D downthesun01 Oct 2009 303 33 Oct 31, 2012 #1 \(\displaystyle s(t)=\frac{a(t)'}{a(t)}\) If \(\displaystyle s(t)=\frac{4}{t+3}\) Find a(t) To me a(t) would just be \(\displaystyle (t+3)^4\) But the book has \(\displaystyle \frac{t+3}{3}^4\) How'd the book get that answer?

\(\displaystyle s(t)=\frac{a(t)'}{a(t)}\) If \(\displaystyle s(t)=\frac{4}{t+3}\) Find a(t) To me a(t) would just be \(\displaystyle (t+3)^4\) But the book has \(\displaystyle \frac{t+3}{3}^4\) How'd the book get that answer?

Prove It MHF Helper Aug 2008 12,897 5,001 Oct 31, 2012 #2 downthesun01 said: \(\displaystyle s(t)=\frac{a(t)'}{a(t)}\) If \(\displaystyle s(t)=\frac{4}{t+3}\) Find a(t) To me a(t) would just be \(\displaystyle (t+3)^4\) But the book has \(\displaystyle \frac{t+3}{3}^4\) How'd the book get that answer? Click to expand... \(\displaystyle \displaystyle \begin{align*} \frac{4}{t + 3} &= \frac{1}{a}\, \frac{da}{dt} \\ \int{\frac{4}{t + 3}\,dt} &= \int{\frac{1}{a}\,\frac{da}{dt}\,dt} \\ 4\ln{|t + 3|} + C_1 &= \int{\frac{1}{a}\,da} \\ 4\ln{|t + 3|} + C_1 &= \ln{|a|} + C_2 \\ 4\ln{|t + 3|} + C_1 - C_2 &= \ln{|a|} \\ e^{4\ln{|t+3|} + C_1 - C_2} &= |a| \\ e^{C_1 - C_2} e^{\ln{\left|t +3 \right|^4}} &= |a| \\ a &= C \left( t + 3 \right) ^4 \textrm{ where }C = \pm e^{C_1 - C_2} \end{align*}\) Reactions: downthesun01 and (deleted member)

downthesun01 said: \(\displaystyle s(t)=\frac{a(t)'}{a(t)}\) If \(\displaystyle s(t)=\frac{4}{t+3}\) Find a(t) To me a(t) would just be \(\displaystyle (t+3)^4\) But the book has \(\displaystyle \frac{t+3}{3}^4\) How'd the book get that answer? Click to expand... \(\displaystyle \displaystyle \begin{align*} \frac{4}{t + 3} &= \frac{1}{a}\, \frac{da}{dt} \\ \int{\frac{4}{t + 3}\,dt} &= \int{\frac{1}{a}\,\frac{da}{dt}\,dt} \\ 4\ln{|t + 3|} + C_1 &= \int{\frac{1}{a}\,da} \\ 4\ln{|t + 3|} + C_1 &= \ln{|a|} + C_2 \\ 4\ln{|t + 3|} + C_1 - C_2 &= \ln{|a|} \\ e^{4\ln{|t+3|} + C_1 - C_2} &= |a| \\ e^{C_1 - C_2} e^{\ln{\left|t +3 \right|^4}} &= |a| \\ a &= C \left( t + 3 \right) ^4 \textrm{ where }C = \pm e^{C_1 - C_2} \end{align*}\)