Page:Calculus Made Easy.pdf/183

(3) Plot the curve ${\displaystyle y=100(1-\epsilon ^{-2t})}$.

(4) The following equations give very similar curves:

Draw all three curves, taking ${\displaystyle a=100}$ millimetres; ${\displaystyle b=30}$ millimetres.

(5) Find the differential coefficient of ${\displaystyle y}$ with respect to ${\displaystyle x}$, if

(6) For “Thorium ${\displaystyle A}$,” the value of ${\displaystyle \lambda }$ is ${\displaystyle 5}$; find the “mean life,” that is, the time taken by the transformation of a quantity ${\displaystyle Q}$ of “Thorium ${\displaystyle A}$” equal to half the initial quantity ${\displaystyle Q_{0}}$ in the expression

${\displaystyle t}$ being in seconds.

(7) A condenser of capacity ${\displaystyle K=4\times 10-6}$, charged to a potential ${\displaystyle V_{0}=20}$, is discharging through a resistance of ${\displaystyle 10,000}$ ohms. Find the potential ${\displaystyle V}$ after (a) ${\displaystyle 0.1}$ second; (b) ${\displaystyle 0.01}$ second; assuming that the fall of potential follows the rule ${\displaystyle V=V_{0}\epsilon ^{-{\frac {t}{KR}}}}$.

(8) The charge ${\displaystyle Q}$ of an electrified insulated metal sphere is reduced from ${\displaystyle 20}$ to ${\displaystyle 16}$ units in ${\displaystyle 10}$ minutes. Find the coefficient ${\displaystyle \mu }$ of leakage, if ${\displaystyle Q=Q_{0}\times \epsilon ^{-\mu t}}$; ${\displaystyle Q_{0}}$ being the initial charge and ${\displaystyle t}$ being in seconds. Hence find the time taken by half the charge to leak away.