Calculus 241, section 12.1 Vector-Valued Functions ...tjp/241 12.1 lecture notes.pdfCalculus 241,...
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Calculus 241, section 12.1 Vector-Valued Functions Introduction notes by Tim Pilachowski In Algebra and Calculus I and II we had real-value functions y = f (x) [rectangular coordinates] and r = f (θ ) [polar coordinates]. In Calculus III (so far) we have encountered z = f (x, y) [multivariable functions], expressed as coordinate points (x, y, z) and as a vector k c j b i a r r r + + [directed line segment from the origin to a point]. Now we introduce a different idea, vector-valued functions, ( ) ( ) ( ) ( ) k t f j t f i t f t F r r r r 3 2 1 + + = . [See Definition 12.1 in your text.] Domain Range real-valued functions z = f (x, y) Real numbers Real numbers vector-valued functions ( ) ( ) ( ) ( ) k t f j t f i t f t F r r r r 3 2 1 + + = Real numbers Vectors We’ll call 2 1 , f f and 3 f the component functions of F r , i.e. ( ) ( ) ( ) t f z t f y t f x 3 2 1 , , = = = . (Think “parametric equations”.) We’ve already seen a vector-valued function, in section 11.5, although we didn’t call it that. A line in 3-D space can be expressed as a function of a real number t: ( ) ( ) ( ) k ct z j bt y i at x L t r r r r r r r r + + + + + = + = 0 0 0 0 [vector equation] ct z z bt y y at x x + = + = + = 0 0 0 , , [parametric equations]. Example A. Determine the domain of ( ) ( ) k t j t i e t F t r r r r - + + + = 5 2 ln . Technically, to graph a vector-valued function ( ) ( ) ( ) ( ) k t f j t f i t f t F r r r r 3 2 1 + + = we’d need a 4-D grid to plot points (t, x, y, z). What we can do in 3-D space is to sketch only the range of F r , in the form of coordinate points (x, y, z) whose values are determined by the component functions. Additionally, we’ll need to think “movement in terms of t ”: As t increases, the component functions will trace out vectors, so that the resulting curve C has an arrow indicating direction of movement. [Analogy in 2-D: polar graphs] Example B. Sketch the curve C traced out by ( ) ( ) j t i t t G r r r - + - = 1 2 . The only remaining question is, “What direction?”
Transcript of Calculus 241, section 12.1 Vector-Valued Functions ...tjp/241 12.1 lecture notes.pdfCalculus 241,...
Calculus 241, section 12.1 Vector-Valued Functions Introduction notes by Tim Pilachowski
In Algebra and Calculus I and II we had real-value functions y = f (x) [rectangular coordinates] and r = f (θ )
[polar coordinates]. In Calculus III (so far) we have encountered z = f (x, y) [multivariable functions], expressed
as coordinate points (x, y, z) and as a vector kcjbiarrr
++ [directed line segment from the origin to a point].
Now we introduce a different idea, vector-valued functions, ( ) ( ) ( ) ( )ktfjtfitftFrrrr
321 ++= .
[See Definition 12.1 in your text.]
Domain Range
real-valued functions z = f (x, y) Real numbers Real numbers
vector-valued functions ( ) ( ) ( ) ( )ktfjtfitftFrrrr
321 ++= Real numbers Vectors
We’ll call 21, ff and 3f the component functions of Fr
, i.e. ( ) ( ) ( )tfztfytfx 321 ,, === .
(Think “parametric equations”.)
We’ve already seen a vector-valued function, in section 11.5, although we didn’t call it that. A line in 3-D space
can be expressed as a function of a real number t:
( ) ( ) ( )kctzjbtyiatxLtrrrrrrrr
+++++=+= 0000 [vector equation]
ctzzbtyyatxx +=+=+= 000 ,, [parametric equations].
Example A. Determine the domain of ( ) ( ) ktjtietFt
rrrr−+++= 52ln .
Technically, to graph a vector-valued function ( ) ( ) ( ) ( )ktfjtfitftFrrrr
321 ++= we’d need a 4-D grid to plot
points (t, x, y, z). What we can do in 3-D space is to sketch only the range of Fr
, in the form of coordinate
points (x, y, z) whose values are determined by the component functions. Additionally, we’ll need to think
“movement in terms of t ”: As t increases, the component functions will trace out vectors, so that the resulting
curve C has an arrow indicating direction of movement. [Analogy in 2-D: polar graphs]
Example B. Sketch the curve C traced out by ( ) ( ) jtittGrrr
−+−= 12 .
The only remaining question is, “What direction?”
Example C [11.5 Example B]. Sketch the curve C traced out by ( ) ( ) ( ) ( )ktjtittHrrrr
612243 −++++= .
Example D. Sketch the curve C traced out by ( ) ktjttFrrr
sin2cos2 −= .
Your text defines the standard unit circle as the curve traced out by the vector-valued function
( ) jtittFrrr
sincos += .
Definition 12.2: Vector-valued functions have the following (not entirely unexpected) properties.
( )( ) ( ) ( )tGtFtGFrrrr
+=+ ( )( ) ( ) ( )tGtFtGFrrrr
−=−
( )( ) ( ) ( )tFtftFfrr
∗=∗
( )( ) ( ) ( )tGtFtGFrrrr
•=• ( )( ) ( ) ( )tGtFtGFrrrr
×=×
( )( ) ( )( )tgFtgFr
or
=
Example D extended. Sketch the curve C traced out by ( ) ktjtittGrrrr
sin2cos2 −+= .
x
y
z
Example E. Sketch the curve C traced out by ( ) ktjtittFrrrr
sin5cos2cos ++= .
Note: Text questions 32 and 33 ask you to find intersections.
32. cylinder 422=+ yx and curve traced out by ( ) ( ) ( ) ktjttitttF
rrrr++= ππ sincos
33. sphere 10222=++ zyx and curve traced out by ( ) ( ) ( ) ktjtittF
rrrr++= ππ sincos
Suggestions?
