t) are orthogonal.
Practice: Show that the position and tangent vectors for the curve traced out by r (t) = 3sin( t) cos( 2 t) ,3sin( t) sin( 2 t) ,3cos ( t ) are always orthogonal. (Note: the curve is always on a sphere of radius 3…) Calc 3 Lecture Notes Section 11.2 Page 6 of 7
Third Calculus Notion: Antiderivatives/Integrals
Definition 2.4: Antiderivative of a Vector-Valued Function The vector-valued function R(t) is an antiderivative of the vector-valued function r(t) if R(t) = r(t).
Definition 2.5: Indefinite Integral of a Vector-Valued Function If R(t) is any antiderivative of the vector-valued function r(t), then the indefinite integral of r(t) is defined as r(t) dt= R( t ) + c , where c = c1, c 2 , c 3 is an arbitrary constant vector.
Note that this simply corresponds to
蝌r(t ) dt= f ( t ) dt , 蝌 g ( t ) dt , h ( t ) dt
Practice:
2 1 tet , , sin 4 t dt = 1+ t 2
Definition 2.6: Definite Integral of a Vector-Valued Function For the vector-valued function r (t) = f( t), g( t) , h( t) , the definite integral of r(t) on the interval [a, b] is defined as:
b b b b 蝌r (t) dt= f( t) dt, 蝌 g( t) dt , h( t) dt , a a a a provided all three integrals on the right exist.
Note that this results in a constant vector. Calc 3 Lecture Notes Section 11.2 Page 7 of 7
Practice: 4 1 sin(t) ,2 , ln( 2 t) dt = 2 1- t
Theorem 2.5: Extension of the Fundamental Theorem of Calculus to Vector-Valued Functions If R(t) is an antiderivative of a vector-valued function r(t) on the interval [a, b], then:
b r(t) dt= R( b) - R ( a) . a