ME 210 Applied Mathematics for Mechanical Engineers

Transcription

1 Note that the unit vectos, T, N and B ae thee unit vectos pependicula to each othe whose diections ae dictated by the local behavio of the cuve C at its point P. They fom a moving ight handed vecto fame called as the TNB fame that plays a significant ole in calculating flight paths of space vehicles, missiles and aiplanes. Rectifying plane B Binomal Cuve Nomal plane N Pincipal nomal T Unit Osculating plane Next to osculating plane, two othe planes shown in this figue ae nomal plane (defined by pincipal nomal and binomal) and ectifying plane (defined by unit tangent and binomal). Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 1/0

2 Simila to T and B, the ate of change in N by the paamete s can be found by using the fact that N(s) = B(s) T(s), leading to dn(s) d ( ) db dt = B T = T + B = τn T + B κn= τb κt Hence, the following diffeential equations called the Fenet-Seet fomulas ae obtained to epesent any cuve in space: dt(s) = dn(s) = κ(s)t(s) db(s) = κ(s)n(s) τ(s)n(s) τ(s)b(s) Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 /0

3 To define the TNB-fame fo any cuve in space as a function of s, only the cuvatue κ(s) and tosion τ(s) functions of that cuve need to be specified to pefom the integation of these equations along the cuve. Note that fo plana cuves τ(s) 0, and fo staight lines both τ(s) 0 and κ(s) 0. Note also that the acceleation vecto can be expessed in TNB-fame as follows. dv d d s dt a = = T = T + dt dt dt dt dt dt d s dt a = d s T + ( dt = T + κn) dt dt dt dt d s a = T + N κ dt dt tangential component nomal component indicating that the acceleation of a moving paticle always lies in the osculating plane. Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 3/0

4 The following two expessions ae useful when a cuve is epesented by a position vecto (t), whee t is any paamete: (t) (t) κ(t) = [ (t) (t)] 3/ (t) (t) (t) τ(t) = (t) (t) Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 4/0

5 Example: (Cicula helix) Find the cuvatue κ(s) and tosion τ(s) functions of a cicula helix which is paametically epesented as (s) = [ acos( s K) ] i + [ asin( s K) ] j + c. ( s K) whee K = a + c and s is the ac length. [ ]k Fist we need to find the expession of unit tangent vecto d(s) T(s) = = [ (a/k).sin( s K) ] i + (a/k).cos( s K) j + [ ] [ c K]k Its ate of change with the paamete s: dt(s) [ ( )] [ = (a/k ).cos s K i + (a/k ).sin( s K) ]j Note that dt(s)/, hence N has no component in z diection. Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 5/0

6 which is a constant. The cuvatue function can be found as dt(s) a κ(s) = = (a/k ).cos s K. + (a/k ).sin s K = K ( ) ( ) Let us now detemine the expession of the pincipal nomal as 1 dt(s) N(s) = = cos(s/k) i sin(s/k) j κ(s) and the expession of binomal as B(s) = T(s) N(s) = (c/k).sin and its deivative db(s)/ as B(s) = (c/k ).cos s K [ ( s K) ] i [(c/k).cos ( s K) ] j + [ a K]k [ ]j [ ( )] i + (c/k ).sin( s K) d Then the tosion function can be obtained by using the 3 d Fenet-Seet fomula as τ(s) = c/k Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 6/0

7 Diectional Deivative; Gadient of a Scala function If a 3-D scala field given by a scala function f(x,y,z) is consideed, its fist patial deivatives, ( f/ x), ( f/ y), and ( f/ z) epesent the ates of change of f(x,y,z) in the x, y, and z diections, espectively. If the ate of change of f(x,y,z) in any given diection is consideed, then the concept of diectional deivative nee to be used. To define this deivative, choose a point P with a position vecto a and a diection C at P towa Q chaacteized by a unit vecto b. a P b (s) s Q C Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 7/0

8 a P b (s) df(x,y,z) s Q C The diectional deivative at the point P in the diection of b is then defined as df f(q) f(p) Dbf = = lim s 0 s Using the chain ule, the diectional deivative can be witten as f(x,y,z) = x dx f(x,y,z) + y The ay C can be epesented by the following paametic fom: (s) = a + sb = x(s) i + y(s)j + z(s)k ; s 0, b = 1 d(s) dx(s) dy(s) dz(s) = b = i + j + k = x (s) i + y (s)j + z (s)k Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 8/0 dy f(x,y,z) + z Hence, the expession fo D b f can be consideed as the dot poduct of two vectos df dx dy dz f f f Dbf = = i + j + k i + j + k = b gadf x y z b gadient of scala gadf function f(x,y,z). dz

9 By intoducing the diffeential opeato called as del o nabla, = i + x j + y k z the gadient and diectional deivative can be ewitten as gad f D b = f = df f f f = i + j + k x y z = b Note that the gadient of a scala field is a vecto field, which is independent of the paticula choice of coodinate system. Similaly, the diectional deivative of a scala field at a point in a paticula diection is also independent of the paticula choice of coodinate system. Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 9/0

10 Consideing the dot poduct featue of the diectional deivative, df D b f = = b = b cosθ = cosθ whee θ is the angle between b and ; it can be said that the gadient of a scala function f at a point indicates the diection in which a maximum ate of change (incease) in f is encounteed, the size (magnitude) of the gadient is the measue of this maximum ate of change in f. the opposite diection of is anothe diection in which a maximum ate of change (decease) in f is encounteed, in negative diection though. along all diections pependicula to (lying on a plane pependicula to ) the gadient of f is zeo; that is, f emains constant along these diections. Some popeties of the gadient opeato (kf) = k( ) (fg) = ( )g + f( g) (f n ) = nf n 1 (k is ( ) a constant) (f + g) = + g (f/g) = [( )g f( g)]/g Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 10/0

11 Example: Conside the scala function f(x,y,z) = x y z + 4 x z. Find its maximum ate of incease, and its deivative at (1,, 1) along the diection i j k. = i + j + k (x yz + 4xz ) = (xyz + 4z ) i + (x z)j + (x y + 8xz)k x y z at (1,, 1): = 8 i j 10k Its maximum ate of incease at (1,, 1) can be found as = (8) + (1) + (10) = = The unit vecto in the diection i j k is found as i - j - k 1 b = = i - j - k Hence, the desied diectional deivative is calculated as 1 37 D b f = b = i j k. 8 i j 10k = 1.33 < ( ) 1.85 Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 11/0

12 Example: In what diections at the point (,0,0) does the function f(x,y,z) = xy have a ate of change of 1? Ae thee diections in which the ate is 3? How about? The gadient of f can be found as = i + j + k (xy) = x y z Its value at (,0,0) is j. y i + xj Diections at which the diectional deivative is 1 at the point (,0,0) can be found as o D b f = b = b j = cosθ = 1 θ = ± 10 whee θ is the angle between y-axis and the diection of. Thee exists no diections at which the ate is 3 at point (,0,0) because the size of the gadient of f(x,y,z) at the point (,0,0) is limited to. The diection at which the ate is is negative y-axis, since in size it is equal to the gadient of the function and the sign indicates a diection, which is opposite to the gadient. Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 1/0 b

13 Gadient and Nomal Vecto to a Level Suface Conside a diffeentiable function f(x,y,z) in space. Then, fo each constant K, the equation f(x,y,z) = K epesents a suface S in 3D space. The family of sufaces obtained by using vaious values fo K ae called as the level sufaces of the function f. Suface S 3 f(x,y,z)=k 3 Suface S f(x,y,z)=k K 3 > K > K 1 Suface S 1 f(x,y,z)=k 1 Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 13/0

14 Gadient and Nomal Vecto to a Level Suface An impotant geometical intepetation of the gadient is as follows: If f is a scala function such that though a point P in space, thee passes pecisely one level suface S of f, and if 0 at P, then is pependicula to S at P, that is, it has the diection of the nomal of S at P. K 3 > K > K 1 P Suface S 3 f(x,y,z)=k 3 Suface S f(x,y,z)=k Suface S 1 f(x,y,z)=k 1 Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 14/0

15 Conside a cuve C in space epesented paametically as (t) = x(t) i + y(t)j + z(t)k If one equies that C lie on S, then x(t), y(t), and z(t) must also satisfy f[x(t),y(t),z(t)] = K Diffeentiating it with espect to t and using the chain ule one gets df[x(t), y(t),z(t)] dt f dx(t) f dy(t) f dz(t) dk = + + = 0 = 0 x dt y dt z dt dt P C Suface S 3 f(x,y,z)=k 3 Suface S f(x,y,z)=k K 3 > K > K 1 Suface S 1 f(x,y,z)=k 1 Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 15/0

16 Since is the tangent to an abitay cuve C lying on S at a point P, it also lies on the tangent plane to the suface S at the point P. Hence will be will be nomal to the suface S at the point P and the unit nomal vecto n = n P C Suface S 3 f(x,y,z)=k 3 Suface S f(x,y,z)=k K 3 > K > K 1 Suface S 1 f(x,y,z)=k 1 Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 16/0

17 Example: Find a unit nomal vecto of the cone of evolution z = 4 (x + y ) at the point P(1,0,). f(x,y,z) as which is f at (1,0,). n The suface of the cone may be consideed as one of the level sufaces f(x,y,z)=4(x +y ) z =K coesponding to K=0. The nomal vecto of this suface can be found by the gadient of function = 8x i + 8yj zk = 8 i 4k Hence, the unit nomal vecto of this cone is found as 8 i 4k 1 n = = = i k Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 17/0 x n P 1 z 0 y

18 Consevative Vecto Fiel Vecto fiel that ae obtained as gadients of scala fiel such as v(x,y,z) = (x,y,z) ae called consevative vecto fiel, and the coesponding scala fiel ae efeed to as potential functions (o simply potentials) of these consevative vecto fiel. Note that, fo a given abitay vecto field, thee is no guaantee that it is a consevative one. In othe wo, fo a given abitay vecto field, thee is no guaantee that thee will be a coesponding scala field (potential function), which can be used to geneate this vecto field by pefoming a gadient opeation. Note also that, like antideivatives, potentials ae not detemined uniquely; abitay constants can be added to them. Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 18/0

19 Example: x x The vecto function v(x, y,z) = e siny i + e cosyj is suspected to be epesenting a consevative vecto field. Detemine its potential function. x x f f f v(x,y,z) = e siny i + e cosyj = i + j + k gives x y z f x f x f = e siny = e cosy = 0 x y z Integating f/ x with espect to x by assuming y and z constant, one gets f(x,y,z)=e x siny + P(y,z) The use of this expession of f in f/ y gives x P(y,z) x P(y,z) e cosy + = e cosy y y The use of Q(z) in f and then foming f/ z gives = 0 P(y,z) = Q(z) (e x siny) dq(z) dq(z) + = 0 = 0 Q(z) C z dz dz = Hence, the potential function of the vecto field given is f(x,y,z)=e x siny + C Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 19/0

20 END OF WEEK 8 Pof. D. Bülent E. Platin Sping 014 Sections 0 & 03 0/0