All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

\frac{d \theta}{dy} = k(20-\theta) where k is a positive constant.

When \theta = 10, \frac{d\theta}{dt}=1

\Rightarrow k = 0.1

\int \frac{1}{20-\theta} d\theta = \int 0.1 dt

-\mathrm{ln}|20-\theta| = 0.1t + C

Since \theta \le 20, -\mathrm{ln}(20-\theta) = 0.1t + C

When t = 0, \theta = 10

\Rightarrow C = -\mathrm{ln}10

\mathrm{ln}(20-\theta) = -0.1t + ln10

\theta = 20 - 10 e^{-0.1t}

When \theta = 15, t = 6.93\mathrm{~min}

KS Comments:

Not a tough question, just be care when you draw the graph, consider only the positive t values. Students can omit the module too since \theta \le 20 always.

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