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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.

#### 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)

#### 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.

#### 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

#### 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