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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 𝐹 is taken out where the temperature is 20 𝐹. The reading is
30 𝐹 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 𝐹 to within half a degree of the air
temperature or 20.5 𝐹.
In newtons law of cooling we will this kind of relation:
𝑑𝑇
𝑑𝑑
= π‘˜(𝑇0 − 𝑇𝑠 )
π‘€β„Žπ‘–π‘β„Ž 𝑖𝑠 𝑇0 𝑖𝑠 π‘‘β„Žπ‘’the initial temperature of the object at time
𝑇𝑠 = the temperature of the surrounding environment
𝑇𝑏 = the temperature of the object after time t
𝑑𝑇
𝑑𝑑
= π‘˜(𝑇 − 20)
𝑑𝑇
𝑇−20
= π‘˜π‘‘π‘‘ π‘–π‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘‘π‘’ π‘π‘œπ‘‘β„Ž 𝑠𝑖𝑑𝑒𝑠
We got
ln(𝑇 − 20) = π‘˜π‘‘ + 𝐢
π‘Ÿπ‘Žπ‘–π‘ π‘–π‘›π‘” π‘π‘œπ‘‘β„Ž 𝑠𝑖𝑑𝑒 π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘‘π‘œ π‘‘β„Žπ‘’ 𝑒π‘₯π‘π‘œπ‘›π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘π‘œπ‘€π‘’π‘Ÿ π‘‘π‘œ π‘π‘Žπ‘›π‘π‘’π‘™ 𝑙𝑛
𝑇 − 20 = 𝑒 π‘˜π‘‘ . 𝑒 𝐢
𝑇 = 20 + 𝐢𝑒 π‘˜π‘‘
To Determine C , we need to use the initial condition that when t=0 the leading thermometer was
75degrees
𝑇 = 20 + 𝐢𝑒 π‘˜π‘‘
75 = 20 + 𝐢𝑒 π‘˜0
75 = 20 + 𝐢
𝐢 = 55
𝑇 = 20 + 55𝑒 π‘˜π‘‘
To find k , we need to use second initial condition
π‘Žπ‘‘ 𝑇 = 30℉ π‘Žπ‘“π‘‘π‘’π‘Ÿ 4 π‘šπ‘–π‘›π‘’π‘‘π‘’π‘  π‘œπ‘‘ 𝑑 = 4 π‘œπ‘“ 𝑒π‘₯π‘π‘œπ‘ π‘’π‘Ÿπ‘’ π‘‘π‘œ π‘‘β„Žπ‘’ π‘œπ‘’π‘‘π‘ π‘–π‘‘π‘’
30 = 20 + 55𝑒 π‘˜(4)
30 − 20 = 55𝑒 4π‘˜
30−20
55
= 𝑒 4π‘˜
30−20
)
55
ln (
Ln(
30−20
)
55
4
π‘‘π‘Žπ‘˜π‘’ ln
π‘œπ‘“ π‘π‘œπ‘‘β„Ž π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘‘π‘œ π‘π‘Žπ‘›π‘π‘’π‘™ 𝑒
= 4π‘˜
=π‘˜
π‘˜ = −0.426
𝑇 = 20 + 55𝑒 −0.426𝑑
π‘ π‘’π‘π‘π‘œπ‘ π‘’ π‘‘β„Žπ‘Žπ‘‘ π‘Žπ‘“π‘‘π‘’π‘Ÿ 7 π‘šπ‘–π‘›π‘  π‘‘β„Žπ‘’ π‘‘β„Žπ‘’π‘Ÿπ‘šπ‘œπ‘šπ‘’π‘‘π‘’π‘Ÿ 𝑖𝑠 π‘π‘Ÿπ‘–π‘›π‘”π‘–π‘›π‘” π‘œπ‘’π‘‘π‘ π‘–π‘‘π‘’ π‘‘β„Žπ‘’ π‘‘β„Žπ‘’π‘Ÿπ‘šπ‘œπ‘šπ‘’π‘‘π‘’π‘Ÿ 𝑖𝑠 (𝑑)π‘‘π‘’π‘”π‘Ÿπ‘’π‘’
π‘Žπ‘‘ 𝑑 = 7
𝑇 = 20 + 55𝑒 −0.426(7)
𝑇 = 22.788℉ π‘‘β„Žπ‘’π‘Ÿπ‘šπ‘œπ‘šπ‘’π‘‘π‘’π‘Ÿ π‘Ÿπ‘’π‘Žπ‘‘π‘–π‘›π‘” 7 π‘šπ‘–π‘›π‘  π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘‘β„Žπ‘’ π‘‘β„Žπ‘’π‘Ÿπ‘šπ‘œπ‘šπ‘’π‘‘π‘’π‘Ÿ π‘€π‘Žπ‘  π‘π‘Ÿπ‘œπ‘’π‘”β„Žπ‘‘ π‘œπ‘’π‘‘π‘ π‘–π‘‘π‘’
b. the time taken for the reading to drop from 750 𝐹 to within half a degree of the air temperature or
20.5 0 𝐹.
𝑙𝑒𝑑 π‘‘β„Žπ‘’ π‘‘π‘–π‘šπ‘’ π‘‘π‘Žπ‘˜π‘’π‘› π‘“π‘Žπ‘™π‘™ π‘“π‘Ÿπ‘œπ‘š 75℉ π‘‘π‘œ20.5℉ 𝑏𝑒 π‘Žπ‘‘ 𝑑(min
)
𝑙𝑒𝑑𝑠 π‘Žπ‘π‘π‘™π‘¦ 𝑖𝑑 π‘‘π‘œ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘π‘œ. 4
20.5 = 20 + 55𝑒 −0.426(𝑑)
0.5 = 55𝑒 −0.426(𝑑) π‘‘π‘Žπ‘˜π‘’ ln π‘œπ‘“ π‘π‘œπ‘‘β„Ž π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›
0.5
ln ( 55 ) = −0.426𝑑
Ln(
0.5
)
55
−0.426
=𝑑
𝑑 = 11.03 π‘šπ‘–π‘›π‘’π‘‘π‘’π‘ 
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?
π‘Žπ‘‘ π‘‘β„Žπ‘–π‘  π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘›
π‘Š = π‘˜π‘£ π‘π‘’π‘π‘Žπ‘’π‘ π‘’ π‘“π‘œπ‘Ÿπ‘π‘’ π‘”π‘Ÿπ‘Žπ‘£π‘–π‘‘π‘¦ π‘‘π‘œπ‘€π‘›π‘€π‘Žπ‘Ÿπ‘‘ 𝑖𝑠 π‘šπ‘Žπ‘¦π‘π‘’ π‘’π‘žπ‘’π‘Žπ‘™ 𝑑𝑝 π‘“π‘œπ‘Ÿπ‘π‘’ π‘œπ‘“π‘Žπ‘–π‘Ÿ π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘’π‘π‘€π‘Žπ‘Ÿπ‘‘
π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑣 = 𝑖𝑠 π‘‘β„Žπ‘’ 𝑠𝑝𝑒𝑒𝑑 π‘œπ‘“ π‘ π‘œπ‘šπ‘’ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘˜ , π‘ π‘œ π‘Š = π‘˜π‘£
𝑏𝑒𝑑 𝑖𝑑 π‘ π‘Žπ‘–π‘‘ π‘‘β„Žπ‘Žπ‘‘ π‘™π‘–π‘šπ‘–π‘‘π‘–π‘›π‘” 𝑠𝑝𝑒𝑒𝑑 𝑖𝑠 π‘‘π‘œπ‘’π‘π‘™π‘’ π‘œπ‘“ π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ 𝑠𝑝𝑒𝑒𝑑 π‘€β„Žπ‘’π‘Ÿπ‘’ 4.6
π‘š
𝑠
𝑖𝑠 π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘–π‘’π‘‘ 𝑏𝑦 2
45
45
π‘˜ 𝑖𝑠 9.2 π‘˜ = 9.2
𝑖𝑓 π‘‘β„Žπ‘’ π‘€π‘’π‘–π‘”β„Žπ‘‘ 𝑖𝑠 45 𝑁 π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘‘β„Žπ‘’ π‘œπ‘π‘—π‘’π‘π‘‘
45
𝑖𝑠 𝑔
, π‘ π‘œ π‘‘π‘’π‘π‘’π‘˜π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑑𝑒𝑒 π‘‘π‘œ π‘Žπ‘–π‘Ÿ π‘Ÿπ‘’π‘ π‘–π‘‘π‘Žπ‘›π‘π‘’ π‘šπ‘’π‘ π‘‘
π‘ π‘œ π‘“π‘œπ‘Ÿ π‘‘β„Žπ‘’ π‘€π‘’π‘–π‘”β„Žπ‘‘ 𝑖𝑠 π‘π‘Ÿπ‘œπ‘—π‘’π‘π‘‘π‘’π‘‘ π‘‘π‘œπ‘€π‘›π‘€π‘Žπ‘Ÿπ‘‘π‘ 
We will use this Differential equation
𝑀
𝑔
∑𝐹 =
𝑑𝑉
. 𝑑𝑑
π‘Š − π‘˜π‘‰ =
𝑀
𝑔
.
𝑑𝑉
𝑑𝑑
π‘š
𝑠
π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑉𝑙 = 9.2
𝑠𝑖𝑛𝑐𝑒 𝑀𝑒 β„Žπ‘Žπ‘£π‘’ π‘Žπ‘–π‘Ÿ π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’
, π‘€β„Žπ‘’π‘›
𝑑𝑉
𝑑𝑑
=π‘œ
π‘€β„Žπ‘’π‘› π‘Š − 9.2π‘˜ = 0
π‘Š
π‘˜ = 9.2
And we will substitute it in our equation
𝑑𝑉
9.2π‘”π‘Š − π‘Šπ‘”π‘‰ = 9.2 π‘Š 𝑑𝑑
𝑑𝑉
π‘Šπ‘”(9.2 − 𝑣) = 9.2π‘Š 𝑑𝑑
9.2 − 𝑉 =
𝑑𝑑 =
9.2 𝑑𝑉
𝑔 𝑑𝑑
9.2 𝑑𝑉
𝑔 9.2−𝑉
𝑑=−
π‘–π‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘‘π‘’ π‘π‘œπ‘‘β„Ž 𝑠𝑖𝑑𝑒𝑠
9.2
ln(9.2 −
𝑔
𝑉) + 𝐢
π‘†π‘œ π‘›π‘œπ‘€ π‘‘π‘œ π‘ π‘œπ‘™π‘£π‘’ 𝐢 , 𝑀𝑒 𝑀𝑖𝑙𝑙 𝑒𝑠𝑒 π‘œπ‘’π‘Ÿ π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑑 = 0 π‘Žπ‘›π‘‘ 𝑉𝑖 = 4.6
9.2
0 = − 9.81 ln(9.2 − 4.6) + 𝐢
𝐢 = 1.43
π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ π‘œπ‘’π‘Ÿ π‘šπ‘œπ‘‘π‘Žπ‘™ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘“π‘œπ‘Ÿ π‘‘β„Žπ‘–π‘  π‘π‘Ÿπ‘œπ‘π‘™π‘’π‘š 𝑖𝑠 𝑑 = −
9.2
ln(9.2 − 𝑉) +
𝑔
π‘‘π‘œ π‘ π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ has the body traveled when it has reached a speed of 6
𝑀𝑒 π‘šπ‘’π‘ π‘‘ 𝑛𝑒𝑒𝑑 π‘‘π‘œ 𝑖𝑑𝑒𝑛𝑑𝑖𝑓𝑦 π‘“π‘–π‘Ÿπ‘ π‘‘ π‘‘β„Žπ‘’ (𝑑)
𝐴𝑛𝑑 π‘‘π‘œ π‘ π‘œπ‘™π‘£π‘’ 𝑑 , 𝑀𝑒 𝑀𝑖𝑙𝑙 𝑒𝑠𝑒 π‘œπ‘’π‘Ÿ π‘ π‘’π‘π‘œπ‘›π‘‘ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑑 = 0 π‘Žπ‘›π‘‘ 𝑉 = 6
π‘Žπ‘‘ 𝑉 = 6
π‘š
𝑠
1.43
m
s
45𝑣
9.2
45
𝑔
9.2
𝑑 = − 9.81 ln(9.2 − 6) + 1.43
𝑑 = 0.34 𝑠𝑒𝑐
use the formula of velocity to get the distance
𝑣=
𝑑
𝑑
π‘Žπ‘‘ 𝑣 = 6 π‘Žπ‘›π‘‘ 𝑑 = 0.34
𝑑
6 = 0.34
𝑑 = 6(0.34)
𝑑 = 2.04 π‘š 𝑖𝑠 π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ has the body traveled when it has reached a speed of 6 m/s
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