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# Application of calculus in everyday life

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### Application of calculus in everyday life

1. 1. Application of calculus in everyday life. Newton’s Law of Cooling.
2. 2. What is the differential equation?  A differential equation is an equation involving derivatives of an unknown function and possibly the function itself as well as the independent variable.  Differential equations have many forms and its order is determined based on the highest order of a derivative in it.  First order differential equations are such equations that have the unknown derivative is the first derivative and its own function.  They are divided into separable and 1st order DFE linear.
3. 3. First DFE 1st order DFE linear1st order DFE linear 𝑑𝑦 𝑑𝑥 = 𝐹(𝑥, 𝑦) 𝒅𝒚 𝒅𝒙 = 𝒇 𝒙 ∗ 𝒈 𝒚 So F(x, y) is simply f(x)*g(y)
4. 4. How can we find the solution of the 1st ODE? A first order linear differential equation is an equation of the form ( ) ( ) dy P x y Q x dx   ( ) 0 dy P x y dx   Which can be solved by separating the variables. ( ) dy P x dx y     ln ( )y P x dx c    ( )P x dx c y e    ( )P x dx c y e e   ( )P x dx y Ce   ( )P x dxd ye dx   ( ) ( ) ( ) P x dx P x dxdy e yP x e dx   ( ) ( ) P x dxdy P x y e dx       
5. 5. ( ) ( ) dy P x y Q x dx   If we multiply both sides by ( )P x dx e ( ) ( ) ( ) P x dx P x dxd ye Q x e dx   Now integrate both sides. ( ) ( ) ( ) P x dx P x dx ye Q x e dx   Returning to equation 1,
6. 6. The change in temperature • An object’s temperature over time will approach the temperature of its surroundings (the medium). • The greater the difference between the object’s temperature and the medium’s temperature, the greater the rate of change of the object’s temperature. • This change is a form of exponential decay. T0 Tm
7. 7. Newton’s Law of Cooling  It is a direct application for differential equations.  Formulated by Sir Isaac Newton.  Has many applications in our everyday life.  Sir Isaac Newton found this equation behaves like what is called in Math (differential equations) so his used some techniques to find its general solution.
8. 8. Derivation of Newton’s law of Cooling  Newton’s observations: He observed that observed that the temperature of the body is proportional to the difference between its own temperature and the temperature of the objects in contact with it .  Formulating: First order separable DE  Applying calculus: 𝑑𝑇 𝑑𝑡 = −𝑘(𝑇 − 𝑇𝑒) Where k is the positive proportionality constant
9. 9. Derivation of Newton’s law of Cooling (continued)  By separation of variables we get 𝑑𝑇 (𝑇−𝑇𝑒) = −𝑘 𝑑𝑡  By integrating both sides we get ln 𝑇 − 𝑇𝑒 + 𝐶 = −𝑘𝑡  At time (t=0) the temperature is T0 −ln 𝑇0 − 𝑇𝑒 = 𝐶  By substituting C with −ln 𝑇0 − 𝑇𝑒 we get ln (𝑇 − 𝑇𝑒) (𝑇0 − 𝑇𝑒) = −𝑘𝑡 𝑇 = 𝑇𝑒 + (𝑇0 − 𝑇𝑒)𝑒−𝑘𝑡
10. 10. Applications on Newton’s Law of Cooling: Investigations. • It can be used to determine the time of death. Computer manufacturing. • Processors. • Cooling systems. solar water heater. calculating the surface area of an object.
11. 11. Expressing the applications of Newton’s law of cooling through mathematical problems Investigations in a crime scene Processor manufacturing
12. 12. The police came to a house at 10:23 am were a murder had taken place. The detective measured the temperature of the victim’s body and found that it was 26.7℃. Then he used a thermostat to measure the temperature of the room that was found to be 20℃ through the last three days. After an hour he measured the temperature of the body again and found that the temperature was 25.8℃. Assuming that the body temperature was normal (37℃), what is the time of death?
13. 13. Solution T (t) = Te + (T0 − Te ) e – kt Let the time at which the death took place be x hours before the arrival of the police men. Substitute by the given values T ( x ) = 26.7 = 20 + (37 − 20) e-kx T ( x+1) = 25.8 = 20 + (37 − 20) e - k ( x + 1) Solve the 2 equations simultaneously 0.394= e-kx 0.341= e - k ( x + 1) By taking the logarithmic function ln (0.394)= -kx …(1) ln (0.341)= -k(x+1) …(2)
14. 14. Solution (continued) By dividing (1) by (2) ln(0.394) ln 0.341 = −𝑘𝑥 −𝑘 𝑥+1 0.8657 = 𝑥 𝑥+1 Thus x≃7 hours Therefore the murder took place 7 hours before the arrival of the detective which is at 3:23 pm
15. 15. A global company such as Intel is willing to produce a new cooling system for their processors that can cool the processors from a temperature of 50℃ to 27℃ in just half an hour when the temperature outside is 20℃ but they don’t know what kind of materials they should use or what the surface area and the geometry of the shape are. So what should they do ? Simply they have to use the general formula of Newton’s law of cooling T (t) = Te + (T0 − Te ) e – k And by substituting the numbers they get 27 = 20 + (50 − 20) e-0.5k Solving for k we get k =2.9 so they need a material with k=2.9 (k is a constant that is related to the heat capacity , thermodynamics of the material and also the shape and the geometry of the material)