VECTOR FUNCTIONS. which a space curve proceeds at any point.

Transcription

1 3 VECTOR FUNCTIONS Tangent vectors show the direction in which a space curve proceeds at an point. The functions that we have been using so far have been real-valued functions. We now stud functions whose values are vectors because such functions are needed to describe curves and surfaces in space. We will also use vector-valued functions to describe the motion of objects through space. In particular, we will use them to derive Kepler s laws of planetar motion. 86

2 3. VECTOR FUNCTIONS AND SPACE CURVES In general, a function is a rule that assigns to each element in the domain an element in the range. A vector-valued function, or vector function, is simpl a function whose domain is a set of real numbers and whose range is a set of vectors. We are most interested in vector functions r whose values are three-dimensional vectors. This means that for ever number t in the domain of r there is a unique vector in V 3 denoted b rt. If f t, tt, and ht are the components of the vector rt, then f, t, and h are real-valued functions called the component functions of r and we can write rt f t, tt, ht f t i tt j ht k We usuall use the letter t to denote the independent variable because it represents time in most applications of vector functions. EXAPLE If rt t 3, ln3 t, st then the component functions are f t t 3 tt ln3 t ht st B our usual convention, the domain of r consists of all values of t for which the epression for rt is defined. The epressions t 3, ln3 t, and st are all defined when 3 t and t. Therefore the domain of r is the interval, 3. The limit of a vector function r is defined b taking the limits of its component functions as follows. N If lim t la rt L, this definition is equivalent to saing that the length and direction of the vector rt approach the length and direction of the vector L. If rt f t, tt, ht, then lim rt t l a lim t l a f t, lim tt, lim ht t l a t l a provided the limits of the component functions eist. Equivalentl, we could have used an definition (see Eercise 45). Limits of vector functions obe the same rules as limits of real-valued functions (see Eercise 43). EXAPLE Find lim rt, where rt t 3 i te t j sin t k. t l t - SOLUTION According to Definition, the limit of r is the vector whose components are the limits of the component functions of r: lim rt t l lim t 3 t l i lim t l tet j lim i k (b Equation 3.3.) sin t t l k t 87

3 88 CHAPTER 3 VECTOR FUNCTIONS A vector function r is continuous at a if lim rt ra t l a C P{f(t), g(t), h(t)} r(t)=kf(t), g(t), h(t)l FIGURE C is traced out b the tip of a moving position vector r(t). In view of Definition, we see that r is continuous at a if and onl if its component functions f, t, and h are continuous at a. There is a close connection between continuous vector functions and space curves. Suppose that f, t, and h are continuous real-valued functions on an interval I. Then the set C of all points,, in space, where f t tt ht and t varies throughout the interval I, is called a space curve. The equations in () are called parametric equations of C and t is called a parameter. We can think of C as being traced out b a moving particle whose position at time t is ( f t, tt, ht). If we now consider the vector function rt f t, tt, ht, then rt is the position vector of the point P( f t, tt, ht) on C. Thus an continuous vector function r defines a space curve C that is traced out b the tip of the moving vector rt, as shown in Figure. V EXAPLE 3 Describe the curve defined b the vector function rt t, 5t, 6t TEC Visual 3.A shows several curves being traced out b position vectors, including those in Figures and. SOLUTION The corresponding parametric equations are t 5t 6t which we recognie from Equations.5. as parametric equations of a line passing through the point,, and parallel to the vector, 5, 6. Alternativel, we could observe that the function can be written as r r tv, where r,, and v, 5, 6, and this is the vector equation of a line as given b Equation.5.. Plane curves can also be represented in vector notation. For instance, the curve given b the parametric equations t t and t (see Eample in Section.) could also be described b the vector equation rt t t, t t t i t j where i, and j,. V EXAPLE 4 Sketch the curve whose vector equation is rt cos t i sin t j t k SOLUTION The parametric equations for this curve are cos t sin t t FIGURE (,, ) π,, Since cos t sin t, the curve must lie on the circular clinder. The point,, lies directl above the point,,, which moves counterclockwise around the circle in the -plane. (See Eample in Section..) Since t, the curve spirals upward around the clinder as t increases. The curve, shown in Figure, is called a heli.

4 SECTION 3. VECTOR FUNCTIONS AND SPACE CURVES 89 The corkscrew shape of the heli in Eample 4 is familiar from its occurrence in coiled springs. It also occurs in the model of DNA (deoribonucleic acid, the genetic material of living cells). In 953 James Watson and Francis Crick showed that the structure of the DNA molecule is that of two linked, parallel helies that are intertwined as in Figure 3. In Eamples 3 and 4 we were given vector equations of curves and asked for a geometric description or sketch. In the net two eamples we are given a geometric description of a curve and are asked to find parametric equations for the curve. FIGURE 3 N Figure 4 shows the line segment PQ in Eample 5. Q(, _, 3) EXAPLE 5 Find a vector equation and parametric equations for the line segment that joins the point P, 3, to the point Q,, 3. SOLUTION In Section.5 we found a vector equation for the line segment that joins the tip of the vector to the tip of the vector : r r rt tr tr t (See Equation.5.4.) Here we take r, 3, and r,, 3 to obtain a vector equation of the line segment from P to Q: rt t, 3, t,, 3 t or rt t, 3 4t, 5t t FIGURE 4 P(, 3, _) The corresponding parametric equations are t 3 4t 5t t V EXAPLE 6 Find a vector function that represents the curve of intersection of the clinder and the plane. SOLUTION Figure 5 shows how the plane and the clinder intersect, and Figure 6 shows the curve of intersection C, which is an ellipse. += (, _, 3) C (_,, ) (,, ) + = (,, ) FIGURE 5 FIGURE 6

5 8 CHAPTER 3 VECTOR FUNCTIONS The projection of C onto the -plane is the circle,. So we know from Eample in Section. that we can write cos t sin t t From the equation of the plane, we have sin t So we can write parametric equations for C as cos t sin t sin t t The corresponding vector equation is rt cos t i sin t j sin t k t This equation is called a parametriation of the curve C. The arrows in Figure 6 indicate the direction in which C is traced as the parameter t increases. USING COPUTERS TO DRAW SPACE CURVES Space curves are inherentl more difficult to draw b hand than plane curves; for an accurate representation we need to use technolog. For instance, Figure 7 shows a computergenerated graph of the curve with parametric equations 4 sin t cos t 4 sin t sin t cos t FIGURE 7 A toroidal spiral It s called a toroidal spiral because it lies on a torus. Another interesting curve, the trefoil knot, with equations cos.5t cos t cos.5t sin t sin.5t is graphed in Figure 8. It wouldn t be eas to plot either of these curves b hand. Even when a computer is used to draw a space curve, optical illusions make it difficult to get a good impression of what the curve reall looks like. (This is especiall true in Figure 8. See Eercise 44.) The net eample shows how to cope with this problem. FIGURE 8 A trefoil knot EXAPLE 7 Use a computer to draw the curve with vector equation This curve is called a twisted cubic. rt t, t, t 3. SOLUTION We start b using the computer to plot the curve with parametric equations t, t, t 3 for t. The result is shown in Figure 9(a), but it s hard to see the true nature of the curve from that graph alone. ost three-dimensional computer graphing programs allow the user to enclose a curve or surface in a bo instead of displaing the coordinate aes. When we look at the same curve in a bo in Figure 9(b), we have a much clearer picture of the curve. We can see that it climbs from a lower corner of the bo to the upper corner nearest us, and it twists as it climbs.

6 SECTION 3. VECTOR FUNCTIONS AND SPACE CURVES 8 6 _6 _ 6 6 _6 _ 4 _6 4 4 (a) (b) (c) 8 8 _ 4 4 _4 _4 3 4 (d) _8 (e) _8 3 4 (f) FIGURE 9 Views of the twisted cubic TEC In Visual 3.B ou can rotate the bo in Figure 9 to see the curve from an viewpoint. We get an even better idea of the curve when we view it from different vantage points. Part (c) shows the result of rotating the bo to give another viewpoint. Parts (d), (e), and (f) show the views we get when we look directl at a face of the bo. In particular, part (d) shows the view from directl above the bo. It is the projection of the curve on the -plane, namel, the parabola. Part (e) shows the projection on the -plane, the cubic curve 3. It s now obvious wh the given curve is called a twisted cubic. FIGURE Another method of visualiing a space curve is to draw it on a surface. For instance, the twisted cubic in Eample 7 lies on the parabolic clinder. (Eliminate the parameter from the first two parametric equations, t and t.) Figure shows both the clinder and the twisted cubic, and we see that the curve moves upward from the origin along the surface of the clinder. We also used this method in Eample 4 to visualie the heli ling on the circular clinder (see Figure ). A third method for visualiing the twisted cubic is to realie that it also lies on the clinder 3. So it can be viewed as the curve of intersection of the clinders and 3. (See Figure.) 8 TEC Visual 3.C shows how curves arise as intersections of surfaces. 4 _4 FIGURE _8 _ 4

7 8 CHAPTER 3 VECTOR FUNCTIONS N Some computer algebra sstems provide us with a clearer picture of a space curve b enclosing it in a tube. Such a plot enables us to see whether one part of a curve passes in front of or behind another part of the curve. For eample, Figure 3 shows the curve of Figure (b) as rendered b the tubeplot command in aple. We have seen that an interesting space curve, the heli, occurs in the model of DNA. Another notable eample of a space curve in science is the trajector of a positivel charged particle in orthogonall oriented electric and magnetic fields E and B. Depending on the initial velocit given the particle at the origin, the path of the particle is either a space curve whose projection on the horiontal plane is the ccloid we studied in Section. [Figure (a)] or a curve whose projection is the trochoid investigated in Eercise 4 in Section. [Figure (b)]. B B E E (a) r(t) = kt-sin t, -cos t, tl FIGURE otion of a charged particle in orthogonall oriented electric and magnetic fields t t 3 3 (b) r(t) = kt- sin t, - cos t, tl For further details concerning the phsics involved and animations of the trajectories of the particles, see the following websites: N N FIGURE Beam/ 3. EXERCISES Find the domain of the vector function.. rt s4 t, e 3t, lnt. rt t t i sin t j ln9 t k 9. rt t, cos t, sin t.. rt, cos t, sin t. 3. rt t i t 4 j t 6 k 4. rt cos t i cos t j sin t k rt t, 3t, t rt t i tj k 3 6 Find the limit lim cos t, sin t, t ln t t l lim t l e t s t 3,, t t t lim t l e 3t i t sin t j cos t k lim t, e t l arctan t, ln t t 7 4 Sketch the curve with the given vector equation. Indicate with an arrow the direction in which t increases. 7. rt sin t, t 8. rt t 3, t 5 8 Find a vector equation and parametric equations for the line segment that joins P to Q. 5. P,,, 6. P,,, 7. P,,, 8. P, 4,, Q,, 3 Q, 3, Q4,, 7 Q6,, 9 4 atch the parametric equations with the graphs (labeled I VI). Give reasons for our choices. 9. cos 4t, t,. t, t, e t sin 4t

8 CHAPTER 3 VECTOR FUNCTIONS AND SPACE CURVES 83. t, t, t. e t cos t, e t sin t, 3. cos t, sin t, sin 5t 4. cos t, sin t, ln t I II e t ; 33. Graph the curve with parametric equations cos 6t cos t, cos 6t sin t, cos 6t. Eplain the appearance of the graph b showing that it lies on a cone. ; 34. Graph the curve with parametric equations s.5 cos t cos t s.5 cos t sin t.5 cos t III IV Eplain the appearance of the graph b showing that it lies on a sphere. 35. Show that the curve with parametric equations t, 3t, t 3 passes through the points (, 4, ) and (9, 8, 8) but not through the point (4, 7, 6). V VI Find a vector function that represents the curve of intersection of the two surfaces. 36. The clinder 4 and the surface 37. The cone s and the plane 38. The paraboloid 4 and the parabolic clinder 5. Show that the curve with parametric equations t cos t, t sin t, t lies on the cone, and use this fact to help sketch the curve. 6. Show that the curve with parametric equations sin t, cos t, sin t is the curve of intersection of the surfaces and. Use this fact to help sketch the curve. 7. At what points does the curve rt t i t t k intersect the paraboloid? 8. At what points does the heli rt sin t, cos t, t intersect the sphere 5? ; 9 3 Use a computer to graph the curve with the given vector equation. ake sure ou choose a parameter domain and viewpoints that reveal the true nature of the curve rt cos t sin t, sin t sin t, cos t rt t, ln t, t rt t, t sin t, t cos t 3. rt t, e t, cos t ; 39. Tr to sketch b hand the curve of intersection of the circular clinder 4 and the parabolic clinder. Then find parametric equations for this curve and use these equations and a computer to graph the curve. ; 4. Tr to sketch b hand the curve of intersection of the parabolic clinder and the top half of the ellipsoid Then find parametric equations for this curve and use these equations and a computer to graph the curve. 4. If two objects travel through space along two different curves, it s often important to know whether the will collide. (Will a missile hit its moving target? Will two aircraft collide?) The curves might intersect, but we need to know whether the objects are in the same position at the same time. Suppose the trajectories of two particles are given b the vector functions r t t, 7t, t for t. Do the particles collide? 4. Two particles travel along the space curves r t t, t, t 3 r t t, 6t, 4t Do the particles collide? Do their paths intersect? 43. Suppose u and v are vector functions that possess limits as t l a and let c be a constant. Prove the following properties of limits. (a) lim ut vt lim ut lim vt t l a t l a t l a r t 4t 3, t, 5t 6

9 84 CHAPTER 3 VECTOR FUNCTIONS (b) (c) (d) lim cut c lim ut t l a t l a lim t l a lim t l a ut vt lim ut lim vt t l a t l a ut vt lim ut lim vt t l a t l a 44. The view of the trefoil knot shown in Figure 8 is accurate, but it doesn t reveal the whole stor. Use the parametric equations cos.5t cos t cos.5t sin t sin.5t to sketch the curve b hand as viewed from above, with gaps indicating where the curve passes over itself. Start b showing that the projection of the curve onto the -plane has polar coordinates r cos.5t and, so r varies between and 3. Then show that has maimum and minimum values when the projection is halfwa between r and r 3. ; When ou have finished our sketch, use a computer to draw the curve with viewpoint directl above and compare with our sketch. Then use the computer to draw the curve from several other viewpoints. You can get a better impression of the curve if ou plot a tube with radius. around the curve. (Use the tubeplot command in aple.) 45. Show that lim t l a rt b if and onl if for ever there is a number such that t if t a then rt b 3. DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS Later in this chapter we are going to use vector functions to describe the motion of planets and other objects through space. Here we prepare the wa b developing the calculus of vector functions. DERIVATIVES TEC Visual 3. shows an animation of Figure. C C FIGURE P r(t+h)-r(t) r(t) r(t+h) (a) The secant vector rª(t) P r(t) r(t+h) (b) The tangent vector Q r(t+h)-r(t) h Q The derivative r of a vector function r is defined in much the same wa as for realvalued functions: dr rt h rt rt lim h l h if this limit eists. The geometric significance of this definition is shown in Figure. If the points P and Q have position vectors rt and rt h, then PQ l represents the vector rt h rt, which can therefore be regarded as a secant vector. If h, the scalar multiple hrt h rt has the same direction as rt h rt. As h l, it appears that this vector approaches a vector that lies on the tangent line. For this reason, the vector rt is called the tangent vector to the curve defined b r at the point P, provided that rt eists and rt. The tangent line to C at P is defined to be the line through P parallel to the tangent vector rt. We will also have occasion to consider the unit tangent vector, which is Tt rt rt The following theorem gives us a convenient method for computing the derivative of a vector function r: just differentiate each component of r. THEORE If rt f t, tt, ht f t i tt j ht k, where f, t, and h are differentiable functions, then rt f t, tt, ht f t i tt j ht k

10 SECTION 3. DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS 85 PROOF rt lim rt t rt t l t lim f t t, tt t, ht t f t, tt, ht t l t lim t l lim t l f t t f t, t f t, tt, ht tt t tt, t f t t f t tt t tt, lim t t l t ht t ht t ht t ht, lim t l t V EXAPLE (a) Find the derivative of rt t 3 i te t j sin t k. (b) Find the unit tangent vector at the point where t. SOLUTION (a) According to Theorem, we differentiate each component of r: rt 3t i te t j cos t k (b) Since r i and r j k, the unit tangent vector at the point,, is T r j k r j k s 4 s5 s5 r() (, ) rª() EXAPLE For the curve rt st i t j, find rt and sketch the position vector r and the tangent vector r. SOLUTION We have rt i j st and r i j FIGURE The curve is a plane curve and elimination of the parameter from the equations st, t gives,. In Figure we draw the position vector r i j starting at the origin and the tangent vector r starting at the corresponding point,. V EXAPLE 3 Find parametric equations for the tangent line to the heli with parametric equations cos t sin t t at the point,,. SOLUTION The vector equation of the heli is rt cos t, sin t, t, so rt sin t, cos t,

11 86 CHAPTER 3 VECTOR FUNCTIONS The parameter value corresponding to the point,, is t, so the tangent vector there is r,,. The tangent line is the line through,, parallel to the vector,,, so b Equations.5. its parametric equations are t t N The heli and the tangent line in Eample 3 are shown in Figure FIGURE _ N In Section 3.4 we will see how rt and rt can be interpreted as the velocit and acceleration vectors of a particle moving through space with position vector rt at time t. Just as for real-valued functions, the second derivative of a vector function r is the derivative of r, that is, r r. For instance, the second derivative of the function in Eample 3 is rt cos t, sin t, DIFFERENTIATION RULES The net theorem shows that the differentiation formulas for real-valued functions have their counterparts for vector-valued functions. 3 THEORE Suppose u and v are differentiable vector functions, c is a scalar, and f is a real-valued function. Then d ut vt ut vt d cut cut d f tut f tut f tut d ut vt ut vt ut vt d ut vt ut vt ut vt d 6. u f t f tu f t (Chain Rule) This theorem can be proved either directl from Definition or b using Theorem and the corresponding differentiation formulas for real-valued functions. The proof of Formula 4 follows; the remaining proofs are left as eercises.

12 SECTION 3. DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS 87 PROOF OF FORULA 4 Let ut f t, f t, f 3 t vt t t, t t, t 3 t Then ut vt f t t t f t t t f 3 t t 3 t 3 f i t t i t i so the ordinar Product Rule gives d ut vt d 3 f i t t i t 3 d i i f it t i t 3 i 3 i f i t t i t f i t t i t f i t t i t 3 f i t t i t i ut vt ut vt V EXAPLE 4 Show that if (a constant), then rt is orthogonal to rt for all t. SOLUTION Since and c rt c rt rt rt c is a constant, Formula 4 of Theorem 3 gives d rt rt rt rt rt rt rt rt Thus rt rt, which sas that rt is orthogonal to rt. Geometricall, this result sas that if a curve lies on a sphere with center the origin, then the tangent vector rt is alwas perpendicular to the position vector rt. INTEGRALS The definite integral of a continuous vector function rt can be defined in much the same wa as for real-valued functions ecept that the integral is a vector. But then we can epress the integral of r in terms of the integrals of its component functions f, t, and h as follows. (We use the notation of Chapter 5.) and so b rt lim a n l n rt* i t i lim n l n f t* i t i n tt* i t j n i i i ht* i t k b a rt b f t i b tt j b a a a ht k

13 88 CHAPTER 3 VECTOR FUNCTIONS This means that we can evaluate an integral of a vector function b integrating each component function. We can etend the Fundamental Theorem of Calculus to continuous vector functions as follows: b rt Rt] b a Rb Ra a where R is an antiderivative of r, that is, Rt rt. We use the notation rt for indefinite integrals (antiderivatives). EXAPLE 5 If rt cos t i sin t j t k, then rt cos t i sin t j t k sin t i cos t j t k C where C is a vector constant of integration, and rt [ sin t i cos t j t k] i j 4 k 3. EXERCISES. The figure shows a curve C given b a vector function rt. (a) Draw the vectors r4.5 r4 and r4. r4. (b) Draw the vectors r4.5 r4.5 (c) Write epressions for r4 and the unit tangent vector T(4). (d) Draw the vector T(4). and r(4.5) C r(4.) r(4) r4. r4. R P Q (b) Draw the vector r starting at (, ) and compare it with the vector r. r. Eplain wh these vectors are so close to each other in length and direction. 3 8 (a) Sketch the plane curve with the given vector equation. (b) Find rt. (c) Sketch the position vector rt and the tangent vector rt for the given value of t. 3. rt t, t, t 4. rt t, st, t 5. rt sin t i cos t j, 6. rt e t i e t j, 7. rt e t i e 3t j, t t t 4 8. rt cos t i sin t j, t 6. (a) ake a large sketch of the curve described b the vector function rt t, t, t, and draw the vectors r(), r(.), and r(.) r(). 9 6 Find the derivative of the vector function. 9. rt t sin t, t, t cos t

14 SECTION 3. DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS rt e t i j ln 3t k rt tan t, sec t, t rt i j e 4t k rt sin t i s t j k rt at cos 3t i b sin 3 t j c cos 3 t k rt a t b t c 6. rt t a b t c 7 Find the unit tangent vector Tt at the point with the given value of the parameter t. 7. rt te t, arctan t, e t, 8. rt 4st i t j t k,. rt sin t i cos t j tan t k,. If rt t, t, t 3, find rt, T, rt, and rt rt.. If rt e t, e t, te t, find T, r, and rt rt. 3 6 Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. 3. st, t 3 t, t 3 t; 4. e t, te t, te t ; 6. ln t, st, t ; ; 7 9 Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustrate b graphing both the curve and the tangent line on a common screen. 7. t, e t, t t ; t 8. cos t, sin t, 4 cos t; 9. t cos t, t, t sin t; t 9. rt cos t i 3t j sin t k,,, t 5. e t cos t, e t sin t, e t ;,,,, t 4,, 3,,,, (s3,, ) 3. (a) Find the point of intersection of the tangent lines to the curve rt sin t, sin t, cos t at the points where t and t.5. ; (b) Illustrate b graphing the curve and both tangent lines. 3. The curves r t t, t, t 3 and r t sin t, sin t, t intersect at the origin. Find their angle of intersection correct to the nearest degree. 3. At what point do the curves r t t, t, 3 t and r s 3 s, s, s intersect? Find their angle of intersection correct to the nearest degree Evaluate the integral t 3 i 9t j 5t 4 k 4 t j t t k 3 sin t cos t i 3 sin t cos t j sin t cos t k (t i tst j t sin t k) e t i t j ln t k cos t i sin t j t k 39. Find rt if rt t i 3t j st k and r i j. 4. Find rt if rt t i e t j te t k and r i j k. 4. Prove Formula of Theorem Prove Formula 3 of Theorem Prove Formula 5 of Theorem Prove Formula 6 of Theorem If ut sin t, cos t, t and vt t, cos t, sin t, use Formula 4 of Theorem 3 to find 46. If u and v are the vector functions in Eercise 45, use Formula 5 of Theorem 3 to find 47. Show that if r is a vector function such that r eists, then d 48. Find an epression for ut vt wt. 49. d If rt, show that rt rt. d rt rt rt rt rt rt rt rt rt [Hint: ] d ut vt d ut vt 5. If a curve has the propert that the position vector rt is alwas perpendicular to the tangent vector rt, show that the curve lies on a sphere with center the origin. 5. If ut rt rt rt, show that ut rt rt rt

15 83 CHAPTER 3 VECTOR FUNCTIONS 3.3 ARC LENGTH AND CURVATURE In Section. we defined the length of a plane curve with parametric equations f t, tt, a t b, as the limit of lengths of inscribed polgons and, for the case where f and t are continuous, we arrived at the formula L b s f t tt b a d d The length of a space curve is defined in eactl the same wa (see Figure ). Suppose that the curve has the vector equation rt f t, tt, ht, a t b, or, equivalentl, the parametric equations f t, tt, ht, where f, t, and h are continuous. If the curve is traversed eactl once as t increases from a to b, then it can be shown that its length is a FIGURE The length of a space curve is the limit of lengths of inscribed polgons. L b s f t tt ht a b d d d a Notice that both of the arc length formulas () and () can be put into the more compact form 3 L b a rt because, for plane curves rt f t i tt j, rt f t i tt j s f t tt and for space curves rt f t i tt j ht k, rt f t i tt j ht k s f t tt ht N Figure shows the arc of the heli whose length is computed in Eample. V EXAPLE Find the length of the arc of the circular heli with vector equation rt cos t i sin t j t k from the point,, to the point,,. SOLUTION Since rt sin t i cos t j k, we have (,, π) (,, ) FIGURE rt ssin t cos t s The arc from,, to,, is described b the parameter interval t and so, from Formula 3, we have L rt s s A single curve C can be represented b more than one vector function. For instance, the twisted cubic 4 r t t, t, t 3 t

16 SECTION 3.3 ARC LENGTH AND CURVATURE 83 could also be represented b the function 5 r u e u, e u, e 3u u ln where the connection between the parameters t and u is given b t e u. We sa that Equations 4 and 5 are parametriations of the curve C. If we were to use Equation 3 to compute the length of C using Equations 4 and 5, we would get the same answer. In general, it can be shown that when Equation 3 is used to compute arc length, the answer is independent of the parametriation that is used. Now we suppose that C is a curve given b a vector function rt f ti ttj htk a t b where r is continuous and C is traversed eactl once as t increases from a to b. We define its arc length function s b st t ru du t a du du d du d 6 du a d s(t) r(a) FIGURE 3 r(t) C Thus st is the length of the part of C between ra and rt. (See Figure 3.) If we differentiate both sides of Equation 6 using Part of the Fundamental Theorem of Calculus, we obtain 7 ds rt It is often useful to parametrie a curve with respect to arc length because arc length arises naturall from the shape of the curve and does not depend on a particular coordinate sstem. If a curve rt is alread given in terms of a parameter t and st is the arc length function given b Equation 6, then we ma be able to solve for t as a function of s: t ts. Then the curve can be reparametried in terms of s b substituting for t: r rts. Thus, if s 3 for instance, rt3 is the position vector of the point 3 units of length along the curve from its starting point. EXAPLE Reparametrie the heli rt cos t i sin t j t k with respect to arc length measured from,, in the direction of increasing t. SOLUTION The initial point,, corresponds to the parameter value t. From Eample we have ds rt s and so s st t ru du t s du s t Therefore t ss and the required reparametriation is obtained b substituting for t: rts cos(ss ) i sin(ss ) j (ss ) k CURVATURE A parametriation rt is called smooth on an interval I if r is continuous and rt on I. A curve is called smooth if it has a smooth parametriation. A smooth curve has no sharp corners or cusps; when the tangent vector turns, it does so continuousl.

17 83 CHAPTER 3 VECTOR FUNCTIONS C FIGURE 4 Unit tangent vectors at equall spaced points on C TEC Visual 3.3A shows animated unit tangent vectors, like those in Figure 4, for a variet of plane curves and space curves. If C is a smooth curve defined b the vector function r, recall that the unit tangent vector Tt is given b and indicates the direction of the curve. From Figure 4 ou can see that Tt changes direction ver slowl when C is fairl straight, but it changes direction more quickl when C bends or twists more sharpl. The curvature of C at a given point is a measure of how quickl the curve changes direction at that point. Specificall, we define it to be the magnitude of the rate of change of the unit tangent vector with respect to arc length. (We use arc length so that the curvature will be independent of the parametriation.) 8 DEFINITION The curvature of a curve is where T is the unit tangent vector. Tt rt rt dt ds The curvature is easier to compute if it is epressed in terms of the parameter t instead of s, so we use the Chain Rule (Theorem 3..3, Formula 6) to write dt dt and ds ds dt ds ds But ds rt from Equation 7, so dt 9 t Tt rt V EXAPLE 3 Show that the curvature of a circle of radius a is a. SOLUTION We can take the circle to have center the origin, and then a parametriation is rt a cos t i a sin t j Therefore so and rt a sin t i a cos t j Tt and rt rt sin t i cos t j Tt cos t i sin t j rt a This gives Tt, so using Equation 9, we have t Tt rt a The result of Eample 3 shows that small circles have large curvature and large circles have small curvature, in accordance with our intuition. We can see directl from the defi-

18 SECTION 3.3 ARC LENGTH AND CURVATURE 833 nition of curvature that the curvature of a straight line is alwas because the tangent vector is constant. Although Formula 9 can be used in all cases to compute the curvature, the formula given b the following theorem is often more convenient to appl. THEORE The curvature of the curve given b the vector function r is t rt rt rt 3 PROOF Since and ds, we have T r r r r r ds T T so the Product Rule (Theorem 3..3, Formula 3) gives r d s ds T T Using the fact that T T (see Eample in Section.4), we have rr ds T T Now for all t, so T and T are orthogonal b Eample 4 in Section 3.. Therefore, b Theorem.4.6, Thus and Tt rr ds T T ds T T ds T T rr ds T r rr r rr r 3 EXAPLE 4 Find the curvature of the twisted cubic rt t, t, t 3 at a general point and at,,. SOLUTION We first compute the required ingredients: rt, t, 3t rt,, 6t rt s 4t 9t 4 rt rt i j t k 3t 6t i 6t j k 6t rt rt s36t 4 36t 4 s9t 4 9t

19 834 CHAPTER 3 VECTOR FUNCTIONS Theorem then gives At the origin, where t, the curvature is. For the special case of a plane curve with equation f, we choose as the parameter and write r i f j. Then r i f j and r f j. Since i j k and j j, we have r r f k. We also have s f and so, b Theorem, r t rt rt rt 3 s 9t 9t 4 4t 9t 4 3 f f 3 = EXAPLE 5 Find the curvature of the parabola at the points,,,, and, 4. SOLUTION Since and, Formula gives FIGURE 5 The parabola = and its curvature function =k() The curvature at, is. At, it is At, 4 it is Observe from the epression for or the graph of in Figure 5 that l as l. This corresponds to the fact that the parabola appears to become flatter as l. THE NORAL AND BINORAL VECTORS N We can think of the normal vector as indicating the direction in which the curve is turning at each point. B(t) FIGURE 6 T(t) N(t) At a given point on a smooth space curve rt, there are man vectors that are orthogonal to the unit tangent vector Tt. We single out one b observing that, because Tt for all t, we have Tt Tt b Eample 4 in Section 3., so Tt is orthogonal to Tt. Note that Tt is itself not a unit vector. But if r is also smooth, we can define the principal unit normal vector Nt (or simpl unit normal) as Nt Tt Tt The vector Bt Tt Nt is called the binormal vector. It is perpendicular to both T and N and is also a unit vector. (See Figure 6.) EXAPLE 6 Find the unit normal and binormal vectors for the circular heli rt cos t i sin t j t k

20 SECTION 3.3 ARC LENGTH AND CURVATURE 835 N Figure 7 illustrates Eample 6 b showing the vectors T, N, and B at two locations on the heli. In general, the vectors T, N, and B, starting at the various points on a curve, form a set of orthogonal vectors, called the TNB frame, that moves along the curve as t varies. This TNB frame plas an important role in the branch of mathematics known as differential geometr and in its applications to the motion of spacecraft. N N B FIGURE 7 T B T TEC Visual 3.3B shows how the TNB frame moves along several curves. N Figure 8 shows the heli and the osculating plane in Eample 7. =_+ π FIGURE 8 P SOLUTION We first compute the ingredients needed for the unit normal vector: rt sin t i cos t j k Tt Tt s cos t i sin t j Nt This shows that the normal vector at a point on the heli is horiontal and points toward the -ais. The binormal vector is Bt Tt Nt i j k sin t cos t sin t, cos t, s s cos t sin t The plane determined b the normal and binormal vectors N and B at a point P on a curve C is called the normal plane of C at P. It consists of all lines that are orthogonal to the tangent vector T. The plane determined b the vectors T and N is called the osculating plane of C at P. The name comes from the Latin osculum, meaning kiss. It is the plane that comes closest to containing the part of the curve near P. (For a plane curve, the osculating plane is simpl the plane that contains the curve.) The circle that lies in the osculating plane of C at P, has the same tangent as C at P, lies on the concave side of C (toward which N points), and has radius (the reciprocal of the curvature) is called the osculating circle (or the circle of curvature) of C at P. It is the circle that best describes how C behaves near P; it shares the same tangent, normal, and curvature at P. V EXAPLE 7 Find the equations of the normal plane and osculating plane of the heli in Eample 6 at the point P,,. SOLUTION The normal plane at P has normal vector r,,, so an equation is or The osculating plane at P contains the vectors T and N, so its normal vector is T N B. From Eample 6 we have Bt s rt rt sin t i cos t j k s Tt Tt cos t i sin t j cos t, sin t, sin t, cos t, B s,, s A simpler normal vector is,,, so an equation of the osculating plane is rt s Tt s or

21 836 CHAPTER 3 VECTOR FUNCTIONS osculating circle = EXAPLE 8 Find and graph the osculating circle of the parabola at the origin. SOLUTION From Eample 5 the curvature of the parabola at the origin is. So the radius of the osculating circle at the origin is and its center is (, ). Its equation is therefore ( ) 4 FIGURE 9 For the graph in Figure 9 we use parametric equations of this circle: cos t sin t We summarie here the formulas for unit tangent, unit normal and binormal vectors, and curvature. TEC Visual 3.3C shows how the osculating circle changes as a point moves along a curve. Tt rt rt dt Nt ds Tt rt Tt Tt rt rt rt 3 Bt Tt Nt 3.3 EXERCISES 6 Find the length of the curve.. rt sin t, 5t, cos t,. rt t, t, 3 t 3, 4. rt cos t i sin t j ln cos t k, 5. rt i t j t 3 k, 6. rt t i 8t 3 j 3t k, 7 9 Find the length of the curve correct to four decimal places. (Use our calculator to approimate the integral.) 7. rt st, t, t, t 4 8. rt t, ln t, t ln t, 9. rt sin t, cos t, tan t, t 3. rt st i e t j e t k, t t t t t 4 t 4 t ;. Graph the curve with parametric equations sin t, sin t, sin 3t. Find the total length of this curve correct to four decimal places.. Let C be the curve of intersection of the parabolic clinder and the surface 3. Find the eact length of C from the origin to the point 6, 8, 36.. Find, correct to four decimal places, the length of the curve of intersection of the clinder 4 4 and the plane. 3 4 Reparametrie the curve with respect to arc length measured from the point where t in the direction of increasing t Suppose ou start at the point,, 3 and move 5 units along the curve 3 sin t, 4t, 3 cos t in the positive direction. Where are ou now? 6. Reparametrie the curve with respect to arc length measured from the point (, ) in the direction of increasing t. Epress the reparametriation in its simplest form. What can ou conclude about the curve? 7 (a) Find the unit tangent and unit normal vectors Tt and Nt. (b) Use Formula 9 to find the curvature. 7. rt t i 3t j 5 4t k rt e t cos t i j e t sin t k rt t i rt sin t, 5t, cos t t t j

22 SECTION 3.3 ARC LENGTH AND CURVATURE rt t, sin t t cos t, cos t t sin t, 9. rt s t, e t, e t. rt t, t, t 3 Use Theorem to find the curvature.. rt t i t k. rt t i t j t k 3. rt 3t i 4 sin t j 4 cos t k t Two graphs, a and b, are shown. One is a curve f and the other is the graph of its curvature function. Identif each curve and eplain our choices a a b b 4. Find the curvature of rt e t cos t, e t sin t, t at the point (,, ). 5. Find the curvature of rt t, t, t 3 at the point (,, ). ; 6. Graph the curve with parametric equations and find the curvature at the point, 4,. 7 9 Use Formula to find the curvature. 3 3 At what point does the curve have maimum curvature? What happens to the curvature as l? 3. ln 3. Find an equation of a parabola that has curvature 4 at the origin. t P 4t (a) Is the curvature of the curve C shown in the figure greater at P or at Q? Eplain. (b) Estimate the curvature at P and at Q b sketching the osculating circles at those points. 3. t cos e C CAS CAS 38. (a) Graph the curve rt sin 3t, sin t, sin 3t. At how man points on the curve does it appear that the curvature has a local or absolute maimum? (b) Use a CAS to find and graph the curvature function. Does this graph confirm our conclusion from part (a)? 39. The graph of is shown in Figure (b) in Section 3.. Where do ou think the curvature is largest? Use a CAS to find and graph the curvature function. For which values of t is the curvature largest? 4. Use Theorem to show that the curvature of a plane parametric curve f t, tt is where the dots indicate derivatives with respect to t. 4 4 Use the formula in Eercise 4 to find the curvature. 4. e t cos t, 4. t 3, e t sin t t t Find the vectors T, N, and B at the given point. 43. rt t,, (, 3 t 3, t 3, ) 44. rt cos t, sin t, ln cos t, rt t 3 sin t, 3 cos t, t 3,, Q Find equations of the normal plane and osculating plane of the curve at the given point. 45. sin 3t, t, cos 3t; 46. t, t, t 3 ;,,,, ; Use a graphing calculator or computer to graph both the curve and its curvature function on the same screen. Is the graph of what ou would epect? ; 47. Find equations of the osculating circles of the ellipse at the points, and, 3. Use a graphing calculator or computer to graph the ellipse and both osculating circles on the same screen.

23 838 CHAPTER 3 VECTOR FUNCTIONS ; 48. Find equations of the osculating circles of the parabola at the points and (,, ). Graph both osculating circles and the parabola on the same screen. CAS 49. At what point on the curve t 3, 3t, t 4 is the normal plane parallel to the plane 6 6 8? 5. Is there a point on the curve in Eercise 49 where the osculating plane is parallel to the plane? [Note: You will need a CAS for differentiating, for simplifing, and for computing a cross product.] 5. Show that the curvature of a plane curve is, where is the angle between T and i; that is, is the angle of inclination of the tangent line. (This shows that the definition of curvature is consistent with the definition for plane curves given in Eercise 69 in Section..) 5. Show that the curvature is related to the tangent and normal vectors b the equation dt ds N dds 53. (a) Show that dbds is perpendicular to B. (b) Show that dbds is perpendicular to T. (c) Deduce from parts (a) and (b) that dbds sn for some number s called the torsion of the curve. (The torsion measures the degree of twisting of a curve.) (d) Show that for a plane curve the torsion is s. 54. The following formulas, called the Frenet-Serret formulas, are of fundamental importance in differential geometr:. dtds N. dnds T B 3. dbds N (Formula comes from Eercise 5 and Formula 3 comes from Eercise 53.) Use the fact that N B T to deduce Formula from Formulas and Use the Frenet-Serret formulas to prove each of the following. (Primes denote derivatives with respect to t. Start as in the proof of Theorem.) (a) r st s N (b) r r s 3 B (c) r s s 3 T 3ss s N s 3 B (d) rr r rr 56. Show that the circular heli rt a cos t, a sin t, bt, where a and b are positive constants, has constant curvature and constant torsion. [Use the result of Eercise 55(d).] 57. Use the formula in Eercise 55(d) to find the torsion of the curve rt t, t, 3 t Find the curvature and torsion of the curve sinh t, cosh t, t at the point,,. 59. The DNA molecule has the shape of a double heli (see Figure 3 on page 89). The radius of each heli is about angstroms ( Å 8 cm). Each heli rises about 34 Å during each complete turn, and there are about.9 8 complete turns. Estimate the length of each heli. 6. Let s consider the problem of designing a railroad track to make a smooth transition between sections of straight track. Eisting track along the negative -ais is to be joined smoothl to a track along the line for. (a) Find a polnomial P P of degree 5 such that the function F defined b F P if if if is continuous and has continuous slope and continuous curvature. ; (b) Use a graphing calculator or computer to draw the graph of F. C O FIGURE 3.4 r(t+h)-r(t) h rª(t) Q P r(t) r(t+h) OTION IN SPACE: VELOCITY AND ACCELERATION In this section we show how the ideas of tangent and normal vectors and curvature can be used in phsics to stud the motion of an object, including its velocit and acceleration, along a space curve. In particular, we follow in the footsteps of Newton b using these methods to derive Kepler s First Law of planetar motion. Suppose a particle moves through space so that its position vector at time t is rt. Notice from Figure that, for small values of h, the vector rt h rt h approimates the direction of the particle moving along the curve rt. Its magnitude measures the sie of the displacement vector per unit time. The vector () gives the average

24 SECTION 3.4 OTION IN SPACE: VELOCITY AND ACCELERATION 839 velocit over a time interval of length h and its limit is the velocit vector vt at time t: rt h rt vt lim rt h l h Thus the velocit vector is also the tangent vector and points in the direction of the tangent line. The speed of the particle at time t is the magnitude of the velocit vector, that is, vt. This is appropriate because, from () and from Equation 3.3.7, we have vt rt ds rate of change of distance with respect to time As in the case of one-dimensional motion, the acceleration of the particle is defined as the derivative of the velocit: at vt rt EXAPLE The position vector of an object moving in a plane is given b rt t 3 i t j. Find its velocit, speed, and acceleration when t and illustrate geometricall. SOLUTION The velocit and acceleration at time t are v() vt rt 3t i t j (, ) a() at rt 6t i j and the speed is FIGURE TEC Visual 3.4 shows animated velocit and acceleration vectors for objects moving along various curves. N Figure 3 shows the path of the particle in Eample with the velocit and acceleration vectors when t. a() v() vt s3t t s9t 4 4t When t, we have v 3 i j a 6 i j These velocit and acceleration vectors are shown in Figure. v s3 EXAPLE Find the velocit, acceleration, and speed of a particle with position vector rt t, e t, te t. SOLUTION vt rt t, e t, te t at vt, e t, te t FIGURE 3 vt s4t e t t e t The vector integrals that were introduced in Section 3. can be used to find position vectors when velocit or acceleration vectors are known, as in the net eample.

25 84 CHAPTER 3 VECTOR FUNCTIONS V EXAPLE 3 A moving particle starts at an initial position r,, with initial velocit v i j k. Its acceleration is at 4t i 6t j k. Find its velocit and position at time t. SOLUTION Since at vt, we have vt at 4t i 6t j k t i 3t j t k C To determine the value of the constant vector C, we use the fact that v i j k. The preceding equation gives v C, so C i j k and N The epression for rt that we obtained in Eample 3 was used to plot the path of the particle in Figure 4 for t 3. vt t i 3t j t k i j k t i 3t j t k Since vt rt, we have rt vt t i 3t j t k 6 4 (,, ) 5 5 FIGURE 4 ( 3 t 3 t) i t 3 t j ( t t) k D Putting t, we find that D r i, so the position at time t is given b rt ( 3t 3 t ) i t 3 t j ( t t) k In general, vector integrals allow us to recover velocit when acceleration is known and position when velocit is known: vt vt t t au du rt rt t t vu du If the force that acts on a particle is known, then the acceleration can be found from Newton s Second Law of otion. The vector version of this law states that if, at an time t, a force Ft acts on an object of mass m producing an acceleration at, then Ft mat N The angular speed of the object moving with position P is, where is the angle shown in Figure 5. d P EXAPLE 4 An object with mass m that moves in a circular path with constant angular speed has position vector rt a cos t i a sin t j. Find the force acting on the object and show that it is directed toward the origin. SOLUTION To find the force, we first need to know the acceleration: vt rt a sin t i a cos t j at vt a cos t i a sin t j Therefore Newton s Second Law gives the force as FIGURE 5 Ft mat m a cos t i a sin t j

26 SECTION 3.4 OTION IN SPACE: VELOCITY AND ACCELERATION 84 Notice that Ft m rt. This shows that the force acts in the direction opposite to the radius vector rt and therefore points toward the origin (see Figure 5). Such a force is called a centripetal (center-seeking) force. v a FIGURE 6 d V EXAPLE 5 A projectile is fired with angle of elevation and initial velocit v. (See Figure 6.) Assuming that air resistance is negligible and the onl eternal force is due to gravit, find the position function rt of the projectile. What value of maimies the range (the horiontal distance traveled)? SOLUTION We set up the aes so that the projectile starts at the origin. Since the force due to gravit acts downward, we have where t a 9.8 ms. Thus F ma mt j a t j Since vt a, we have vt tt j C where C v v. Therefore rt vt tt j v Integrating again, we obtain rt tt j t v D But D r, so the position vector of the projectile is given b 3 rt tt j t v If we write v v (the initial speed of the projectile), then v v cos i v sin j and Equation 3 becomes rt v cos t i [v sin t tt ] j The parametric equations of the trajector are therefore N If ou eliminate t from Equations 4, ou will see that is a quadratic function of. So the path of the projectile is part of a parabola. 4 v cos t v sin t tt The horiontal distance d is the value of when. Setting, we obtain t or t v sin t. This second value of t then gives d v cos v sin t v sin cos t v sin t 4 Clearl, d has its maimum value when sin, that is,.

27 84 CHAPTER 3 VECTOR FUNCTIONS V EXAPLE 6 A projectile is fired with mule speed 5 ms and angle of elevation 45 from a position m above ground level. Where does the projectile hit the ground, and with what speed? SOLUTION If we place the origin at ground level, then the initial position of the projectile is (, ) and so we need to adjust Equations 4 b adding to the epression for. With v,, and t 9.8 ms 5 ms, we have Impact occurs when, that is, 4.9t 75st. Solving this quadratic equation (and using onl the positive value of t), we get Then 75s.74 36, so the projectile hits the ground about 36 m awa. The velocit of the projectile is So its speed at impact is 45 5 cos4t 75s t 5 sin4t 9.8t 75s t 4.9t t 75s s, vt rt 75s i (75s 9.8t) j v.74 s(75s ) (75s ) 5 ms TANGENTIAL AND NORAL COPONENTS OF ACCELERATION When we stud the motion of a particle, it is often useful to resolve the acceleration into two components, one in the direction of the tangent and the other in the direction of the normal. If we write for the speed of the particle, then and so v v Tt rt rt v vt vt vt v v If we differentiate both sides of this equation with respect to t, we get 5 a v vt vt If we use the epression for the curvature given b Equation 3.3.9, then we have 6 T r T v T The unit normal vector was defined in the preceding section as N T T, so (6) gives so v and Equation 5 becomes 7 T T N vn a vt v N

28 SECTION 3.4 OTION IN SPACE: VELOCITY AND ACCELERATION 843 Writing and for the tangential and normal components of acceleration, we have a T a N a T T N a where 8 a T v a a T T a N N and a N v FIGURE 7 a N This resolution is illustrated in Figure 7. Let s look at what Formula 7 sas. The first thing to notice is that the binormal vector B is absent. No matter how an object moves through space, its acceleration alwas lies in the plane of T and N (the osculating plane). (Recall that T gives the direction of motion and N points in the direction the curve is turning.) Net we notice that the tangential component of acceleration is v, the rate of change of speed, and the normal component of acceleration is v, the curvature times the square of the speed. This makes sense if we think of a passenger in a car a sharp turn in a road means a large value of the curvature, so the component of the acceleration perpendicular to the motion is large and the passenger is thrown against a car door. High speed around the turn has the same effect; in fact, if ou double our speed, a N is increased b a factor of 4. Although we have epressions for the tangential and normal components of acceleration in Equations 8, it s desirable to have epressions that depend onl on r, r, and r. To this end we take the dot product of v vt with a as given b Equation 7: v a vt vt v N vvt T v 3 T N vv (since T T and T N ) Therefore 9 a T v v a v rt rt rt Using the formula for curvature given b Theorem 3.3., we have a N v rt rt rt rt rt rt 3 rt EXAPLE 7 A particle moves with position function rt t, t, t 3. Find the tangential and normal components of acceleration. SOLUTION rt t i t j t 3 k rt t i t j 3t k rt i j 6t k rt s8t 9t 4 Therefore Equation 9 gives the tangential component as a T rt rt rt 8t 8t 3 s8t 9t 4