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Question Number 6152 by sanusihammed last updated on 16/Jun/16
Differentiate s = ut + (1/2)at^2 from the first principle
$${Differentiate}\:\:\:{s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\:\:\:\:{from}\:{the}\:{first}\:{principle} \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 16/Jun/16
s = ut + (1/2)at^2 s+δs=u(t+δt)+(1/2)a(t+δt)^2 δs=(u(t+δt)+(1/2)a(t+δt)^2 )−(ut + (1/2)at^2 ) =u(t+δt)+(1/2)a(t+δt)^2 −ut − (1/2)at^2 =u(t+δt−t)+(1/2)a(t^2 +2t δt+δt^2 −t^2 ) =u δt+(1/2)a(2t δt+δt^2 ) ((δs)/(δt))=u+at+(1/2)aδt lim_(δt→0) ((δs)/(δt))=lim_(δt→0) (u+at+(1/2)aδt) (ds/dt)=u+at
$${s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\: \\ $$$${s}+\delta{s}={u}\left({t}+\delta{t}\right)+\frac{\mathrm{1}}{\mathrm{2}}{a}\left({t}+\delta{t}\right)^{\mathrm{2}} \\ $$$$\delta{s}=\left({u}\left({t}+\delta{t}\right)+\frac{\mathrm{1}}{\mathrm{2}}{a}\left({t}+\delta{t}\right)^{\mathrm{2}} \right)−\left({ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \right) \\ $$$$\:\:\:={u}\left({t}+\delta{t}\right)+\frac{\mathrm{1}}{\mathrm{2}}{a}\left({t}+\delta{t}\right)^{\mathrm{2}} −{ut}\:−\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$$$\:\:\:\:\:={u}\left({t}+\delta{t}−{t}\right)+\frac{\mathrm{1}}{\mathrm{2}}{a}\left({t}^{\mathrm{2}} +\mathrm{2}{t}\:\delta{t}+\delta{t}^{\mathrm{2}} −{t}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:={u}\:\delta{t}+\frac{\mathrm{1}}{\mathrm{2}}{a}\left(\mathrm{2}{t}\:\delta{t}+\delta{t}^{\mathrm{2}} \right) \\ $$$$\frac{\delta{s}}{\delta{t}}={u}+{at}+\frac{\mathrm{1}}{\mathrm{2}}{a}\delta{t} \\ $$$$\underset{\delta{t}\rightarrow\mathrm{0}} {{lim}}\frac{\delta{s}}{\delta{t}}=\underset{\delta{t}\rightarrow\mathrm{0}} {{lim}}\left({u}+{at}+\frac{\mathrm{1}}{\mathrm{2}}{a}\delta{t}\right) \\ $$$$\frac{{ds}}{{dt}}={u}+{at} \\ $$
Answered by FilupSmith last updated on 16/Jun/16
f(t)=ut+(1/2)at^2 f(t+h)=u(t+h)+(1/2)a(t+h)^2 f ′(t)=lim_(h→0) ((f(t+h)−f(t))/h) f ′(t)=lim_(h→0) (((u(t+h)+(1/2)a(t+h)^2 )−(ut+(1/2)at^2 ))/h) f ′(t)=lim_(h→0) ((u(t+h−t)+(1/2)a((t+h)^2 −t^2 ))/h) f ′(t)=lim_(h→0) ((u(h)+(1/2)a(t^2 +2th+h^2 −t^2 ))/h) f ′(t)=lim_(h→0) ((uh+(1/2)a(2th+h^2 ))/h) f ′(t)=lim_(h→0) ((uh+(1/2)ah(2t+h))/h) f ′(t)=lim_(h→0) (u+(1/2)a(2t+h)) f ′(t)=u+(1/2)a(2t) ∴ f ′(t)=u+at
$${f}\left({t}\right)={ut}+\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$$${f}\left({t}+{h}\right)={u}\left({t}+{h}\right)+\frac{\mathrm{1}}{\mathrm{2}}{a}\left({t}+{h}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${f}\:'\left({t}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({t}+{h}\right)−{f}\left({t}\right)}{{h}} \\ $$$${f}\:'\left({t}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({u}\left({t}+{h}\right)+\frac{\mathrm{1}}{\mathrm{2}}{a}\left({t}+{h}\right)^{\mathrm{2}} \right)−\left({ut}+\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \right)}{{h}} \\ $$$${f}\:'\left({t}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{u}\left({t}+{h}−{t}\right)+\frac{\mathrm{1}}{\mathrm{2}}{a}\left(\left({t}+{h}\right)^{\mathrm{2}} −{t}^{\mathrm{2}} \right)}{{h}} \\ $$$${f}\:'\left({t}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{u}\left({h}\right)+\frac{\mathrm{1}}{\mathrm{2}}{a}\left({t}^{\mathrm{2}} +\mathrm{2}{th}+{h}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)}{{h}} \\ $$$${f}\:'\left({t}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{uh}+\frac{\mathrm{1}}{\mathrm{2}}{a}\left(\mathrm{2}{th}+{h}^{\mathrm{2}} \right)}{{h}} \\ $$$${f}\:'\left({t}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{uh}+\frac{\mathrm{1}}{\mathrm{2}}{ah}\left(\mathrm{2}{t}+{h}\right)}{{h}} \\ $$$${f}\:'\left({t}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left({u}+\frac{\mathrm{1}}{\mathrm{2}}{a}\left(\mathrm{2}{t}+{h}\right)\right) \\ $$$${f}\:'\left({t}\right)={u}+\frac{\mathrm{1}}{\mathrm{2}}{a}\left(\mathrm{2}{t}\right) \\ $$$$ \\ $$$$\therefore\:{f}\:'\left({t}\right)={u}+{at} \\ $$
Commented by FilupSmith last updated on 16/Jun/16
added this as a different notation
$$\mathrm{added}\:\mathrm{this}\:\mathrm{as}\:\mathrm{a}\:\mathrm{different}\:\mathrm{notation} \\ $$
Commented by sanusihammed last updated on 16/Jun/16
Thank you
$${Thank}\:{you}\: \\ $$

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