Question

The velocity vector of an object is given by y(t) = (* sin(at), 1, a cos(at)). Assume that at t = 1, the object is at the point P(1,1,0). (a) Find the position vector F(t) of the object. (b) Find parametric equations of the line which is tangent to r(t) at P. (c) Find the distance that the object traveled from the point t = 0 to t = 1. (d) Find an equation for the normal plane of r(t) at P.

Answer 1

501":- las The velocity of an object is Ū(t) = (asinat, i, reoort), d = K Bing)- + to A(x) { 3 dr z tsinats at it dej treosent)dt Integrating we get Plt) = (-codext) + c) î+ (t+C2) [ + sinre) +(3) @ .ir (1) = (1+4)+(1+6)ì+60+63 ) (6+1) P+(2+1)] + C Again at tæt, Fit) = (1,1,0) jie. Fli) ni nj compairing 3 * 3 & ③ we get G=0, C2=0, C3=0: From ... The position vector of the object setor is YIL) = . Costei + ti+ sin(at) From @ (+) (tarinat, Bacoolet) :: V(!) = (0,1,-*) The tangent line to the vector F(t) at P(1,1,0) (i.e. at t=1) is Rit)= F (1)+ tř!() = (1,1,0) + til) . The parametric equations tangent line xlt) = 1 +tio → Xt)=1 yet) = = 1+ til ² Yt)= tt 2(t) = 0+(-1).t → Elt)=-it , - (6) I Where R(H) - (xt), y(t), 2(t)) of the to 8(t) at P.(110) is

( The distance that the object traveled from the point t=0 point t=0 to t=1 - the distance between two position vectors rol and (!).. Now, 7(0) = (-1,0,0) [ Prom @ 8 (1)=(1, 1, o) ... The required distance is 1(1+1)+(1-0)*(0-ojº 4 + 1 = 5 units ! The equation of the normal plane is (8-7) + = 0 Where F = leve , Ř= <, 4,2). At P زهرا (1) i. . t =) - at P ř F = d = (0,1,-*) ... The equation of the normal plane is (R-8).T=0 → 0.652-1) +1.(4-1)+(-1).(2-0)= 0, y-1-A2 y-tz 1 © of 8(t) is The curvature X*1 és IPP Now, for <T"6086E), 0,-"sin(at)> Flt) = (Binat, i reosat), At tai liieat p) , (t) = 000(0,1,-1) Ñ = <-T", O,

At p k px ... 0 زn = + R"k .: (-x = 16 + 79 .:. ( = ۷ -۰۲۷ Curvature at P روا | x3 |13 باہ+5 (1+) ۱۳۹ ۸ ۱۹۴۷ رام + ۷۱ (۱۰۴) * | +. و )