Aimathic – Exponential models – Free Math Worksheets

Exponential models

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52770512

A metal part is removed from a furnace at \(95\,{}^\circ\text{C}\) and placed in a room maintained at \(20\,{}^\circ\text{C}\). Its temperature satisfies \(T'(t)=k(20-T(t))\), where \(t\) is the number of minutes since removal. Immediately after removal, the cooling rate is \(T'(0)=-4.5\,{}^\circ\text{C}/\text{min}\). Find \(k\) and a function \(T(t)\) for the temperature.

Hints

- Substitute the initial temperature and initial cooling rate into the differential equation. - The temperature difference from the room temperature decays exponentially. - Use the initial condition to determine the coefficient of the exponential term.

Solution

1. The initial temperature is \(T(0)=95\). Substitute \(t=0\) into the differential equation: \(-4.5=k(20-95)=-75k\). 2. Therefore, \(k=0.06\). 3. The solution that approaches the room temperature and satisfies \(T(0)=95\) is \(T(t)=20+(95-20)e^{-0.06t}=20+75e^{-0.06t}\).

Answer

\(k=0.06\); \(T(t)=20+75e^{-0.06t}\)

52770612

A frozen pizza at \(-18\,{}^\circ\text{C}\) is placed in an oven preheated to \(220\,{}^\circ\text{C}\). Its temperature satisfies Newton's law of heating, \(T'(t)=k(220-T(t))\), where \(t\) is measured in minutes. At \(t=0\), the instantaneous heating rate is \(11.9\,{}^\circ\text{C}/\text{min}\). a) Find \(k\) and the corresponding temperature function \(T(t)\). b) Find when the pizza reaches an internal temperature of \(75\,{}^\circ\text{C}\).

Hints

- Substitute the initial temperature and heating rate into the differential equation. - The temperature difference from the oven temperature decays exponentially. - Isolate the exponential expression before taking a logarithm.

Solution

1. Since \(T(0)=-18\) and \(T'(0)=11.9\), the differential equation gives \(11.9=k(220-(-18))=238k\). Thus, \(k=0.05\). 2. The solution approaching the oven temperature is \(T(t)=220-(220-(-18))e^{-0.05t}=220-238e^{-0.05t}\). 3. Set \(T(t)=75\): \(75=220-238e^{-0.05t}\). Then \(e^{-0.05t}=\frac{145}{238}\), so \(t=\frac{\ln(145/238)}{-0.05}\approx 9.91\) minutes.

Answer

a) \(k=0.05\); \(T(t)=220-238e^{-0.05t}\) b) About \(9.91\,\text{minutes}\)

53277712

A cup of hot coffee is placed in a room at a constant temperature of \(70\,^{\circ}\text{F}\). Its temperature is modeled by \(f(t)=ae^{-kt}+c\), where \(t\ge 0\) is measured in minutes and \(f(t)\) is measured in degrees Fahrenheit. Immediately after the coffee is poured, its temperature is \(180\,^{\circ}\text{F}\), and it is cooling at an instantaneous rate of \(11\,^{\circ}\text{F}\) per minute. The graph of the temperature is shown below. a) Find \(a\), \(c\), and \(k\). b) Find the coffee's temperature after \(10\) minutes. c) When will the coffee cool to \(125\,^{\circ}\text{F}\)? Round to two decimal places.

Hints

- Translate the initial temperature, long-term temperature, and initial cooling rate into equations. - Which parameter gives the long-term temperature? - Use the derivative to represent the initial cooling rate. - After finding the complete model, substitute the given time to evaluate it. - To find a time for a specified temperature, set the model equal to that temperature and solve using logarithms.

Solution

1. The coffee approaches room temperature as \(t\to\infty\). Since \(e^{-kt}\to 0\), the model approaches \(c\), so \(c=70\). 2. Use the initial temperature: \(f(0)=a+70=180\), so \(a=110\). 3. Differentiate: \(f'(t)=-110ke^{-kt}\). The initial cooling rate is \(f'(0)=-11\), so \(-110k=-11\), giving \(k=0.1\). Thus \(f(t)=110e^{-0.1t}+70\). 4. After \(10\) minutes, \(f(10)=110e^{-1}+70\approx 110.47\,^{\circ}\text{F}\). 5. Set the temperature equal to \(125\): \(110e^{-0.1t}+70=125\). Then \(e^{-0.1t}=\frac{1}{2}\), so \(t=10\ln(2)\approx 6.93\) minutes.

Answer

a) \(a=110\), \(c=70\), and \(k=0.1\), so \(f(t)=110e^{-0.1t}+70\). b) \(f(10)\approx 110.47\,^{\circ}\text{F}\) c) Approximately \(6.93\) minutes

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