164 CHAPTER 2. VECTOR FUNCTIONS

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1 164 CHAPTER. VECTOR FUNCTIONS.4 Curvture.4.1 Definitions nd Exmples The notion of curvture mesures how shrply curve bends. We would expect the curvture to be 0 for stright line, to be very smll for curves which bend very little nd to be lrge for curves which bend shrply. If we move long curve, we see tht the direction of the tngent vector will not chnge s long s the curve is flt. Its direction will chnge if the curve bends. The more the curve bends, the more the direction of the tngent vector will chnge. So, it mkes sense to study how the tngent vector chnges s we move long curve. But becuse we re only interested in the direction of the tngent vector, not its mgnitude, we will consider the unit tngent vector. Curvture is defined s follows: Definition.4.1 (Curvture) Let C be smooth curve with position vector r (s) where s is the rc length prmeter. The curvture κ of C is defined to be: κ ds (.11) where T is the unit tngent vector. Note the letter used to denote the curvture is the greek letter kpp denoted κ. Remrk.4. The bove formul implies tht T be expressed in terms of s, rc length. Therefore, we must hve the curve prmetrized in terms of rc length. Exmple.4.3 Find the curvture of circle of rdius. We sw erlier tht the prmetriztion of circle of rdius with respect to rc length length ws r (s) cos s, sin s. First, we need to compute T (s). By definition, r (s) T (s) r (s) So, we must first compute r (s). r (s) ( 1 sin s ) sin s, cos s ( 1, cos s We cn see tht r (s) 1 (but we lredy knew tht from theorem Thus T (s) sin s, cos s It follows tht d T ds 1 cos s, 1 sin s )

2 .4. CURVATURE 165 nd therefore κ ds 1 In other words, the curvture of circle is the inverse of its rdius. This grees with our intuition of curvture. Curvture is supposed to mesure how shrply curve bends. The lrger the rdius of circle, the less it will bend, tht is the less its curvture should be. This is indeed the cse. The lrger the rdius, the smller its inverse. Exmple.4.4 Find the curvture of the circulr helix. Erlier, we found tht the prmetriztion of the circulr helix with respect to rc length ws r (s) cos s, sin s s,. As before, we need to compute T (s) which cn be obtined from r (s). 1 r (s) sin s 1, cos s 1, Thus Therefore It follows tht T (s) 1 sin s, 1 cos s, 1 sin s, cos s, 1 1 ds 1 1 cos s, 1 sin s, 0 κ ds 1 The definition for the curvture works well when the curve is prmetrized with respect to rc length, or when this cn be done esily. However, since s cn be expressed s function s t, there should lso be formuls for the curvture in term of t. The next theorems give us vrious formuls for the curvture. Theorem.4.5 Let C be smooth curve with position vector r (t) where t is ny prmeter. Then the following formuls cn be used to compute κ. κ r (.1) (t)

3 166 CHAPTER. VECTOR FUNCTIONS κ r (t) r (t) 3 (.13) Proof. We prove ech formul seprtely. 1. Proof of κ Using the chin rule, we hve d T dt ds ds dt ds r (t) from formul.10. Thus, ds dt T (t) Hence κ ds. Proof of κ r (t) r (t) 3 We express r (t) nd r (t) in terms of T nd T then compute their cross product. Computtion of r. Since r T r nd ds dt r, we get tht r ds T dt Computtion of r Tking the derivtive with respect to t of the previous formul gives us r d s ds T + T dt dt Computtion of r (t) r (t). From the two previous formuls nd using the properties of cross products, we see tht r r ds d s ( T ) T dt dt + ( ) ds ( T ) T dt

4 .4. CURVATURE 167 Since the cross product of vector by itself is lwys the zero vector, we see tht r ( ) ds r T T dt ( ) ds T T sin θ dt where θ is the ngle between T n. Since T 1, by proposition..10, we know tht T T thus r ( ) ds r T dt r (t) T from formul.10. Therefore So T r r κ κ r r 3 Remrk.4.6 Formul.1 is consistent with the definition of curvture. It sys tht if t is ny prmeter used for curve C, then the curvture of C is κ r. In the cse the prmeter is s, then the formul nd using the (t) fct tht r (s) 1, the formul gives us the definition of curvture. Theorem.4.7 If C is curve with eqution y f (x) where f is twice differentible then f (x) κ (.14) ( 1 + (f (x)) )3 Proof. The proof follows esily from formul.13. First, let us remrk tht it is esy to prmetrize the curve given by y f (x) s 3-D prmetric curve. We cn simply use x x y f (x) z 0

5 168 CHAPTER. VECTOR FUNCTIONS Using x s the nme of the prmeter. Thus, the position vector of our curve is r (x) x, f (x), 0. It follows tht r (x) 1, f (x), 0 nd Thus Hence nd Therefore r (x) 0, f (x), 0 r r 0, 0, f (x) r r f (x) r 1 + (f (x)) κ r r 3 f (x) (1 + (f (x)) ) 3 f (x) ( 1 + (f (x)) ) 3 We illustrte these formuls with some exmples. Of course, the formul we use depends on the informtion given. Exmple.4.8 Find the curvture of the curve given by r (t) t, t, 13 t3 s function of t Since we hve the position vector describing the curve but it is not given with respect to rc length, we cn find κ by using either formul.1 or formul.13. Since this is n exmple, we show both methods. Since both methods require r nd r, we compute both here. Therefore r (t), t, t 4 + t + t 4 ( + t ) + t since + t > 0

6 .4. CURVATURE 169 Method 1 Here, we use formul.1. Using this formul, we need to find So, therefore It follows tht r (t) T (t), t, t T (t) + t + t, t + t, t + t 1 4t, 4 t ( + t ), 4t 16t t + 4t t ( + t ) 4 (t4 + 4t + 4) ( + t ) + t κ ( + t ) Method Here, we use formul.13.using this formul, we need to find r (t). r (t) 0,, t Next, we find r (t) r (t). Therefore r (t) r (t), t, t 0,, t t, 4t, 4 r (t) r (t) t, 4t, 4 4t t + 16 ( t + )

7 170 CHAPTER. VECTOR FUNCTIONS Finlly, The sme nswer s bove. κ r (t) r (t) 3 ( t + ) (t + ) 3 ( + t ) Exmple.4.9 Find the curvture of y x s function of x. Then, find the curvture of the sme curve when x 0, x 1. Here, we use formul.14. Let f (x) x. First, we need to find f (x) nd f (x). f (x) x f (x) So κ κ f (x) ( 1 + (f (x)) ) 3 (1 + 4x ) 3 When x 0, we get When x 1, we get κ κ 5 3 Mke sure you hve red, studied nd understood wht ws done bove before ttempting the problems..4. Problems 1. Fin nd κ for r (t) (t, ln (cos t)), π < t < π.. Fin nd κ for r (t) ( t + 3, 5 t ) 3. Using κ f (x) (1+(f (x)) ) 3, nswer the questions below. () Find the curvture of y ln (cos x), π < t < π wht you found in problem bove. nd compre with

8 .4. CURVATURE 171 (b) Show tht the curvture is 0 t point of inflection. 4. Fin nd κ for r (t) (3 sin t, 3 cos t, 4t). 5. Fin nd κ for r (t) (e t cos t, e t sin t, ). 6. Fin nd κ for ( ) t r (t) 33, t, 0 for t > Show tht the curvture of the helix r (t) ( cos t, sin t, bt), 0, b 0 is κ +b. 8. We found tht the curvture of the helix r (t) ( cos t, sin t, bt), 0, b 0 to be κ +b. Wht is the lrgest vlue κ cn hve for given vlue of b? Explin. 9. Find κ (x) for f (x) x then grph f nd κ for x. 10. Find κ (x) for f (x) sin x then grph f nd κ for 0 x π..4.3 Answers 1. Fin nd κ for r (t) (t, ln (cos t)), π < t < π. T (cos t, sin t) It follows tht κ cos t. Fin nd κ for r (t) ( t + 3, 5 t ) ( ) 1 T, t 1 + t 1 + t Finlly, 1 κ (t + 1) 3 3. Using κ f (x) (1+(f (x)) ) 3, nswer the questions below. () Find the curvture of y ln (cos x), π < t < π wht you found in problem bove. nd compre with κ cos x (b) Show tht the curvture is 0 t point of inflection. A point of inflection hppens when y 0 hence κ 0.

9 17 CHAPTER. VECTOR FUNCTIONS 4. Fin nd κ for r (t) (3 sin t, 3 cos t, 4t). T ( cos t, 5 sin t, 4 ) 5 κ Fin nd κ for r (t) (e t cos t, e t sin t, ). ( ) cos t sin t sin t + cos t T,, 0 κ 1 e t 6. Fin nd κ for ( ) t r (t) 33, t, 0 for t > 0. ( ) t T t + 1, 1 t + 1, 0 1 κ t (t + 1) 3 7. Show tht the curvture of the helix r (t) ( cos t, sin t, bt), 0, b 0 is κ +b. Just use the definition. 8. We found tht the curvture of the helix r (t) ( cos t, sin t, bt), 0, b 0 to be κ +b. Wht is the lrgest vlue κ cn hve for given vlue of b? Explin. κ 1 b 9. Using the formul given bove, find κ (x) for f (x) x then grph f nd κ for x. κ (x) (1 + 4x ) 3 The grphs re shown below.

10 .4. CURVATURE 173 y f (x) in red, κ (x) in blue x 10. Using the formul given bove, find κ (x) for f (x) sin x then grph f nd κ for 0 x π. sin x κ (x) (1 + cos x) 3 The grphs re shown below.

11 174 CHAPTER. VECTOR FUNCTIONS y f (x) in red, κ (x) in blue x

12 Bibliogrphy [1] Joel Hss, Murice D. Weir, nd George B. Thoms, University clculus: Erly trnscendentls, Person Addison-Wesley, 01. [] Jmes Stewrt, Clculus, Cengge Lerning, 011. [3] Michel Sullivn nd Kthleen Mirnd, Clculus: Erly trnscendentls, Mcmilln Higher Eduction,