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Newton's Law of Cooling: Calculating Wait Time for Overheated Car

This example demonstrates how to use Newton's Law of Cooling to calculate the wait time for a car to cool down before continuing driving after overheating.

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Newton's Law of Cooling: Calculating Wait Time for Overheated Car

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  1. x log 4 = log 11 4 4 Takelogof each side. 4 x = log 11 x = 11 log 4 x 4 = 11 log 4 x x Solve 4 = 11. log b = x b The solution is about 1.73. Check this in the original equation. ANSWER x 1.73 EXAMPLE 2 Take a logarithm of each side SOLUTION Write original equation. Change-of-base formula Use a calculator.

  2. You are driving on a hot day when your car overheats and stops running. It overheats at 280°F and can be driven again at 230°F. If r = 0.0048 and it is 80°F outside,how long (in minutes) do you have to wait until you can continue driving? Cars EXAMPLE 3 Use an exponential model

  3. Substitute for T, T , T, and r. R ° x x 230 = (280 – 80)e + 80 Ine = log e = x –0.0048t – 0.2877 – 0.0048t e –0.0048t 0.75 =e –0.0048t 150 = 200e –0.0048t ln 0.75 = ln e 60 t –rt T = ( T – T )e + T R ° R EXAMPLE 3 Use an exponential model SOLUTION Newton’s law of cooling Subtract 80 from each side. Divide each side by 200. Take natural log of each side. Divide each side by – 0.0048.

  4. ANSWER You have to wait about 60 minutes until you can continue driving. EXAMPLE 3 Use an exponential model

  5. x Log 2 = log 5 2 2 Takelogof each side. x 2 = 5 2 x = log 5 x = 5 log 2 log x 2 4. 2 = 5 x log b = x b x 2.32 for Examples 2 and 3 GUIDED PRACTICE Solve the equation. SOLUTION Write original equation. Change-of-base formula Use a calculator.

  6. ANSWER The solution is about 2.32. Check this in the original equation. for Examples 2 and 3 GUIDED PRACTICE

  7. 9x 7 = 15 Takelogof each side. 7 log 15 9x 5. 7 = 15 log 7 9x Log 7 = log 15 7 7 9x = 15 log 7 9x = x log b = x b x 0.155 for Examples 2 and 3 GUIDED PRACTICE Solve the equation. SOLUTION Write original equation. Change-of-base formula Use a calculator.

  8. ANSWER The solution is about 0.155. Check this in the original equation. for Examples 2 and 3 GUIDED PRACTICE

  9. –0.3x –0.3x –0.3x 4e –7 = 13 4e = 13 + 7 4e = 20 20 4 –0.3x –0.3x e = = 5 6. 4e –7 = 13 –0.3x log e = log 5 e e for Examples 2 and 3 GUIDED PRACTICE Solve the equation. SOLUTION

  10. x = = – 5.365 – 1.6094 0.3 ANSWER The solution is about 5.365. Check this in the original equation. x x Ine = log e = x e for Examples 2 and 3 GUIDED PRACTICE – 0.3x =In 5 Divide each side by – 0.3x. – 0.3x =1.6094

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