Question

T(r)=5.92×〖10〗^(-3) r+3.32×〖10〗^(-1).

can someone differentiate this with respect to 'r' so ic an find the velcoity at a certain point along the line? (radius is a distance and T is a time period)

Answer 1

Use this log rule:
\[\log_AB = (\log_CB)/(\log_CA)\]temploarily replace the x 10 bit with r in it to:\[x=10^{-3r}\]\[\log _{10}x = \log_{10}10^{-3r}\]\[\log_{10}x = -3r\]\[(\ln x)/(\ln 10) = -3r \]\[\ln x = (-3\ln 10).r\]\[e^{\ln x} = x = e^{(-3\ln 10).r}\]so you have \[T(r)=5.92\times10^{-3r} + 3.32\times10^{-1}\]becomes\[T(r) = 5.92.e^{(-3\ln 10)r} + 3.32\times10^{-1}\]
\[T'(r) = 5.92\times(-3\ln 10).e^{(-3\ln 10)r} = -17.76\times \ln 10\times 10^{-3r} = -40.8939 \times 10^{-3r}\]