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Further Applications Of Differential Equations

Learn application of differential equation using newton’s law of cooling differential equation, compound interest & falling Body Problem.

Practical Application of Differential Equation ( Newton's Law )

  • Newton's law of cooling says that time rate of change of temperature of a body is proportional to temperature difference between the body and its surrounding medium. 

            \(\dfrac{dT}{dt} = k \;(T_m - T)\)

        where k > 0, T is the temperature of body at any instance 't' , Tm is the temperature of surrounding.

Illustration Questions

A metal bar at a temperature of 100° F is placed in a room at a constant temperature of 0° F. If after 20 minutes the temperature of the bar is 50° F, find the time it will take the bar to reach 25° F.  [ \(ln\dfrac{1}{4}=-1.3862\) ]

A 53.2 min

B 40.8 min

C 39.6 min

D 28.2 min

×

According to Newton's law of cooling 

\(\dfrac{dT}{dt} = k \;(T_m - T)\)

In this case, \(Tm = 0\)

\(\therefore\;\dfrac{dT}{dt} = - kT \)

\(\Rightarrow \dfrac{dT}{T} = - k dt\)

Integrate both sides

 \(ln \;T = - k\; t + C\)

Now \(T = 100\) where \(t = 0 \Rightarrow C = ln \;100\)

\(\therefore l n \; T = - k \;t + ln \;100 \Rightarrow T = 100 \;e^{-k\; t}\)

 

\(T(20) = 50 \; \Rightarrow 50 = 100 \; e^{-20 \; k}\)

\(\Rightarrow k = - \dfrac{1}{20} \; ln \; \dfrac{1}{2}=0.035\)

\(\Rightarrow T = 100 \; e^{-0.035 \; t}\)

\(T(t)=25\)\(\Rightarrow25 = 100 \; e ^{-.035 t}\;\)

\(\Rightarrow - .035 \; t = ln \dfrac{1}{4}\)

\(\Rightarrow \; t = \dfrac{ln\; 1/4}{-.035} = 39.6 \; min\)

 

A metal bar at a temperature of 100° F is placed in a room at a constant temperature of 0° F. If after 20 minutes the temperature of the bar is 50° F, find the time it will take the bar to reach 25° F.  [ \(ln\dfrac{1}{4}=-1.3862\) ]

A

53.2 min

.

B

40.8 min

C

39.6 min

D

28.2 min

Option C is Correct

Application of Differential Equation (Falling Body Problem)

  • Consider a vertically falling body of mass \(m\) that is being influenced by gravity \('g'\) and air resistance that is proportional to velocity of body.
  • Newton's law of motion says that net force acting on body is equal to the time rate of change of momentum of body.

         \(F=m\,\dfrac{dv}{dt}\) (\(m\) is constant)

\(\therefore\;mg-kv=m\dfrac{dv}{dt}\)

\(\Rightarrow\;\dfrac{dv}{dt}+\dfrac{kv}{m}=g\;\to\) This is a separable differential equation and can be solved.

  • The limiting velocity is defined as \(v_\ell=\dfrac{mg}{k}\) (where resultant force is 0)

Illustration Questions

A body weighing \(50\,kg\) is dropped from a height of \(100\,m\) with an initial velocity of \(5\,m/sec.\) If air resistance is proportional to velocity of the body and limiting velocity is \(65\,m/sec.\) Find velocity at time \(t=5\,sec.\) (Take \(g=9.8\,m/s^2\))

A \(28.12\,m/s\)

B \(36.76\,m/s\)

C \(42.13\,m/s\)

D \(52.2\,m/s\)

×

If air resistance is proportional to velocity then

\(\dfrac{dv}{dt}+\dfrac{kv}{m}=g\) and \(v_\ell=\dfrac{mg}{k}\)

In this case \(\dfrac{mg}{k}=65\,m/s\)

\(\Rightarrow\;\dfrac{50×9.8}{k}=65\)

\(\Rightarrow\;k=7.54\)

\(\therefore\;\dfrac{dv}{dt}+\dfrac{7.54v}{50}=9.8\)

\(\Rightarrow\;\dfrac{dv}{dt}=9.8-0.1508\,v\)

\(\Rightarrow\;\displaystyle\int\dfrac{dv}{9.8-0.1508\,v}=\int dt\)

\(\Rightarrow\;\dfrac{ln\,|9.8-0.1508\,v|}{-0.1508}=t+c\)

\(\Rightarrow\;9.8-0.1508v=ce^{-0.1508\,t}\)

\(\Rightarrow\;v=\left(\dfrac{9.8-ce^{-0.1508\,t}}{0.1508}\right)\)

At \(t=0,\;v=5\,m/s\)

\(\Rightarrow\;5=\dfrac{9.8-c}{0.1508}\)

\(\Rightarrow\;c=9.046\)

\(\therefore\;v(t)=\left(\dfrac{9.8-9.046\,e^{-0.1508\,t}}{0.1508}\right)\)

\(\therefore\;v(5)=\dfrac{9.8-9.046\,e^{-0.1508×5}}{0.1508}=36.76\,m/s\)

A body weighing \(50\,kg\) is dropped from a height of \(100\,m\) with an initial velocity of \(5\,m/sec.\) If air resistance is proportional to velocity of the body and limiting velocity is \(65\,m/sec.\) Find velocity at time \(t=5\,sec.\) (Take \(g=9.8\,m/s^2\))

A

\(28.12\,m/s\)

.

B

\(36.76\,m/s\)

C

\(42.13\,m/s\)

D

\(52.2\,m/s\)

Option B is Correct

Practical Application of Differential Equation (Compound Interest Problem)

  • Suppose a person places an amount \(N_0\) in a saving account which pays \(P\)% interest per annum compounded continuously, then \(\dfrac{dN}{dt}=\dfrac{P}{100}N\)

where \(N(t)\) is the balance in account at time \('t'\)

\(\Rightarrow\;\dfrac{dN}{N}=\dfrac{P\,dt}{100}\;\to\) Separate the variables

\(\Rightarrow\;ln\,N=\dfrac{P}{100}t+C\)

This equation can be used to find one of the parameter given the value of other. 

Illustration Questions

A person places \($\;20000\) in a saving account which pays 5% interest per annum compounded continuously. Find the amount in the account after \(3\) years .

A \($\,22512.12\;\)

B \($\,23236.38\;\)

C \($\,18192.02\;\)

D \($\,25918.23\;\)

×

\(\dfrac{dN}{dt}=\dfrac{P}{100}N\) when \(N(t)=\text{amount at time}\;t\)

In this case \(P=5,\;N_0=20000\)

\(\therefore\;\dfrac{dN}{dt}=\dfrac{5}{100}N\)

\(\Rightarrow\;\dfrac{dN}{N}=0.05\,dt\)

\(\therefore\;ln\,N=0.05\,t+C\)

\(\Rightarrow\;N=Ce^{0.05\,t}\)

At \(t=0,\;N=20000=N_0\)

\(\therefore\;20000=Ce^0\)

\(\Rightarrow C=20000\)

\(\therefore\;N(t)=20000\,e^{0.05\,t}\)

\(\therefore\;N(3)=20000×e^{0.05×3}\)

\(=$\,23236.68\;\)

A person places \($\;20000\) in a saving account which pays 5% interest per annum compounded continuously. Find the amount in the account after \(3\) years .

A

\($\,22512.12\;\)

.

B

\($\,23236.38\;\)

C

\($\,18192.02\;\)

D

\($\,25918.23\;\)

Option B is Correct

Illustration Questions

What constant rate is required if an initial deposit placed into an account that accrues interest compounded continuously is to double its value in \(6\) years ?

A \(11.55\,\text{%}\)

B \(9.55\,\text{%}\)

C \(10.21\,\text{%}\)

D \(12.25\) %

×

\(\dfrac{dN}{dt}=\dfrac{P}{100}N\) when \(N(t)=\text{amount at time}\;t\)

 \(\dfrac{dN}{N}=\dfrac{P}{100}dt\)

\(\Rightarrow\;ln\,N=\dfrac{P}{100}t+c\)

At \(t=0,\;N=N_0\)

\(\Rightarrow\;c=ln\,N_0\)

\(\therefore\;\dfrac{ln\,N}{N_0}=\dfrac{P}{100}\,t\)

\(\therefore\;N(t)=N_0\,e^{Pt/100}\)

Now \(N(6)=2N_0\)

\(\Rightarrow\;2N_0=N_0\,e^{Pt/100}\)

\(\Rightarrow\;2=\,e^{6P/100}\)

\(\Rightarrow \dfrac{6P}{100}=ln\,2\)

\(\Rightarrow\;P=\dfrac{100×ln\,2}{6}\)

\(=11.55\,\text{%}\)

What constant rate is required if an initial deposit placed into an account that accrues interest compounded continuously is to double its value in \(6\) years ?

A

\(11.55\,\text{%}\)

.

B

\(9.55\,\text{%}\)

C

\(10.21\,\text{%}\)

D

\(12.25\) %

Option A is Correct

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