A spherical snowball melts so that its volume changes at a rate proportional to

Question

A spherical snowball melts so that its volume changes at a rate proportional to its
surface area. (This is reasonable since the heat comes in only through the surface.)
If the initial radius is 8 cm, and and in 1 hour the radius becomes 5 cm, when will
the snowball disappear? Hint: The volume of a sphere is
4/3 * pi * r^3
and the surface area
is 4 * pi * r^2 , where r is the radius. Find an equation expressing the rate of change of the
volume of the snowball in terms of the volume.

Explanation / Answer

1.) Since ln(x) = - ln(1x) you can rewrite the differential equation as: dP/dt = -k·ln(P/M)·P Substitute x = ln(P/M) => dx/dt = (1/M) · (1/(P/M)) ·dP/dt = (1/P)·dP/dt = - k·ln(M/P) dx/dt = -k·x Separation of variables leads to. (1/x) dx = - k dx By integration ones finds: ? (1/x) dx = ? - k dx ln(x) = c - k·t (c is the constant of integration) => x = e^(c - k·t ) = e^(c) · e^(-k·t) = C·e^(-k·t) => ln(P/M) = C·e^(-k·t) Let P0 be the initial population, i.e at t=0. Apply this initial condition to compute C: ln(P0/M) = C·e^(-k·0) = C Hence ln(P/M) = ln(P0/M)·e^(-k·t) => P(t) = M·e^( ln(P0/M)·e^(-k·t) ) because a·ln(b) = ln(b^a) you can rewrite as: P(t) = M·e^( ln( (P0/M)^e^(-k·t) ) ) = M · (P0/M)^e^(-k·t) Here M = 60 P0 = 8 P(t) = 60· (2/15)^e^(-k·t) The k can be computed with the given population after 3 years P(3) = 15 15 = 60 · (2/15)^e^(-k·3) (1/4) = (2/15)^e^(-k·t) e^(-k·3) = ln(1/4)/ln(2/15) = ln(4)/ln(15/2) => k = - (1/3)·ln( ln(4)/ln(15/2) ) ˜ 0.1246456 2. P(6) = 60· (2/15)^e^(-k·6) = 23.1 P(10) = 60· (2/15)^e^(-k·10) = 59.9995 ˜ 60 P(100) = 60· (2/15)^e^(-k·100) = 59.9995 ˜ 60 3: As t approaches infinity e^(-k·t) approaches zero. therefore lim|t?8| P(t) = 60· (2/15)^0 = 60 5) When growth rate dP/dt reaches maximum for a certain value of P, its derivative w.r.t. P at this values is zero: d/dP( -k·ln(P/M)·P ) = 0 - k· ( (1/M)·(M/P)·P + ln(P/M) ) = 0 - k· (1 + ln(P/M)) = 0 ln(P/M) = -1 P/M = e?¹ = (1/e) => P = M/e 2nd derivative test: d²/dP²( -k·ln(P/M)·P ) = d/dP( -k·(1 + ln(P/M)) = -k·(1 + 1/P) Since population P is positive 2nd derivative is negative for all populations particularly when P=M/e.

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