(a)
Let's use the [ ] instead of the < > such that e-t
r(t)=⎣⎡e−tt−t3ln(t)⎦⎤ so that the derivative of r (t) will be;
r′(t)=dtdr(t)=⎣⎡dtde−tdtdt−t3dtdln(t)⎦⎤
=⎣⎡−e−t1−3t3t1⎦⎤
(b)
r(t)=1+t1i+1+ttj+1+tt2k here, we apply the formula;
r′(t)=Δt→0limΔtr(t+Δt)−r(t)
=Δt→0limΔt1+t+Δt1i+1+t+Δtt+Δtj+1+t+Δt(t+Δt)2k−(1+t1i+1+ttj+1+tt2k)
=Δt→0limΔt1+t+Δt1i+1+t+Δtt+Δtj+1+t+Δt(t2+2tΔt+(Δt)2k−(1+t1i+1+ttj+1+tt2k)
=Δt→0limΔt[1+t+Δt1−1+t1]i+[1+t+Δtt+Δt−1+tt]j+[1+t+Δt(t2+2tΔt+(Δt)2−1+tt2]k
=Δt→0limΔt[(1+t)(1+t+Δt)1+t−1−t−Δt]i+[(1+t)(1+t+Δt)(1+t)−t(1+t+Δt)]j+[(1+t)(1+t+Δt)(t+1)(t2+2tΔt+(Δt)2)−t2(1+t+Δt)]k
=Δt→0limΔt[(1+t)(1+t+Δt)−Δt]i+[(1+t)(1+t+Δt)Δt]j+[(1+t)(1+t+Δt)Δt(2tΔt+t2+2t)]k
=Δt→0limΔt[Δt((1+t)(1+t+Δt)−1)]i+[Δt((1+t)(1+t+Δt)1)]j+[Δt((1+t)(1+t+Δt)2tΔt+t2+2t)]k Factorize Δt outside so that we have;
=Δt→0limΔtΔt[(1+t)(1+t+Δt)−1i+(1+t)(1+t+Δt)1j+(1+t)(1+t+Δt)2tΔt+t2+2tk] and it cancels out, remaining with;
=Δt→0lim(1+t)(1+t+Δt)−1i+(1+t)(1+t+Δt)1j+(1+t)(1+t+Δt)2tΔt+t2+2tk Now substitute the value of Δt
So,therefore,r′(t)=(1+2t+t2)−1i+(1+2t+t2)1j+(1+2t+t2)t2+2tk
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