huynh (lth436) – HW07 – meth – (91845)4Explanation:The acceleration of a particle having posi-tion functionr(t) =x(t)i+y(t)j+z(t)kis given byr′′(t) =x′′(t)i+y′′(t)j+z′′(t)k.Thus, sincer′(t) = 9 costi+ 5j−9 sintkwe have thata(t) =r′′(t) =−9 sinti−9 costk.keywords:Stewart5e, vector function, posi-tion function, acceleration, curve in 3-space00810.0pointsA particle moving in 3-space has positionfunctionr(t) = (5√2)ti+e5tj+e−5tk.What is its speed at timet?1.speed = 5(e5t+e−5t)correct2.speed =radicalbig10 + 5et+ 5e−t3.speed =radicalbig10 +e10t+e−10t4.speed = 5(√2 +e5t−e−5t)5.speed =e5t+e−5tExplanation:A particle with position functionr(t) =x(t)i+y(t)j+z(t)k,has velocityr′(t) =x′(t)i+y′(t)j+z′(t)k,andspeed =|r′(t)|.Now whenr(t) = (5√2)ti+e5tj+e−5tk,we see thatr′(t) = (5√2)i+ 5e5tj−5e−5tk,while|r′(t)|2= 50 + 25e5t+ 25e−5t= 25(e5t+e−5t)2.Consequently, the particle hasspeed = 5(e5t+e−5t).keywords: position function, vector function,derivative, speed,00910.0pointsFind the position vector,b, at timet= 1for a particle moving in 3-space with acceler-ationa(t) =i+ 3tjif its initial velocity and position are given byv0=−5k,r0=i,respectively.1. b=−12i−j+ 5k2. b=32i−12j+ 5k3. b=32i+12j−5kcorrect4. b=−12i+j+ 5k5. b=32i−12j−5k6. b=−12i+j−5kExplanation:

huynh (lth436) – HW07 – meth – (91845)5The velocity vector,v(t), of the particlesatisfies the equationdvdt=a(t) =i+ 3tj.Sov(t) =ti+32t2j+CwhereCis a constant vector such thatv(0) =v0=C=−5k.But then the position vector,r(t) of the par-ticle satisfies the equationdrdt=v(t) =ti+32t2j−5kThusr(t) =12t2i+12t3j−5tk+DwhereDis a constant vector such thatr(0) =r0=D=i,and sor(t) =parenleftBig12t2+ 1parenrightBigi+12t3j−5tk.Consequently, att= 1,r(1) =b=32i+12j−5k.keywords: vector equation, acceleration, ve-locity, position, initial velocity, initial posi-