Question

Please Help me!

Answer 1

what school

Answer 2

hahahahahaha
that's not a question lol

Answer 3

it's not?

Answer 4

no. you gave an equation, and a description of the meaning.

Answer 5

you could've asked for a solution to the equation, or something more specific.

Answer 6

Solve this differential equation to find a function that can describe the object’s temperature at any time t.

Answer 7

okay. what course is this? calculus 2, or diff eq?

Answer 8

that is what it said. I forgot to post the last part

Answer 9

This is calculus 2

Answer 10

you can choose to solve this equation as a separable differential equation, or a first order linear diff eq

Answer 11

Any way I can get the answer is good. Haha. What is the easiest way>

Answer 12

so you know what the first step to solving a separable diff eq is correct?

Answer 13

Yes I know how to solve differential equations

Answer 14

first order linear diff eqs aren't necessarily harder, but they are considered more advanced usually and not considered part of the standard calculus curriculum

Answer 15

I am taking AP calculus and it is killing me. Haha

Answer 16

and the statement you made is actually not very accurate. what you must recognize first is that we are dealing with a separable differential equation here, so you must use what you learned in class about SEPARABLE differential equations to solve this problem

Answer 17

do you remember the first step to solving a separable diff eq?

Answer 18

yes you seperate it into two different equations and you then integrate them

Answer 19

okay, so are you having trouble with this for this equation?

Answer 20

yes

Answer 21

dT/dt =k(T-m)
essentially you seek to put the equation in some form
dT/dt*f(T)=g(t)
where f(T) is some function of T (or expression of T if you want to think of it that way) and g(t) is some function or expression of t.

Answer 22

any ideas now?

Answer 23

I already had that written down. All of the letters are confusing. I wish they were numbers.

Answer 24

don't wish they were numbers lol. you gotta start being able to recognize in your math classes what variables serve what purpose, otherwise you will struggle continuously throughout any STEM field.

recall that in any ordinary (and we are only dealing with ordinary diff eqs here) differential equation you are interested in the rate of change of one variable with respect to another.
Let us identify the variables here at play, and their significance:
we have
T, t, k, and m
which variable(s) is the dependent variable? which variable(s) is the independent variable? which variable(s) is constant?

Answer 25

dependent variable is T I think and t

Answer 26

Independent variable is k and m?

Answer 27

think again. when you have some equation y=f(x), y is a function of x, meaning that y is dependent on x. x is independent. in other words, when you take dy/dx you are finding the rate of change of ONE variable (y) with respect to ANOTHER variable (x)

Answer 28

it might seem like I am going on a tangent here, but it's far more important for you to undersatnd this concept than to work through this problem with a partial understanding of what's going on (that's probably what led you here in the first place).

if you look at the equation y=mx+b where b is the y-intercept and m is the slope of a line, which variables are the dependent, which are the independent, and which are constants?

Answer 29

I just solved one of these problems with numbers and I got it right.

Answer 30

okay I am trying to understand it is just hard.

Answer 31

yes, just pretend we aren't in calculus for a minute here, and go back to algebra.

Answer 32

y= dependent variable b=constant

Answer 33

x= independent variable

Answer 34

when you were first introduced to the concept of a function, you were given some
y=f(x)=mx+b probably.
in any case, y serves as a function of x, which is why y=f(x).
m and b are arbitrary constants.
the reason they are are arbitrary constants and *NOT* independent variables is because when you graph, you graph y on the vertical (dependent) and x on the horizontal (independent).
does this make sense?

Answer 35

yes that makes sense

Answer 36

now in calculus, you learned how to find the rate of change of one variable with respect to another.
a basic question is dy/dx for y=mx+b *assuming* that m and b are constant. Note the EXTREME importance of that assumption. if m and b were variables not held constant, we would not actually be able to find dy/dx.
in other words, dy/dx=m.
in fact, you can find dm/dx, and dy/dm, and db/dx in similar fashions, but only if you take ALL OTHER VARIABLES TO BE CONSTANT.. it should be clear that the variable in the top of the differentiation operator is the DEPENDENT variable and the variable on the bottom is the independent

Answer 37

I had this issue too, and to be honest the complete resolution of constants, variables, etc is covered in multivariate calculus which is outside of the scope of your course (but many parts are vital to understanding your material).
so let's look back at our equation

dT/dt =k(T-m)

can you identify which variable is what (dependent, independent, or constant)?

Answer 38

dT=dependent dt= independent in my equation?

Answer 39

close. T is the dependent, and t is the independent. dT and dt are not actually variables

Answer 40

oh yes that is what i meant. haha

Answer 41

what about k and m?

Answer 42

I read your thing how to top was dependent and the bottom one was independent

Answer 43

are they the contants

Answer 44

yes, they are constants. which means that ALL rules for constants in DIFFERENTIATION AND INTEGRATION apply.

Answer 45

they must be constants. what else could they be?

Answer 46

exactly. In an ordinary differential equation, there is really only one dependent variable and one independent variable, of which are easily identified with the differentition operator (note that in later courses you may experience equations like y'+y-3mx=0. then it is not so clear what variable is the independent anymore, and you may have to use context)

Answer 47

okay I am still not sure how to set up the equation to solve this question.

Answer 48

so let's work out this problem. we seek to seperate the equation so the expression of T is on the same side as the dT/dt.
dT/dt =k(T-m)
so let's just divide both sides by (T-m)
\(\Huge \frac{dT}{dt}*\frac{1}{T-m}=k\)

Answer 49

1/T-m dT = k dt???? Is that how I seperate them?

Answer 50

without knowing the solution process to this equation, you should be able to get this step: the reasoning is simple. All you are trying to do at this point is to move all the dependent variables to the left, and leave whatever is left on the right

Answer 51

yup.

Answer 52

then it is as easy as integrating both sides, and you have your answer.

Answer 53

okay now do integrate both sides?

Answer 54

k just one minute I want to make sure I do it right...

Answer 55

k for the left side I got ln(T-m)+C

Answer 56

yeah, I got \(\large ln(T-m)+C_1=kt+C_2\)

Answer 57

that can't be right let me try that again

Answer 58

why not?

Answer 59

that is what I got too but it is not one of my answers

Answer 60

no, because you aren't finished.

Answer 61

we got some simplifying and rewriting to do...

Answer 62

oh okay... I thought I was finished

Answer 63

well I am doing it right!

Answer 64

:)

Answer 65

mm. so as with all functions, you can safely assume that you want to put it in the form y=f(x)

Answer 66

so with your equation
\(\large ln(T-m)+C_1=kt+C_2\)
you want to rewrite it so your dependent variable is isolated on the left, and an expression in the independent on the r gith.

Answer 67

i don't get what to do...

Answer 68

well, what's your dependent variable?

Answer 69

T

Answer 70

you want that on the left hand side, and everything else on the right.

Answer 71

so the first thing I would do it put C_1 on the right hand side, and then I would take e to the power of both sides to cancel out the remaining logarithm...

Answer 72

is C_1 T?

Answer 73

no.

Answer 74

look, all you are doing is solving for T.

Answer 75

oh okay

Answer 76

so T is what I am trying to solve

Answer 77

yes. you aren't going to get an answer in nice numbers like you are used to, but you gotta get accustomed to dealing with variables otherwise you will have a bad time lol

Answer 78

So would I combine C_2 and C_1 on the right side and get C_3?

Answer 79

yes, I would personally rename it C because you won't have to deal with anymore constants of integration

Answer 80

T = (C - m)e^kt when you use e on both sides.

Answer 81

wait...

Answer 82

no that is not right

Answer 83

I am confused. I think I just confused myself even more. i multiply each side by e, right?

Answer 84

you are doing this in steps right?
\(\Huge \ln(T-m)+C_1=kt+C_2\)
\(\Huge \ln(T-m)=kt+C\)
call C_2-C_1 C
\(\Huge e^{\ln(T-m)}=e^{kt+C}\)
\(\Huge T-m=e^{kt+C}\)
\(\Huge T=e^{kt+C}+m\)

Answer 85

the real homework you should be getting is not calculus homework, but algebra :)
if you want to get a better understanding of algebra (but it will take you a few hours), go over the solutions of the cubic (cardano's first)
http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method
and perhaps the solution of the quartic if you are into that kind of thing. Though you probably want to get that calculus coursework done, if you find a good summer day to go over the solution of the cubic, you will gain far more algebra knowledge than what most of your teachers have taught you probably.
I will outright say that I give this suggestion to everybody I help with algebra as a friend and everyone I know ignores me. I will also outright say that it was only after going over the solution that I truly grasped algebra, and much more so than the peers I have met as of yet.

Answer 86

I will look over that. Thanks for helping me! Should I just practice on a bunch of algebra questions?

Answer 87

Would it be T = m + Ce^kt

Answer 88

well, to be honest, the thing with the algebra is not doing practice questions, as some teachers might think. You got to have an understanding for what algebra is, and what you are doing... a good way is to check for harder precalculus level problems (the cubic and quartic are a crapton of algebra and if you can get through the solutions you will definitely have a better understanding of algebra)

the thing is though, these deficiencies aren't built over a day, and are thus not fixed in a day. my advice to you right now is to finish the work you have for calculus, but definitely find some time in the summer to go over algebra: not the busywork and review questions, but look for some real challenging ones that make you think

and yes, that is the answer that I got.

Answer 89

Do you know what C is?

Answer 90

Constant

Answer 91

no, in significance to your equation

Answer 92

I don't know anything. I thought I knew things until now. haha

Answer 93

I went through the lesson that my teacher gave me and it didn't have any problems like this yet there is one in the little test that I was given.

Answer 94

your solution is T = m + Ce^(kt)
to find the meaning of that constant take t=0
T = m+C*1
C=T-m
now m is the temperature of the medium, and T is the temperature at time 0 (remember we set t or time=0), so C is really just the difference between the medium and initial temperature.

Answer 95

Do you have time to help me with one more?

Answer 96

probably not unfortunately, but most of the high level users are more than capable of helping with calculu2 problems

Answer 97

feel free to close this question, and post another

Answer 98

good luck!