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(a) Given that \(\log _{10} p=a, \log _{10} q=b\) and \(\log _{10} s=c\), express \(\log _{10}\left(\frac{p^{\frac{1}{3}} q^{4}}{s^{2}}\right.\) in terms of \(a, b\) and \(c\).

(b) The radius of a circle is \(6 \mathrm{~cm}\). If the area is increasing at the rate of \(20 \mathrm{~cm}^{2} s^{-1}\), find, leaving the answer in terms of \(\pi\), the rate at which the radius is increasing.
in Mathematics by Platinum (102k points) | 440 views

1 Answer

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\[
\begin{aligned}
&(a) \log _{10} p=a ; \log _{10} q=b ; \log _{10} s=c \\
&\log _{10}\left(\frac{p^{\frac{1}{3}} q^{4}}{s^{2}}=\log _{10} p^{\frac{1}{3}}+\log _{10} q^{4}-\log _{10} s^{2}\right. \\
&=\frac{1}{3} \log _{10} p+4 \log _{10} q-2 \log _{10} s \\
&=\frac{1}{3} a+4 b-2 c
\end{aligned}
\]
(b) Area of circle \(=\pi r^{2}\)
Given \(r=6 \mathrm{~cm}\)
\[
\begin{aligned}
&\frac{\mathrm{d} A}{\mathrm{~d} r}=2 \pi r \\
&\frac{\mathrm{d} A}{\mathrm{~d} t}=\frac{\mathrm{d} A}{\mathrm{~d} r} \times \frac{\mathrm{d} r}{\mathrm{~d} t} \\
&20=2 \pi(6) \times \frac{\mathrm{d} r}{\mathrm{~d} t} \\
&\frac{\mathrm{d} r}{\mathrm{~d} t}=\frac{20}{12 \pi} \\
&=\frac{5}{3 \pi}
\end{aligned}
\]
by Platinum (102k points)

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