Vijandre, Jorrelle Lanz B. 01/13/21 DIFFERENTIAL EQUATION 2CE-1 MW(2:30PM-4:00PM) ASSIGNMENT 10 OR ASSIGNMENT 3 IN FINALS Assignment: 1. A thermometer reading 75 πΉ is taken out where the temperature is 20 πΉ. The reading is 30 πΉ 4 minutes later. Find a) the thermometer reading 7 minutes after the thermometer was brought outside, and b) the time taken for the reading to drop from 75 πΉ to within half a degree of the air temperature or 20.5 πΉ. In newtons law of cooling we will this kind of relation: ππ ππ‘ = π(π0 − ππ ) π€βππβ ππ π0 ππ π‘βπthe initial temperature of the object at time ππ = the temperature of the surrounding environment ππ = the temperature of the object after time t ππ ππ‘ = π(π − 20) ππ π−20 = πππ‘ πππ‘πππππ‘π πππ‘β π ππππ We got ln(π − 20) = ππ‘ + πΆ ππππ πππ πππ‘β π πππ πππ’ππ‘πππ π‘π π‘βπ ππ₯ππππππ‘πππ πππ€ππ π‘π ππππππ ππ π − 20 = π ππ‘ . π πΆ π = 20 + πΆπ ππ‘ To Determine C , we need to use the initial condition that when t=0 the leading thermometer was 75degrees π = 20 + πΆπ ππ‘ 75 = 20 + πΆπ π0 75 = 20 + πΆ πΆ = 55 π = 20 + 55π ππ‘ To find k , we need to use second initial condition ππ‘ π = 30β πππ‘ππ 4 ππππ’π‘ππ ππ‘ π‘ = 4 ππ ππ₯πππ π’ππ π‘π π‘βπ ππ’π‘π πππ 30 = 20 + 55π π(4) 30 − 20 = 55π 4π 30−20 55 = π 4π 30−20 ) 55 ln ( Ln( 30−20 ) 55 4 π‘πππ ln ππ πππ‘β πππ’ππ‘πππ π‘π ππππππ π = 4π =π π = −0.426 π = 20 + 55π −0.426π‘ π π’ππππ π π‘βππ‘ πππ‘ππ 7 ππππ π‘βπ π‘βπππππππ‘ππ ππ ππππππππ ππ’π‘π πππ π‘βπ π‘βπππππππ‘ππ ππ (π‘)ππππππ ππ‘ π‘ = 7 π = 20 + 55π −0.426(7) π = 22.788β π‘βπππππππ‘ππ πππππππ 7 ππππ πππ‘ππ π‘βπ π‘βπππππππ‘ππ π€ππ ππππ’πβπ‘ ππ’π‘π πππ b. the time taken for the reading to drop from 750 πΉ to within half a degree of the air temperature or 20.5 0 πΉ. πππ‘ π‘βπ π‘πππ π‘ππππ ππππ ππππ 75β π‘π20.5β ππ ππ‘ π‘(min ) πππ‘π πππππ¦ ππ‘ π‘π πππ’ππ‘πππ ππ. 4 20.5 = 20 + 55π −0.426(π‘) 0.5 = 55π −0.426(π‘) π‘πππ ln ππ πππ‘β πππ’ππ‘πππ 0.5 ln ( 55 ) = −0.426π‘ Ln( 0.5 ) 55 −0.426 =π‘ π‘ = 11.03 ππππ’π‘ππ 2. A 45 N weight is projected downward with an initial speed of 4.6 m/s. If the air resistance is proportional to the speed and if the limiting speed is double the initial speed, how far has the body traveled when it has reached a speed of 6 m/s? ππ‘ π‘βππ ππππππ‘πππ π = ππ£ πππππ’π π πππππ ππππ£ππ‘π¦ πππ€ππ€πππ ππ πππ¦ππ πππ’ππ π‘π πππππ πππππ πππ ππ π‘ππππ π’ππ€πππ π€βπππ π£ = ππ π‘βπ π ππππ ππ π πππ ππππ π‘πππ‘ π , π π π = ππ£ ππ’π‘ ππ‘ π πππ π‘βππ‘ πππππ‘πππ π ππππ ππ πππ’πππ ππ ππππ‘πππ π ππππ π€βπππ 4.6 π π ππ ππ’ππ‘ππππππ ππ¦ 2 45 45 π ππ 9.2 π = 9.2 ππ π‘βπ π€πππβπ‘ ππ 45 π πππ π ππ π‘βπ ππππππ‘ 45 ππ π , π π πππππππππ‘πππ ππ’π π‘π πππ πππ ππ‘ππππ ππ’π π‘ π π πππ π‘βπ π€πππβπ‘ ππ πππππππ‘ππ πππ€ππ€ππππ We will use this Differential equation π€ π ∑πΉ = ππ . ππ‘ π − ππ = π€ π . ππ ππ‘ π π π€βπππ ππ = 9.2 π ππππ π€π βππ£π πππ πππ ππ π‘ππππ , π€βππ ππ ππ‘ =π π€βππ π − 9.2π = 0 π π = 9.2 And we will substitute it in our equation ππ 9.2ππ − πππ = 9.2 π ππ‘ ππ ππ(9.2 − π£) = 9.2π ππ‘ 9.2 − π = ππ‘ = 9.2 ππ π ππ‘ 9.2 ππ π 9.2−π π‘=− πππ‘πππππ‘π πππ‘β π ππππ 9.2 ln(9.2 − π π) + πΆ ππ πππ€ π‘π π πππ£π πΆ , π€π π€πππ π’π π ππ’π ππππ‘πππ ππππππ‘πππ π€βπππ π‘ = 0 πππ ππ = 4.6 9.2 0 = − 9.81 ln(9.2 − 4.6) + πΆ πΆ = 1.43 π‘βπππππππ ππ’π πππππ πππ’ππ‘πππ πππ π‘βππ πππππππ ππ π‘ = − 9.2 ln(9.2 − π) + π π‘π π πππ£π π‘βπ πππ π‘ππππ has the body traveled when it has reached a speed of 6 π€π ππ’π π‘ ππππ π‘π πππππ‘πππ¦ ππππ π‘ π‘βπ (π‘) π΄ππ π‘π π πππ£π π‘ , π€π π€πππ π’π π ππ’π π πππππ ππππππ‘πππ π€βπππ π‘ = 0 πππ π = 6 ππ‘ π = 6 π π 1.43 m s 45π£ 9.2 45 π 9.2 π‘ = − 9.81 ln(9.2 − 6) + 1.43 π‘ = 0.34 π ππ use the formula of velocity to get the distance π£= π π‘ ππ‘ π£ = 6 πππ π‘ = 0.34 π 6 = 0.34 π = 6(0.34) π = 2.04 π ππ π‘βπ πππ π‘ππππ has the body traveled when it has reached a speed of 6 m/s
