1 / 10

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.

Download Presentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related