Section 14.3 Arc Length and Curvature


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1 Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in terms of prmetric equtions (in spce) ws determined. A similr formul holds for 3spce. Result.. If r(t) = f(t) i + g(t) j + h(t) k defines smooth curve C nd f (t), g (t) nd h (t) re ll continuous, then the rc length long the portion of the curve with t b (provided it is trversed only is) is given by the formul or L = b [f (t)] + [g (t)] + [h (t)] dt L = b r (t) dt. The formul is stright forwrd to work with: Exmple.. Find the length of for t. r(t) = i + t j + t 3 k This is stright forwrd clcultions: L = + 4t + 9t 4 dt = t 4 + 9t dt. Substituting u = 4 + 9t, we hve du/dt = 8t, dt = du/8t, nd when t =, u = 4, nd t = gives u = 3, t t dt = t u 4 8t du = 3 [ ] 3 udu = u 3 4 = 7 [ Often we re not interested in specific distnce prticle hs trveled, but rther formul to determine how fr prticle hs trveled in terms of me vrible (this is useful in things like irline flight or spce trvel), we need to determine wy to do this. Suppose r(u) is vector function for curve C which trverses C only once for u t ].
2 (notice we hve chnged the prmeter to u nd we re using t s time vrible ). We define the rclength function s for C by r (u) du = [f (u)] + [g (u)] + [h (u)] du. Notice tht the integrl defines function in the vrible t. Al note tht by the fundmentl Theorem of Clculus, this implies ds dt = r (t). The function s(t) mesures the distnce from the point r() to the point r(t), s on s we plug in vlue for t, it will provide us with this numericl distnce. We illustrte. Exmple.3. Find the distnce formul for r(u) = u i + ( 3u) j + (5 + 4u) k from u =. Use the formul to determine the distnce trveled fter seconds, nd how long it tkes for the prticle to trvel 8 units. The formul is stright forwrd: + ( 3) + 4 du = 9t This mens fter seconds, the prticle will hve trveled 9 units. In order to trvel 8 units, we need 9t = 8, or t = 9/8.67 seconds. Notice tht in the lst exmple, we obtined formul for s in terms of t. In prticulr, we could lve this formul for t in terms of s, t = s/ 9, nd then subsitute into the originl prmeteriztion for r nd thus we would hve prmeteriztion for r in terms of s insted of t. We cll such prmetriztion prmeteriztion with respect to rclength. This is often preferred method of prmeteriztion since it depends only upon the curve itself (or the length of the curve), nd not prticulr coordinte system. In order to determine prmeteriztion with respect to rclength of curve with vector eqution r(t), we do the following: (i) Solve the distnce formul r (u) du for the prmeteriztion you re given. You should hve function of s in terms of t. (ii) Cn you lve the function s(t) for t? If not, you cnnot reprmeterize this wy. Otherwise, lve for t ( t will be function of s, t(s)).
3 (iii) Substitute the function t(s) in for u (nd renme s s u) in the originl prmetriztion  this is now prmeteriztion with respect to rclength. Exmple.4. Find prmeteriztion with respect to rclength for r(u) = u i + ( 3u) j + (5 + 4u) k from u =. Recll, we hve t = s/ 9. Therefore, + ( 3) + 4 du = 9t r(t) = u ( 3u) (5 + 4u) i + j + k is prmeteriztion with respect to rclength.. Curvture Recll tht if C is smooth curve defined by the vector function r(t), nd r (t), then the unit tngent vector is given by T(t) = r(t)/ r (t) which indictes the direction of the curve. Since T(t) provides the direction of r(t), the rte of chnge of T with respect to s, the distnce function, mesures how quickly the direction of C is chnging (see figure below)  when d T/ds is lrge, it mens the direction of C is chnging quickly over distnce, nd when it is smll, it mens it is not chnging drmticlly. 3 This motivtes the following definition: Definition.. The curvture of curve is κ = d T ds where T is the unit tngent vector with prmeteriztion with respect to rclength s. By the wy we hve defined κ, it seems difficult to clculte  first we need to determine the prmeteriztion with respect to rclength of T (which s we sw previously, is not esy). However, by using the chin rule, we cn void doing this.
4 4 Result.. κ(t) cn be clculted using the following formul: κ(t) = T (t) r (t). Observe tht this completely voids hving to find s. We illustrte. Exmple.3. Find the curvture of r(t) = 3t i+4 sin (t) j +4 cos (t) k. First we find the unit norml vector, T(t) = r (t) r (t) = 3 i + 4 cos(t) j 4 sin (t) k (3 + 4 ) Then we hve Then we hve T (t) = 4 5 sin (t) j 4 5 cos (t) k. = 3 5 i+ 4 5 cos (t) j 4 5 sin (t) k. κ(t) = 4 sin (t) j 4 cos (t) 4 k i + 4 cos (t) j 4 sin(t) k = 5 5 = 4 5. Notice tht this is constnt (it does not depend upon t) which mens tht the curvture is constnt. This is l pprent from the grph below where we cn see the tngent vectors re chnging t constnt rte: There re other wys to clculte curvture which do not rely upon finding the tngent vector nd insted use crossproduct. Result.4. The curvture of the curve C given by r(t) is κ(t) = r (t) r (t) r (t) 3. Exmple.5. Find the curvture of r(t) = t i + e t j + e t k. We shll pply the recent formul: we hve r (t) = i + e t j e t j, nd r (t) = e t j + e t k, r (t) r (t) = i e t j + e t k r (t) r (t) = 4 + e t + e t = (e t + e t ) = (e t + e t )
5 We l hve r (t) 3 = (4 + e t + e t ) 3/ = ( (e t + e t )) 3 κ(t) = (e t + e t ). For the specil cse of plne curve y = f(x), ( dimensionl curve), the zcoordinte is lwys zero, we cn tke the prmeter to be x vector eqution will be Then we shll hve r(x) = x i + f(x) j. r (x) r (x) = ( i + f (x) j) (f (x) j) = f (x) k = f (x), nd r (x) 3 = ( i + f (x) j) 3 = + (f (x)) 3 = + (f (x)) 3/. Thus we hve Result.6. If y = f(x), then the curvture t ny point x is given by the formul f (x) κ(x) = + (f (x)) 3 Exmple.7. Find the curvture of y = x 3 t (, ). We just pply the formul, κ() = 6 3/ 3. Norml nd Binorml Vectors We hve lredy seen tht t ny point on curve, there is vector clled the unit tngent vector which tells us the direction the curve is going. There re two other vectors closely relted to this vector which l provide informtion bout the curve C. Specificlly, provided the unit tngent vector is nonzero, we cn find two other vectors which re perpendiculr to it nd re mutully perpendiculr to ech other (giving mething like coordinte xis t the point). We define them s follows: Definition 3.. Suppose C is curve with vector eqution r(t) nd let T(t) be its unit tngent vector defined s Then we define: T(t) = r (t) r (t). 5
6 6 (i) The principl unit norml vector N(t) defined s N(t) = T (t) T (t). (ii) The binorml vector B(t) defined s B(t) = T(t) B(t). All three vectors re mutully perpendiculr. Clcultion of these vectors though cumberme, is firly stright forwrd. We give n exmple. Exmple 3.. Find the vectors T, N nd B for r(t) = t i+t 3 /3 j+t k t the point (, /3, ). We hve T(t) = t i + t j + k (4t4 + 4t + = t i + t j + k (t + ) To clculte T (t), we use the generlized product rule: = t i + t j + k. (t + ) T 4t (t) = (t + ) (t i + t j + k) + (t + ) ( i + 4t j) = i + 8t 3 j + 4t (t + ) k) + + ) i + (8t 3 + 4t) j) (8t (t + ) ((4t = (t + ) (( 4t ) i+4t j 4t k) = Then we l hve N(t) = T (t) = (t + ) (( t ) i+t j t k). (t + ) ( 4t + 4t 4 + 4t + 4t ) = (t +) (( t ) i + t j t k) (t +) At (, /3, ) we hve t =, T() = i + j + k 9 = (t + ) t + (( t ) i+t j t k). = 3 i + 3 j + 3 k nd N() = 3 i + 3 j 3 k, B() = ( 3 i + 3 j + 3 k) ( 3 i + 3 j 3 k) = 3 i + 3 j + 3 k.
7 Recll tht given pir of (nonprllel) vectors nd point, there is plne which contins both vectors nd the point. The plnes scited to the vectors we hve introduced bove provide useful informtion bout curve C with vector r(t), we introduce me terminology for these plnes: Definition 3.3. Suppose tht r(t) is vector eqution for curve C. For fixed vlue, t =, we define the following: (i) The plne contining the point r() nd vectors N nd B is clled the norml plne to C t P = r(). It contins ll vectors orthogonl to C t t =. (ii) The plne contining the point r() nd vectors N nd T is clled the osculting plne to C t P = r(). It is the plne which C most closely lies in t P. Exmple 3.4. Find the equtions for the norml plne nd the osculting plne to r(t) = t i + t 3 /3 j + t k t the point (, /3, ). We hve lredy determined the three required vectors: T() = 3 i + 3 j + 3 k 7 nd nd N() = 3 i + 3 j 3 k, B() = 3 i + 3 j + 3 k. The norml plne contins B nd N, norml vector to the plne will be T (which is perpendiculr to both B nd N). This n eqution will be 3 (x ) + 3 (y 3 ) + (z ) =. 3 The osculting plne will hve B s norml vector, will hve eqution 3 (x ) + 3 (y 3 ) + (z ) =. 3
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