Section 1.2 – Solutions of Some Differential Equations

Directions. The following are review problems for the section. It is recommended you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it.
1. The instantaneous rate of change of the temperature $$T$$ of coffee at time $$t$$ is proportional to the difference between the temperature $$M$$ of the air and the temperature $$T$$ at time $$t$$.
1. Find the mathematical model for the problem.
2. Given that the room temperature is $$75^{\circ}$$ and $$k=0.08$$, find the solutions to the differential equation.
3. The initial temperature of the coffee is $$200^{\circ}$$F. Find the solution to the problem.​

1. $$\dfrac{dT}{dt}=-k(T-M) \quad k>0$$
2. $$T(t)=75+Ce^{-0.08t}$$
3. $$T(t)=75+125e^{-0.08t}$$

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2. Your swimming pool containing 60,000 gal of water has been contaminated by 5 kg of a non toxic dye that leaves a swimmer's skin an unattractive green. The pool's filtering system can take water from the pool, remove the dye, and return the water to the pool at a flow rate of 200 gal/min.
1. Write down the initial value problem for the filtering process; let $$q(t)$$ be the amount of dye in the pool at any time $$t$$.
2. Solve the problem.
3. You have invited several dozen friends to a pool party that is scheduled to begin in 4 hours. You have also determined that the effect of the dye is imperceptible if its concentration is less than 0.02 g/gal. Is your filtering system capable of reducing the dye concentration to this level within 4 hours?​

1. $$\dfrac{dq}{dt}=-\dfrac{1}{300}q(t),\quad$$$$q(0)=5$$
2. $$q(t)=5e^{-\frac{1}{300}t}$$
3. Four hours is not enough time.

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3. In curling, the player has to slide a stone of mass $$m$$ on a smooth surface with friction coefficient $$\mu$$. If the air friction is proportional to the velocity of the stone with a drag coefficient of $$\gamma$$, find the stopping time and the stopping distance of the stone if the initial velocity is $$v_0$$.​

The stopping time of the stone is $$T=\dfrac{m}{\gamma}\ln(1 +\dfrac{\gamma v_0}{\mu mg})$$ and the stopping distance of the stone is $$D=\dfrac{m}{\gamma}\left(\dfrac{\mu mg}{\gamma}+v_0\right)\left[1-\dfrac{\mu mg}{\mu mg+\gamma v_0}\right]-\mu g (\dfrac{m}{\gamma})^2\ln(1+\dfrac{\gamma v_0}{\mu mg})$$

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