Answer to Question #133036 in Calculus for Promise Omiponle

Question

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

Answer 1

(a)

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

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

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

$=\begin{bmatrix} -e^{-t} \\ 1- 3t^3 \\ \frac{1}{t} \end{bmatrix}$

(b)

$𝐫(t)=\frac{1}{1+t}i + \frac{t}{1+t}j + \frac{t^2}{1+t}k$ here, we apply the formula;

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

$=\lim\limits_{Δt→0} \frac{\frac{1}{1+t + Δt}i + \frac{t + Δt}{1+t + Δt}j + \frac{(t + Δt)^2}{1+t + Δt}k- (\frac{1}{1+t}i + \frac{t}{1+t}j + \frac{t^2}{1+t}k)} {Δt}$

$=\lim\limits_{Δt→0} \frac{\frac{1}{1+t + Δt}i + \frac{t + Δt}{1+t + Δt}j + \frac{(t^2 + 2tΔt +(Δt)^2}{1+t + Δt}k- (\frac{1}{1+t}i + \frac{t}{1+t}j + \frac{t^2}{1+t}k)} {Δt}$

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

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

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

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

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

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

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