MHB Modeling with Differential Equations

[ineedhelpnow]

#1
Which of the following FUNCTIONS are SOLUTIONS (meaning PLURAL) of the differential equation $y''+y=sin(x)$?
a. $y=sin(x)$
b. $y=cos(x)$
c. $y=\frac{1}{2}xsin(x)$
d. $y=\frac{-1}{2}xcos(x)$#2
Suppose you have just poured a cup of freshly brewed coffee with temperature $95^oC$ in a room where the temperature is $20^oC$.

a. When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain.

b. Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton's Law of Cooling for this particular situation. What is the initial condition? In view of answer to part (a), do you think this differential equation is an appropriate model for cooking?

Please Please help me. I have a test tomorrow and I really need to understand this stuff.

 

[chisigma]

#1
Which of the following FUNCTIONS are SOLUTIONS (meaning PLURAL) of the differential equation $y''+y=sin(x)$?
a. $y=sin(x)$
b. $y=cos(x)$
c. $y=\frac{1}{2}xsin(x)$
d. $y=\frac{-1}{2}xcos(x)$#2
Suppose you have just poured a cup of freshly brewed coffee with temperature $95^oC$ in a room where the temperature is $20^oC$.

a. When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain.

b. Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton's Law of Cooling for this particular situation. What is the initial condition? In view of answer to part (a), do you think this differential equation is an appropriate model for cooking?

Please Please help me. I have a test tomorrow and I really need to understand this stuff.

A simple inspection shows that a [particular] solution of the DE $\displaystyle y^{\ ''} + y = \sin x$ is $\displaystyle y= - \frac {x}{2}\ \cos x$...

Kind regards

$\chi$ $\sigma$

 

[I like Serena]

A nickle for your thoughts on these problems.

 

[ineedhelpnow]

i want to be able to use my calculator do these but every time i put it into my calculator i get a different answer.

 

[I like Serena]

i want to be able to use my calculator do these but every time i put it into my calculator i get a different answer.

What do you get with your calculator?

 

[ineedhelpnow]

$y=c1*cos(x)+c2*sin(x)-\frac{xcos(x)}{2}$

 

[M R]

$y=c1*cos(x)+c2*sin(x)-\frac{xcos(x)}{2}$

and in the case where $c_1=c_2=0$?

 

[ineedhelpnow]

idk. i guess that would make sense. kinda.

 

[MarkFL]

For the first problem, you could try taking each given function, compute its second derivative, and then substitute the function and its second derivative into the given ODE to see it an identity results. :D

 

[ineedhelpnow]

what about the second question?

 

[MarkFL]

Newton's Law of Cooling says the time rate of change of the temperature of the object is proportional to the temperature difference between the object and its environment, or the ambient temperature.

So, the change will be greatest when the difference is greatest, right? When is this?

 

[ineedhelpnow]

...wheeeen it's placed in the room? (Tmi)

 

[MarkFL]

...wheeeen it's placed in the room? (Tmi)

Yes, when $t=0$, because thereafter the object's temperature moves towards the ambient temperature, and so the difference, and hence the rate of change, will thereafter decrease. :D

 

[I like Serena]

...wheeeen it's placed in the room? (Tmi)

That was easy, wasn't it? ;)
How about the rest of the problem?
Any thoughts?

 

[ineedhelpnow]

no laughing or i WILL (Wait) kick you but here's what I am thinking. (Nerd)

$\frac{dT}{dt}=k(T-T_{monkey})$

$T'(t)=-k(T-20)$ and $T(0)=95$

$\frac{dT}{T-20}=-kdt$

$\ln\left({T-20}\right)=-kt+C$

$\ln\left({95-20}\right)=-k(0)+C$

$C=\ln\left({75}\right)$

$\ln\left({T-20}\right)=-kt+ \ln\left({75}\right)$

$\ln\left({\frac{T-20}{75}}\right)=-kt$

$\frac{T-20}{75}=e^{-kt}$

$T=20+75e^{-kt}$

did it based of another question i did yesterday but i don't know if its right...at all.

 

[ineedhelpnow]

It's right?! (Dance)

 

[I like Serena]

no laughing or i WILL (Wait) kick you but here's what I am thinking. (Nerd)

$\frac{dT}{dt}=k(T-T_{monkey})$

$T'(t)=-k(T-20)$ and $T(0)=95$

$\frac{dT}{T-20}=-kdt$

$\ln\left({T-20}\right)=-kt+C$

$\ln\left({95-20}\right)=-k(0)+C$

$C=\ln\left({75}\right)$

$\ln\left({T-20}\right)=-kt+ \ln\left({75}\right)$

$\ln\left({\frac{T-20}{75}}\right)=-kt$

$\frac{T-20}{75}=e^{-kt}$

$T=20+75e^{-kt}$

did it based of another question i did yesterday but i don't know if its right...at all.

I like it. I think it is all correct!
(Although I would write $\frac{dT}{dt}=-k(T-T_{monkey})$ to be consistent.)

Btw, what would be the temperature of the bunny? (Wondering)
(Need to be complete!)

 

[ineedhelpnow]

~1.5 degrees F. Because the monkey will DESTROY the bunny, therefore causing the bunny's temperature to drop to approx 1.5 F. don't take me wrong. I love bunnies but we all know the monkey is going to win.

- - - Updated - - -

View attachment 2780 or View attachment 2781

Newton's Law of Cooling states that the rate of change of the temperature of the bunny is proportional to the difference between its own temperature and hole in which the monkey is putting it in (due to loss of battle versus monkey).