Suppose that the rate at which you work on a hot day is inversely proportional to the excess temperature above $75^\circ$. One day the temperature is rising steadily, and you start studying at $2$ p.m. You cover $20$ pages the first hour and $10$ pages the second hour. At what time was the temperature $75^\circ$?
My initial model for a differential equation is: $$\frac{dw}{dt}=\frac{k}{\frac{dT}{dt}+75},$$ where $w$ goes for work. But I can't find the correct way to develop it to the right answer. Please help.
For time $t$ hours after $2$pm, let the excess temperature above $75^0$ be $T+at$. (In other words $T$ is the excess temperature at $2$pm)
Then $$\frac{dw}{dt}=\frac{k}{T+at}$$ Integrating $$w=\frac{k}{a}\ln (T+at)+C.$$
Then $$20=\frac{k}{a}\ln (\frac{T+a}{T})\text { and } 30=\frac{k}{a}\ln (\frac{T+2a}{T})$$ and therefore $$(1+\frac {a}{T})^3=(1+\frac {2a}{T})^2.$$
Multiplying out and solving gives $\frac{a}{T}=1.618$, the Golden ratio.
We now know that the temperature above $75^0$ is $2.618T$ at 3pm and $4.236T$ at 4pm. In other words, a change of $1.618T$ takes an hour. So, for the temperature to rise from an excess of $0$ to an excess of $2.618T$ takes $$\frac{2.618}{1.618}= 1.618\text { hours}.$$
The temperature was $75^0$ at a time of $1$ hours $37$ minutes before $3$pm i.e. at $1.23$pm.