Section 3.1 Calculus of Vector-Functions

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1 Section 3.1 Calculus of Vector-Functions De nition. A vector-valued function is a rule that assigns a vector to each member in a subset of R 1. In other words, a vector-valued function is an ordered triple of functions, say f (t), g (t), h (t), and can be expressed as For instance, r (t) = hf (t), g (t), h (t)i. r (t) = h1 + t, 2t, 2 ti 1 q (t) = t 1, ln (t), p À 2 t are vector-valued functions. The domain of a vector-valued function is a subset of all real number at which the function is well-de ned, i.e., Domain of r (t) = ft j r (t) = hf (t), g (t), h (t)i is de nedg = ft j each of f (t), g (t), h (t) is de nedg So = ft j f (t) is de nedg \ ft j g (t) is de nedg \ ft j g (t) is de nedg. D ( r) = D (f) \ D (g) \ D (h). Any vector-valued function r (t) = hx, y, zi may be written in terms of its components as x = f (t) y = g (t) z = h (t). Thus, the graph of a vector-valued function is a parametric curve in space. For instance, the function r (t) = h1 + t, 2t, 2 ti is de ned for all t. Its component form is x = 1 + t y = 2t z = 2 t. 1

2 The graph is a straight line with a direction h1, 2, 1i passing through (1, 0, 2). Example 1.1. Find the domain of 1 r (t) = t 1, ln (t), p À 2 t. Sol: We know that µ 1 D = ft 6= 1g = ( 1, 1) [ (1, 1) t 1 D (ln (t)) = ft > 0g = (0, 1) D p 2 t = ft 2g = ( 1, 2]. So µ 1 D ( r) = D \ D (ln (t)) \ D p 2 t t 1 = (( 1, 1) [ (1, 1)) \ (0, 1) \ ( 1, 2] = (( 1, 1) [ (1, 1)) \ (0, 2] = (0, 1) [ (1, 2]. ) ( O 1 2 Limits of vector-valued functions are de ned through components: For any vector-valued function r (t) = hf (t), g (t), h (t)i, the limit D E lim r (t) = lim f (t), lim g (t), lim h (t) t!a t!a t!a t!a exists if and only if the limits of all three components exist. Example 1.2. Consider r (t) = h2 cos t, sin t, ti. 2

3 (a) Find (b) Discuss and sketch its graph. Solution: (a) lim r (t) = t!0 lim r (t) = t!π/2 lim r (t), lim r (t). t!0 t!π /2 D lim E t t!0 (2 cos t), lim sin t, lim t!0 t!0 = h2, 0, 0i lim (2 cos t), lim t!π /2 t!π /2 D = 0, 1, π E. 2 À sin t, lim t t!π/2 (b) Let us rst take a look at the projection of the curve onto xy plane We know that its graph is a ellipse x = 2 cos t y = sin t. t = π 4 is the angle to x axis In 3D, as t increases from t = 0, the curve starting at (2, 0, 0) on xy-plane, moves in the way that its rst two component (x, y) moving along the ellipse in the above gure counter-clockwise while its z component increases linearly, as if we raise vertically the ellipse. The curve is on the elliptic cylinder, and is called elliptic helix. 3

4 De nition. For any vector-valued function r (t) = hf (t), g (t), h (t)i, if the limit of the di erence quotation r (t 0 + h) r (t 0 ) lim h!0 h exists, we say r (t) is di erentiable at t = t 0. In this case, we call the limit the derivative at t = t 0 and denote it by r 0 (t 0 ) or d r dt (t 0) = r 0 r (t 0 + h) r (t 0 ) (t 0 ) = lim. h!0 h We can show that r (t) is di erentiable at t = t 0 if and only if all three components are di erentiable and r 0 (t 0 ) = hf 0 (t 0 ), g 0 (t 0 ), h 0 (t 0 )i. The derivative vector for any t, r 0 (t), is again a vector-valued function. Higher order derivatives are then de ned accordingly. For instance, Geometrically, r 00 (t) = hf 00 (t), g 00 (t), h 00 (t)i r (t 0 + h) r (t 0 ) represents the vector from r (t 0 ) to r (t 0 + h). So for any small h > 0, r (t 0 + h) r (t 0 ) h 4

5 r (t 0 +h) r(t 0 ) r (t 0 ) r (t 0 +h) is a normalized (otherwise, the length of r (t 0 + h) r (t 0 ) would be a very small) secant direction. Therefore, the limit vector is "tangent" to the curve at t = t 0. De nition. We call r 0 (t 0 ) = hf 0 (t 0 ), g 0 (t 0 ), h 0 (t 0 )i the tangent vector of the parametric curve r (t) at t = t 0, and T (t 0 ) = r0 (t 0 ) j r 0 (t 0 )j the unit tangent vector. A curve r (t) is called smooth if r 0 (t) exists and r 0 (t) 6= 0. Example 1.3. Consider a circular helix r (t) = hcos t, sin t, ti. Find r 0 (t), T (t), and r 00 (t). Find also r 0 (0), T (0). 5

6 Solution: r 0 (t) = h sin t, cos t, 1i r 00 (t) = h cos t, sin t, 0i T (t) = 1 j r 0 (t)j r0 (t) 1 1 = p h sin t, cos t, 1i = p h sin t, cos t, 1i sin 2 t + cos 2 t r 0 (0) = h0, 1, 1i T (0) = 1 p 2 h0, 1, 1i. Properties of derivatives: function. λ is a scalar constant, f (t) is a scalar 1. Addition: ( u (t) + v (t)) 0 = u 0 (t) + v 0 (t) 2. Scalar function multiplication: 3. Scalar (constant) multiplication: 4. Dot product: 5. Cross product: 6. Chain rule: (f (t) u (t)) 0 = f (t) u 0 (t) + f 0 (t) u (t) (λ u (t)) 0 = λ u 0 (t) ( u (t) v (t)) 0 = u 0 (t) v (t) + u (t) v 0 (t) ( u (t) v (t)) 0 = u 0 (t) v (t) + u (t) v 0 (t) d u (f (t)) = dt µ d u (f (t)) df dt dt (t) = u0 (f (t)) f 0 (t). 6

7 All above properties can be veri ed by direction computations. As in the case of one variable functions, derivative r 0 (t 0 ) measures the rate (vector) at which function r (t) changes across t = t 0. Thus Note that since j r 0 (t 0 )j is the magnitude of the rate of change T (t 0 ) is the direction of change. j r (t)j = p r (t) r (t) = ( r (t) r (t)) 1 2 we have d dt j r (t)j = 1 2 ( r (t) r (t)) 1 2 ( r (t) r (t)) 0 = 1 2 j r (t)j 1 ( r 0 (t) r (t) + r (t) r 0 (t)) = r0 (t) r (t). j r (t)j This shows that in general, i.e., d dt j r (t)j 6= j r0 (t)j, Rate of change for j r (t)j 6= Magnitude of rate of change for r (t). In physics, if r (t) describes the position of a moving object, then v (t) = r 0 (t) υ (t) = j v (t)j is velocity is speed a (t) = v 0 (t) = r 00 (t) is acceleration. De nition. Integrals, inde nite and de nite, are de ned accordingly: Z Z Z Z À r (t) dt = f (t) dt, g (t) dt, h (t) dt Z b Z b Z b Z b À r (t) dt = f (t) dt, g (t) dt, h (t) dt. a a a a 7

8 Note that for inde nite integrals, we always end up a constant vector C = hc 1, C 2, C 3 i: Z Z r (t) dt = Example 1.4. Consider Z f (t) dt, Z g (t) dt, r (t) = 1 + t 3, te t, sin (2t). À h (t) dt + C. Find (a) r 0 (t), and (b) equations of the tangent at t = 0. Solution: (a) r 0 (t) = 3t 2, e t te t, 2 cos (2t). (b) The tangent line passes through the terminal point of the vector r (0) = h1, 0, 0i,i.e., passing through (1, 0, 0) with direction So the equations are r 0 (0) = h0, 1, 2i. x = 1 y = t z = 2t. Example 1.5. Find (a) R r (t) dt and (b) R π 0 r (t) = 2 cos t, sin t, 3t 2. r (t) dt if Solution: (a) Z Z Z Z À r (t) dt = 2 cos tdt, sin tdt, 3t 2 dt = 2 sin t + C 1, cos t + C 2, t 3 + C 3 = 2 sin t, cos t, t 3 + C where C = hc 1, C 2, C 3 i is an arbitrary constant vector. 8

9 (b) According to Fundamental Theorem of Calculus, Z π 0 r (t) dt = 2 sin t, cos t, t 3 j t=π t=0 = 2 sin π, cos π, π 3 h2 sin 0, cos 0, 0i = 0, 2, π 3 Homework: 1. Find domain and limit. t 1 (a) r (t) = t + 1, p À t, sin (π (t 2 + 1)) (b) r (t) = arctan t, e t2, ln t t, lim t!1 r (t) =? À, lim t!1 r (t) =? 2. Sketch the curve. Indicate with an arrow the direction in which t increases. (a) r (t) = hcos 2t, t, sin 2ti (b) r (t) = h1 + 2t, t, 3ti 3. Find a vector equation for the curve of intersection of two surfaces. (a) x 2 + y 2 = 4 and z = xy (b) z = 2x 2 + y 2 and y = x 2 4. Find the derivative. E (a) r (t) = Dt 2, cos 3t, e t3 (b) r (t) = t 2 a ³e t b + 2t c, a, b, and c are three constant vectors. 5. Find the integral. (a) R hsin πt, cos πt, e 2t i dt (b) R 1 ³8t 2 i + 9t 2 j + 25t 4 0 k dt 9

10 6. Find (i) unit tangent T at given point and (ii) equation of tangent line to the curve at that point. (a) r (t) = h2e t cos t, e t sin t, e t i ; (2, 0, 1) (b) r (t) = ln t, 2 p t, t 2 ; (0, 2, 1) 7. The angle between two curves at a point of intersection is deined as the angle between their tangents. Find the point of intersection and the angle between r 1 = ht, 1 t, 3 + t 2 i and r 2 = h3 s, s 2, s 2 i. 10