An ice ball melts at a rate proportional to t | vedclass

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(D) Let $x(t)$ be the amount of ice at time $t$. The rate of melting is proportional to the amount present,so $\frac{dx}{dt} = -cx$ for some constant

Vedclass · Accepted Answer

$1/4$

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(D) Let $x(t)$ be the amount of ice at time $t$. The rate of melting is proportional to the amount present,so $\frac{dx}{dt} = -cx$ for some constant $c > 0$.Integrating this,we get $\ln(x) = -ct + C$,which implies $x(t) = x_0 e^{-ct}$.Given that half the ice melts in $20 	ext{ minutes}$,at $t = 20$,$x(20) = \frac{x_0}{2}$.So,$\frac{x_0}{2} = x_0 e^{-20c}$,which means $e^{-20c} = \frac{1}{2}$.We want to find the amount of ice left after $40 	ext{ minutes}$,which is $x(40) = x_0 e^{-40c}$.Since $e^{-40c} = (e^{-20c})^2 = (\frac{1}{2})^2 = \frac{1}{4}$.Thus,$x(40) = \frac{1}{4} x_0$.Comparing this with $k x_0$,we get $k = \frac{1}{4}$.

An ice ball melts at a rate proportional to the amount of ice present at that instant. Half the quantity of ice melts in $20 \text{ minutes}$. Let $x_0$ be the initial quantity of ice. If after $40 \text{ minutes}$ the amount of ice left is $k x_0$,then the value of $k$ is:

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