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Water is being heated in a kettle. At time tt seconds, the temperature of the water is θ∘C\theta^{\circ}C. The rate of increase of the temperature of the water at time tt is modelled by the differential equation\newlinedθdt=λ(120−θ)θ≤100\frac{d\theta}{dt}=\lambda(120-\theta)\quad \theta \leq 100\newlinewhere λ\lambda is a positive constant. Given that θ=20\theta=20 when t=0t=0\newline(a) solve this differential equation to show that\newlineθ=120−100e−λt\theta=120-100e^{-\lambda t}\newline(88)

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Q. Water is being heated in a kettle. At time tt seconds, the temperature of the water is θ∘C\theta^{\circ}C. The rate of increase of the temperature of the water at time tt is modelled by the differential equation\newlinedθdt=λ(120−θ)θ≤100\frac{d\theta}{dt}=\lambda(120-\theta)\quad \theta \leq 100\newlinewhere λ\lambda is a positive constant. Given that θ=20\theta=20 when t=0t=0\newline(a) solve this differential equation to show that\newlineθ=120−100e−λt\theta=120-100e^{-\lambda t}\newline(88)
  1. Identify Given Equation: Identify the given differential equation and the initial condition. The differential equation is given by dθdt=λ(120−θ)\frac{d\theta}{dt} = \lambda(120 - \theta), and the initial condition is θ=20\theta = 20 when t=0t = 0.
  2. Separate Variables: Separate the variables θ\theta and tt in the differential equation.\newlineWe can rewrite the equation as dθ120−θ=λdt\frac{d\theta}{120 - \theta} = \lambda dt.
  3. Integrate Both Sides: Integrate both sides of the equation with respect to their respective variables.\newlineThe integral of dθ120−θ\frac{d\theta}{120 - \theta} with respect to θ\theta is −ln⁡∣120−θ∣-\ln|120 - \theta|, and the integral of λdt\lambda dt with respect to tt is λt\lambda t. We add the constant of integration CC on the right side.
  4. Solve for Constant: Solve for the constant of integration using the initial condition.\newlineWhen t=0t = 0, θ=20\theta = 20. Plugging these values into the integrated equation gives us −ln⁡∣120−20∣=λ⋅0+C-\ln|120 - 20| = \lambda \cdot 0 + C, which simplifies to −ln⁡(100)=C-\ln(100) = C.
  5. Express Constant in Terms: Express the constant CC in terms of the initial condition.C=−ln⁡(100).C = -\ln(100).
  6. Substitute Value of C: Substitute the value of C back into the integrated equation.\newline−ln⁡∣120−θ∣=λt−ln⁡(100).-\ln|120 - \theta| = \lambda t - \ln(100).
  7. Simplify by Exponentiating: Simplify the equation by exponentiating both sides to remove the natural logarithm. e(−ln⁡∣120−θ∣)=e(λt−ln⁡(100))e^{(-\ln|120 - \theta|)} = e^{(\lambda t - \ln(100))}, which simplifies to ∣120−θ∣=100e(−λt)|120 - \theta| = 100e^{(-\lambda t)}.
  8. Solve for Theta: Solve for theta by removing the absolute value, considering that θ≤100\theta \leq 100. Since theta is always less than or equal to 100100, we have 120−θ=100e(−λt)120 - \theta = 100e^{(-\lambda t)}.
  9. Isolate Theta: Isolate theta to find the temperature as a function of time. θ=120−100e−λt.\theta = 120 - 100e^{-\lambda t}.

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