Question #133036
Find the derivative of the vector function.
a)r(t) =<e^(-t),t-t^3,ln t>
b)r(t) =(1/(1 +t)i+(t/(1 +t))j+((t^2)/(1 +t))k
1
Expert's answer
2020-09-17T15:28:05-0400

(a)

Let's use the [ ] instead of the < > such that e-t

𝐫(t)=[ettt3ln(t)]𝐫 (t)=\begin{bmatrix} e^{-t} \\ t-t^3\\ ln(t) \end{bmatrix} so that the derivative of r (t) will be;


𝐫(t)=d𝐫dt(t)=[ddtetddttt3ddtln(t)]𝐫'(t)=\frac{d𝐫}{dt} (t) = \begin{bmatrix} \frac{d}{dt} e^{-t} \\ \frac{d}{dt} t-t^3 \\ \frac{d}{dt}ln (t) \end{bmatrix}

=[et13t31t]=\begin{bmatrix} -e^{-t} \\ 1- 3t^3 \\ \frac{1}{t} \end{bmatrix}


(b)


𝐫(t)=11+ti+t1+tj+t21+tk𝐫(t)=\frac{1}{1+t}i + \frac{t}{1+t}j + \frac{t^2}{1+t}k here, we apply the formula;


𝐫(t)=limΔt0𝐫(t+Δt)𝐫(t)Δt𝐫'(t)= \lim\limits_{Δt→0} \frac{𝐫(t + Δt) - 𝐫(t)}{Δt}


=limΔt011+t+Δti+t+Δt1+t+Δtj+(t+Δt)21+t+Δtk(11+ti+t1+tj+t21+tk)Δt=\lim\limits_{Δt→0} \frac{\frac{1}{1+t + Δt}i + \frac{t + Δt}{1+t + Δt}j + \frac{(t + Δt)^2}{1+t + Δt}k- (\frac{1}{1+t}i + \frac{t}{1+t}j + \frac{t^2}{1+t}k)} {Δt}


=limΔt011+t+Δti+t+Δt1+t+Δtj+(t2+2tΔt+(Δt)21+t+Δtk(11+ti+t1+tj+t21+tk)Δt=\lim\limits_{Δt→0} \frac{\frac{1}{1+t + Δt}i + \frac{t + Δt}{1+t + Δt}j + \frac{(t^2 + 2tΔt +(Δt)^2}{1+t + Δt}k- (\frac{1}{1+t}i + \frac{t}{1+t}j + \frac{t^2}{1+t}k)} {Δt}


=limΔt0[11+t+Δt11+t]i+[t+Δt1+t+Δtt1+t]j+[(t2+2tΔt+(Δt)21+t+Δtt21+t]kΔt=\lim\limits_{Δt→0}\frac{[\frac{1}{1+t + Δt}- \frac{1}{1+t}] i + [\frac{t + Δt}{1+t + Δt}-\frac{t}{1+t}]j + [\frac{(t^2 + 2tΔt +(Δt)^2}{1+t + Δt}-\frac{t^2}{1+t}] k} {Δt}


=limΔt0[1+t1tΔt(1+t)(1+t+Δt)]i+[(1+t)t(1+t+Δt)(1+t)(1+t+Δt)]j+[(t+1)(t2+2tΔt+(Δt)2)t2(1+t+Δt)(1+t)(1+t+Δt)]kΔt=\lim\limits_{Δt→0}\frac{[\frac{1+t-1-t-Δt} {(1+t) (1+t+ Δt)}] i + [\frac{(1+t)-t(1+t+Δt)}{(1+t)(1+t + Δt)} ]j + [\frac{(t+1)(t^2+2tΔt+(Δt)^2) -t^2(1+t+Δt)} {(1+t)(1+t + Δt)}] k} {Δt}


=limΔt0[Δt(1+t)(1+t+Δt)]i+[Δt(1+t)(1+t+Δt)]j+[Δt(2tΔt+t2+2t)(1+t)(1+t+Δt)]kΔt=\lim\limits_{Δt→0}\frac{[\frac{-Δt} {(1+t) (1+t+ Δt)}] i + [\frac{Δt} {(1+t)(1+t + Δt)} ]j + [\frac{ Δt (2tΔt+t^2+2t)} {(1+t)(1+t + Δt)}] k} {Δt}


=limΔt0[Δt(1(1+t)(1+t+Δt))]i+[Δt(1(1+t)(1+t+Δt))]j+[Δt(2tΔt+t2+2t(1+t)(1+t+Δt))]kΔt=\lim\limits_{Δt→0}\frac{[Δt (\frac{-1} {(1+t) (1+t+ Δt)}) ] i + [ Δt(\frac{1} {(1+t)(1+t + Δt)}) ]j + [ Δt(\frac{ 2tΔt+t^2+2t} {(1+t)(1+t + Δt)}) ] k} {Δt} Factorize Δt outside so that we have;


=limΔt0Δt[1(1+t)(1+t+Δt)i+1(1+t)(1+t+Δt)j+2tΔt+t2+2t(1+t)(1+t+Δt)k]Δt= \lim\limits_{Δt→0}Δt\frac{ [ \frac{-1} {(1+t) (1+t+ Δt)} i + \frac{1} {(1+t)(1+t + Δt)} j + \frac{ 2tΔt+t^2+2t} {(1+t)(1+t + Δt)} k]} {Δt} and it cancels out, remaining with;



=limΔt01(1+t)(1+t+Δt)i+1(1+t)(1+t+Δt)j+2tΔt+t2+2t(1+t)(1+t+Δt)k= \lim\limits_{Δt→0}{ \frac{-1} {(1+t) (1+t+ Δt)} i + \frac{1} {(1+t)(1+t + Δt)} j + \frac{ 2tΔt+t^2+2t} {(1+t)(1+t + Δt)} k} Now substitute the value of Δt


So,therefore,𝐫(t)=1(1+2t+t2)i+1(1+2t+t2)j+t2+2t(1+2t+t2)kSo, therefore, 𝐫'(t)= {\frac{-1} {(1+2t+t^2)} i + \frac{1} {(1+2t+t^2)} j + \frac{ t^2+2t} {(1+2t+t^2)} k}



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