From scalar functions to vector-valued functions
A scalar function f : R → R f : \mathbb{R} \to \mathbb{R} f : R → R t t t f ( t ) f(t) f ( t ) x y xy x y
A vector-valued function generalises this: instead of returning a single number, it returns a vector . More precisely, it is a map
r : R → R n . \mathbf{r} : \mathbb{R} \to \mathbb{R}^n. r : R → R n . We are most interested in the case n = 3 n = 3 n = 3
r : R → R 3 . \mathbf{r} : \mathbb{R} \to \mathbb{R}^3. r : R → R 3 . Scalar function Vector-valued function Input t ∈ R t \in \mathbb{R} t ∈ R t ∈ R t \in \mathbb{R} t ∈ R Output f ( t ) ∈ R f(t) \in \mathbb{R} f ( t ) ∈ R r ( t ) ∈ R 3 \mathbf{r}(t) \in \mathbb{R}^3 r ( t ) ∈ R 3 Graph curve in R 2 \mathbb{R}^2 R 2 curve in R 3 \mathbb{R}^3 R 3 Calculus f ′ ( t ) f'(t) f ′ ( t ) ∫ f d t \int f\,dt ∫ f d t r ′ ( t ) \mathbf{r}'(t) r ′ ( t ) ∫ r d t \int \mathbf{r}\,dt ∫ r d t
The parameter t t t time , and r ( t ) \mathbf{r}(t) r ( t )
Component form and the connection to parametric curves
A vector-valued function r : R → R 3 \mathbf{r} : \mathbb{R} \to \mathbb{R}^3 r : R → R 3 component form as
r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) = f ( t ) i + g ( t ) j + h ( t ) k , \boxed{\mathbf{r}(t) = \bigl(f(t),\, g(t),\, h(t)\bigr) = f(t)\,\mathbf{i} + g(t)\,\mathbf{j} + h(t)\,\mathbf{k},} r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) = f ( t ) i + g ( t ) j + h ( t ) k , where f f f g g g h : R → R h : \mathbb{R} \to \mathbb{R} h : R → R component functions .
This is the 3D analogue of parametric curves : just as a 2D parametric curve
( x ( t ) , y ( t ) ) \bigl(x(t), y(t)\bigr) ( x ( t ) , y ( t ) ) traces a path in the plane, the vector-valued function r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) \mathbf{r}(t) = (f(t), g(t), h(t)) r ( t ) = ( f ( t ) , g ( t ) , h ( t )) space curve in R 3 \mathbb{R}^3 R 3 R 3 \mathbb{R}^3 R 3
In short: the component functions f f f g g g h h h x = f ( t ) x = f(t) x = f ( t ) y = g ( t ) y = g(t) y = g ( t ) z = h ( t ) z = h(t) z = h ( t )
▼ Exercise 1: Identify component functions Write out the three component functions of
r ( t ) = ( t 2 , e t , sin t ) . \mathbf{r}(t) = (t^2,\; e^t,\; \sin t). r ( t ) = ( t 2 , e t , sin t ) . Answer:
f ( t ) = t 2 , g ( t ) = e t , h ( t ) = sin t . f(t) = t^2, \quad g(t) = e^t, \quad h(t) = \sin t. f ( t ) = t 2 , g ( t ) = e t , h ( t ) = sin t . The corresponding parametric equations are x = t 2 x = t^2 x = t 2 y = e t y = e^t y = e t z = sin t z = \sin t z = sin t
▼ Exercise 2: Evaluate a vector-valued function Let r ( t ) = ( cos t , sin t , t ) \mathbf{r}(t) = (\cos t,\, \sin t,\, t) r ( t ) = ( cos t , sin t , t ) r ( 0 ) \mathbf{r}(0) r ( 0 ) r ( π ) \mathbf{r}(\pi) r ( π )
Answer:
r ( 0 ) = ( 1 , 0 , 0 ) , r ( π ) = ( − 1 , 0 , π ) . \mathbf{r}(0) = (1, 0, 0), \qquad \mathbf{r}(\pi) = (-1, 0, \pi). r ( 0 ) = ( 1 , 0 , 0 ) , r ( π ) = ( − 1 , 0 , π ) . Space curves: examples
The graph of a vector-valued function r ( t ) \mathbf{r}(t) r ( t ) space curve — a one-dimensional path winding through three-dimensional space.
Example 1: Circular helix
r ( t ) = ( cos t , sin t , t ) . \mathbf{r}(t) = (\cos t,\; \sin t,\; t). r ( t ) = ( cos t , sin t , t ) .
The x x x y y y unit circle x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1
The z z z
The result is a helix — a spiral staircase shape.
r(t) = (cos t, sin t, t), t ∈ [0, 4π]
Example 2: Toroidal wave
r ( t ) = ( ( 4 + sin 20 t ) cos t , ( 4 + sin 20 t ) sin t , cos 20 t ) . \mathbf{r}(t) = \bigl((4 + \sin 20t)\cos t,\;\; (4 + \sin 20t)\sin t,\;\; \cos 20t\bigr). r ( t ) = ( ( 4 + sin 20 t ) cos t , ( 4 + sin 20 t ) sin t , cos 20 t ) . The slowly-varying angle t t t sin 20 t \sin 20t sin 20 t cos 20 t \cos 20t cos 20 t
r(t) = ((4+sin 20t) cos t, (4+sin 20t) sin t, cos 20t)
Example 3: Torus knot
r ( t ) = ( ( 2 + cos 1.5 t ) cos t , ( 2 + cos 1.5 t ) sin t , sin 1.5 t ) . \mathbf{r}(t) = \bigl((2 + \cos 1.5t)\cos t,\;\; (2 + \cos 1.5t)\sin t,\;\; \sin 1.5t\bigr). r ( t ) = ( ( 2 + cos 1.5 t ) cos t , ( 2 + cos 1.5 t ) sin t , sin 1.5 t ) . Here the angular frequency 1.5 1.5 1.5 not close after one revolution. It closes after t t t 4 π 4\pi 4 π
r(t) = ((2+cos 1.5t) cos t, (2+cos 1.5t) sin t, sin 1.5t)
Limits and continuity
Limit
The limit of a vector-valued function is taken component-wise. If r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) \mathbf{r}(t) = (f(t), g(t), h(t)) r ( t ) = ( f ( t ) , g ( t ) , h ( t ))
lim t → a r ( t ) = ( lim t → a f ( t ) , lim t → a g ( t ) , lim t → a h ( t ) ) , \lim_{t \to a} \mathbf{r}(t)
=
\Bigl(\lim_{t \to a} f(t),\;\; \lim_{t \to a} g(t),\;\; \lim_{t \to a} h(t)\Bigr), t → a lim r ( t ) = ( t → a lim f ( t ) , t → a lim g ( t ) , t → a lim h ( t ) ) , provided all three component limits exist.
Example. Let r ( t ) = ( sin t t , e t , t 2 ) \mathbf{r}(t) = \left(\dfrac{\sin t}{t},\; e^t,\; t^2\right) r ( t ) = ( t sin t , e t , t 2 )
lim t → 0 r ( t ) = ( lim t → 0 sin t t , e 0 , 0 2 ) = ( 1 , 1 , 0 ) . \lim_{t \to 0} \mathbf{r}(t) = \left(\lim_{t \to 0}\frac{\sin t}{t},\; e^0,\; 0^2\right) = (1, 1, 0). t → 0 lim r ( t ) = ( t → 0 lim t sin t , e 0 , 0 2 ) = ( 1 , 1 , 0 ) .
Continuity
r \mathbf{r} r continuous at t = a t = a t = a
lim t → a r ( t ) = r ( a ) . \lim_{t \to a} \mathbf{r}(t) = \mathbf{r}(a). t → a lim r ( t ) = r ( a ) . Equivalently, r \mathbf{r} r a a a each component function f f f g g g h h h a a a
▼ Exercise 1: Compute a limit Find
lim t → 1 ( t 2 , ln t , t ) . \lim_{t \to 1} \bigl(t^2,\; \ln t,\; \sqrt{t}\bigr). t → 1 lim ( t 2 , ln t , t ) . Answer:
lim t → 1 t 2 = 1 , lim t → 1 ln t = 0 , lim t → 1 t = 1. \lim_{t \to 1} t^2 = 1, \quad
\lim_{t \to 1} \ln t = 0, \quad
\lim_{t \to 1} \sqrt{t} = 1. t → 1 lim t 2 = 1 , t → 1 lim ln t = 0 , t → 1 lim t = 1. So the limit is ( 1 , 0 , 1 ) (1, 0, 1) ( 1 , 0 , 1 )
▼ Exercise 2: Check continuity Is
r ( t ) = ( t 2 − 1 t − 1 , cos t , t ) \mathbf{r}(t) = \left(\frac{t^2 - 1}{t - 1},\; \cos t,\; t\right) r ( t ) = ( t − 1 t 2 − 1 , cos t , t ) continuous at t = 1 t = 1 t = 1
Answer:
The first component is t 2 − 1 t − 1 = t + 1 \dfrac{t^2-1}{t-1} = t+1 t − 1 t 2 − 1 = t + 1 t ≠ 1 t \neq 1 t = 1 r ( 1 ) \mathbf{r}(1) r ( 1 ) not defined at t = 1 t=1 t = 1 not continuous there.
(If we define r ( 1 ) = ( 2 , cos 1 , 1 ) \mathbf{r}(1) = (2, \cos 1, 1) r ( 1 ) = ( 2 , cos 1 , 1 )
▼ Exercise 3: Identify where continuity breaks For
r ( t ) = ( t , ln ( t − 1 ) , e t ) , \mathbf{r}(t) = \Bigl(\sqrt{t},\;\; \ln(t-1),\;\; e^t\Bigr), r ( t ) = ( t , ln ( t − 1 ) , e t ) , on what interval is r \mathbf{r} r
Answer:
t \sqrt{t} t t ≥ 0 t \ge 0 t ≥ 0 ln ( t − 1 ) \ln(t-1) ln ( t − 1 ) t > 1 t > 1 t > 1 e t e^t e t The intersection is t > 1 t > 1 t > 1 ( 1 , ∞ ) (1, \infty) ( 1 , ∞ )
Example: intersection curve from two surfaces
Problem. Find a vector-valued function r ( t ) \mathbf{r}(t) r ( t )
x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1 and the plane
y + z = 2. y + z = 2. y + z = 2. Solution.
The cylinder x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1 ( x , y ) (x, y) ( x , y ) unit circle at every height z z z parametric curve :
x = cos t , y = sin t . x = \cos t, \qquad y = \sin t. x = cos t , y = sin t . Now use the plane equation y + z = 2 y + z = 2 y + z = 2 z z z
z = 2 − y = 2 − sin t . z = 2 - y = 2 - \sin t. z = 2 − y = 2 − sin t . Therefore the intersection curve is
r ( t ) = ( cos t , sin t , 2 − sin t ) , t ∈ [ 0 , 2 π ] . \boxed{\mathbf{r}(t) = (\cos t,\;\; \sin t,\;\; 2 - \sin t), \quad t \in [0, 2\pi].} r ( t ) = ( cos t , sin t , 2 − sin t ) , t ∈ [ 0 , 2 π ] . Verification. For all t t t
x 2 + y 2 = cos 2 t + sin 2 t = 1 ✓ x^2 + y^2 = \cos^2 t + \sin^2 t = 1 \checkmark x 2 + y 2 = cos 2 t + sin 2 t = 1 ✓ y + z = sin t + ( 2 − sin t ) = 2 ✓ y + z = \sin t + (2 - \sin t) = 2 \checkmark y + z = sin t + ( 2 − sin t ) = 2 ✓ The graph below shows the cylinder x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1 y + z = 2 y+z=2 y + z = 2
Cylinder x²+y²=1 (blue) ∩ Plane y+z=2 (yellow) → r(t) = (cos t, sin t, 2−sin t)
Reading the components. The three component functions tell the whole story:
Component Formula Role x = f ( t ) x = f(t) x = f ( t ) cos t \cos t cos t moves left–right, period 2 π 2\pi 2 π y = g ( t ) y = g(t) y = g ( t ) sin t \sin t sin t moves front–back, period 2 π 2\pi 2 π z = h ( t ) z = h(t) z = h ( t ) 2 − sin t 2 - \sin t 2 − sin t altitude, highest when sin t = − 1 \sin t = -1 sin t = − 1 sin t = 1 \sin t = 1 sin t = 1
Notice that z z z y y y z = 2 z=2 z = 2 y y y z z z
▼ Exercise: Intersection of a cylinder with a different plane Find r ( t ) \mathbf{r}(t) r ( t ) x 2 + y 2 = 4 x^2 + y^2 = 4 x 2 + y 2 = 4 x + z = 3 x + z = 3 x + z = 3
Answer:
Parametrize the circle of radius 2: x = 2 cos t x = 2\cos t x = 2 cos t y = 2 sin t y = 2\sin t y = 2 sin t
From the plane: z = 3 − x = 3 − 2 cos t z = 3 - x = 3 - 2\cos t z = 3 − x = 3 − 2 cos t
r ( t ) = ( 2 cos t , 2 sin t , 3 − 2 cos t ) , t ∈ [ 0 , 2 π ] . \mathbf{r}(t) = (2\cos t,\;\; 2\sin t,\;\; 3 - 2\cos t), \quad t \in [0, 2\pi]. r ( t ) = ( 2 cos t , 2 sin t , 3 − 2 cos t ) , t ∈ [ 0 , 2 π ] . Check:
x 2 + y 2 = 4 cos 2 t + 4 sin 2 t = 4 ✓ x^2 + y^2 = 4\cos^2 t + 4\sin^2 t = 4 \checkmark x 2 + y 2 = 4 cos 2 t + 4 sin 2 t = 4 ✓ x + z = 2 cos t + 3 − 2 cos t = 3 ✓ x + z = 2\cos t + 3 - 2\cos t = 3 \checkmark x + z = 2 cos t + 3 − 2 cos t = 3 ✓ Mixed practice
▼ 🎯 Practice 1: Write a vector-valued function for a line Write a vector-valued function for the line through ( 1 , 2 , 3 ) (1, 2, 3) ( 1 , 2 , 3 ) ( 4 , − 1 , 0 ) (4, -1, 0) ( 4 , − 1 , 0 )
▼ ✅ Solution r ( t ) = ( 1 + 4 t , 2 − t , 3 ) , t ∈ R . \mathbf{r}(t) = (1 + 4t,\;\; 2 - t,\;\; 3), \quad t \in \mathbb{R}. r ( t ) = ( 1 + 4 t , 2 − t , 3 ) , t ∈ R . ▼ 🎯 Practice 2: Identify the space curve What familiar surface does the curve r ( t ) = ( 3 cos t , 3 sin t , 5 ) \mathbf{r}(t) = (3\cos t, 3\sin t, 5) r ( t ) = ( 3 cos t , 3 sin t , 5 )
▼ 💡 Hint Consider x 2 + y 2 x^2 + y^2 x 2 + y 2 z z z
▼ ✅ Solution x 2 + y 2 = 9 cos 2 t + 9 sin 2 t = 9 x^2 + y^2 = 9\cos^2 t + 9\sin^2 t = 9 x 2 + y 2 = 9 cos 2 t + 9 sin 2 t = 9 z = 5 z = 5 z = 5 circular cylinder x 2 + y 2 = 9 x^2 + y^2 = 9 x 2 + y 2 = 9 z = 5 z = 5 z = 5
▼ 🎯 Practice 3: Limit of a vector-valued function Evaluate
lim t → 0 ( 1 − cos t t 2 , sin 2 t t , e 3 t ) . \lim_{t \to 0} \left(\frac{1 - \cos t}{t^2},\;\; \frac{\sin 2t}{t},\;\; e^{3t}\right). t → 0 lim ( t 2 1 − cos t , t sin 2 t , e 3 t ) . ▼ 💡 Hint Use standard limits: lim t → 0 1 − cos t t 2 = 1 2 \lim_{t\to 0}\dfrac{1-\cos t}{t^2} = \dfrac{1}{2} lim t → 0 t 2 1 − cos t = 2 1 lim t → 0 sin 2 t t = 2 \lim_{t\to 0}\dfrac{\sin 2t}{t} = 2 lim t → 0 t sin 2 t = 2
▼ ✅ Solution lim t → 0 1 − cos t t 2 = 1 2 , lim t → 0 sin 2 t t = 2 , e 0 = 1. \lim_{t\to 0}\frac{1-\cos t}{t^2}=\frac{1}{2}, \quad
\lim_{t\to 0}\frac{\sin 2t}{t}=2, \quad
e^0=1. t → 0 lim t 2 1 − cos t = 2 1 , t → 0 lim t sin 2 t = 2 , e 0 = 1. So the limit is ( 1 2 , 2 , 1 ) \left(\dfrac{1}{2}, 2, 1\right) ( 2 1 , 2 , 1 )
▼ 🎯 Practice 4: Intersection curve Find r ( t ) \mathbf{r}(t) r ( t ) y 2 + z 2 = 9 y^2 + z^2 = 9 y 2 + z 2 = 9 x + y = 5 x + y = 5 x + y = 5
▼ 💡 Hint Parametrize the circle y 2 + z 2 = 9 y^2 + z^2 = 9 y 2 + z 2 = 9 y y y z z z x x x
▼ ✅ Solution Parametrize: y = 3 cos t y = 3\cos t y = 3 cos t z = 3 sin t z = 3\sin t z = 3 sin t
From the plane: x = 5 − y = 5 − 3 cos t x = 5 - y = 5 - 3\cos t x = 5 − y = 5 − 3 cos t
r ( t ) = ( 5 − 3 cos t , 3 cos t , 3 sin t ) , t ∈ [ 0 , 2 π ] . \mathbf{r}(t) = (5 - 3\cos t,\;\; 3\cos t,\;\; 3\sin t), \quad t \in [0, 2\pi]. r ( t ) = ( 5 − 3 cos t , 3 cos t , 3 sin t ) , t ∈ [ 0 , 2 π ] .