Application of calculus in everyday life

Application of calculus in everyday life

• 1.
  Application of calculusin everyday life. Newton’s Law of Cooling.
• 2.
  What is thedifferential 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.
  First DFE 1st orderDFE 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.
  How can wefind 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.
  ( ) () 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.
  The change intemperature • 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.
  Newton’s Law ofCooling  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.
  Derivation of Newton’slaw 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.
  Derivation of Newton’slaw 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.
  Applications on Newton’sLaw 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.
  Expressing the applications of Newton’slaw of cooling through mathematical problems Investigations in a crime scene Processor manufacturing
• 12.
  The police cameto 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.
  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.
  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.
  A global companysuch 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)