Need help solving: Velocity v(t) = -e^{kt}(k cos(2pi f t) + 2pi f sin(2pi f t)), k = 0.5s^{-1}, f = 2Hz.

[DCE123]

Hi all,

Hope you can help,

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Calculate the value of ? after the start at which the velocity of the tip first becomes zero

The expression for Velocity is:

?(?) = -?^{??}(????(????) + ??????(????))

Where ? = ?. ??^{-?} and ? = ???

Firstly, let’s make the equation 0 and solve for t
? = -?^{-?.??}(?. ? ∗ ???(????) + ? ∗ ? ∗ ? ∗ ???(? ∗ ? ∗ ?? ∗))

Bring ???(????)onto the LHS of the equation

-???(????) = -?^{-?.??}(?. ? ∗ ? ∗ ? ∗ ? ∗ ???(? ∗ ? ∗ ? ∗ ?))

Using Trig Identity:
???? / ???? = ????

Divide both sides by???(????). Here is am trying to remove one of the trig values so that were only
dealing with one trig value

-[???(????)]/[???(????)] = -?^{-?.??}(?. ? + ??? ∗ [???(????)/???(????)])

Do the inverse of tan to remove it from RHS of equation

-? = -?^{-?.??}(?. ? ∗ ??? ∗ ???(????))

I get stuck here. I tried to go down the route of taking the In of -?^{-?.??} so that later on, I can write it as -?. ?? ?? - ?. ( I did this for the next question) With the question being complex, I’m not sure how to tackle the exponential, especially with ???^{?} -? in there. I might be going off on a tangent with this question to be honest.

???^{-?} -? = -?^{-?.??}(?. ? ∗ ??? ∗ ????)

Hopefully the above make some sense. Am I going in the right direction with this question? It’s only worth 4 marks and I feel I am doing too much work to justify 4 marks. I might be missing something simple?

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I might be missing something simple as the question is only work 4 marks and i feel this alot of work for 4 marks.

Thanks

J

 

[Dr.Peterson]

One error is that the exponential multiplies the entire parenthetical expression, so you can't move the cosine out of it as you did. Instead, recognize that the exponential can't be zero, so you can ignore it. In fact, that whole line where you "bring the cosine onto the LHS" is utter nonsense for several reasons. Do you see why?

When you get stuck, go back and check each line of your work to see if you have over-complicated things.

 

[HallsofIvy]

I have always disliked the phrase "move ____ to the other side of the equation". There are many ways to get a value from one side of the equation to the other: add it to both sides of the equation, subtract it from both sides of the equation, multiply both sides of the equation by it, divide both sides of the equation by it, depending on what it is doing on the first side of the equation. The letters in an equation represent numbers and you have to do some arithmetic operation too them. "Moving" is not an arithmetic operation!

 

[DCE123]

Thanks for the responses

Dr Peterson - What you say makes sense, looking now of course I cant move the cosine to the LHS.

I'm don't understand why we can simply ignore the exponential though?

Sorry for the lack of understanding. I've had some time away from studies and have pick calculus back up. I am taking the time to go back through my algebra and trig notes, but there is a lot to recap on.

Thanks

J

 

[HallsofIvy]

Your original equation is \(\displaystyle -e^{-kt}(k cos(2\pi ft)+ 2\pi f sin(2\pi ft))= 0\)
There are two crucial points you might be missing. First, the exponential is multiplying everything in the parentheses, both the "cosine" and the sine. You simply "moved the cosine term over leaving everything else, even the number multiplying "cosine" behind. You can't do that! As I said before, there has to be an arithmetic operation that results in the "move". Which arithmetic operation did you use here?

The other point is that if AB= 0 then either A= 0 or B= 0 (or both). If A is not 0, we can divide both sides by it to get B= 0. If B is not 0, we can divide both sides by it to get A= 0. An exponential, like \(\displaystyle -e^{-kt}\) is never 0 so we can divide both sides by it.

 

[DCE123]

Thanks for your response

the 'move' was me adding 'cos(2*pi*2*t)' to both sides, so the LHS side of the equation became '-cos(2*pi*2*t). I understand now I can't do this, I should of spotted that to be honest. I'll bear in mind to use the correct terminology in future.

I understand that we have we have a value for 'k' and clearly some time must have passed for the velocity of the tip to become 0, so I can see that -e^(-kt) can never be 0.

So if am to divide both sides by -e^(-kt), then I can simply cancel out the -e^(-kt) out as of course anything divided by 0 = 0. This would then leave me to solve the rest of the equation in similar fashion to how I did in the originally attached document.

I'll give this a go tonight.

Thanks a lot for your help.

 

[Dr.Peterson]

I understand that we have we have a value for 'k' and clearly some time must have passed for the velocity of the tip to become 0, so I can see that -e^(-kt) can never be 0.

Just to clarify, the reason -e^(-kt) can never be 0 is not based on the physical problem you are solving (which you never fully stated for us anyway), but on the way the exponential function works. No power of a non-zero base can be zero; the range of f(x) = e^x is the positive real numbers only. This fact will be worth knowing for many future problems.

And, yes, when I informally stated that you can "ignore" the exponential, I was referring to this fact that you can divide both sides by it because it is non-zero; or rather, in the factored form, you can set aside this factor that can't be zero, knowing that therefore the other factor must be zero.

 

[DCE123]

Cheers for getting my back to me. I'm feeling confident i can finish this question now.

I will be revising exponentials as well, i need a better understanding of them to be honest.

Thanks

 

[DCE123]

Hi all, i'm afraid i have marginally progressed with it but am faced with another obstacle.

Please see attached.

I understand that in some equations you sometimes use trig identities to convert your trig variables so that you either have COSINE or SINE for example. This makes them easier to deal with. In the attached document, what I've done in red i understand is wrong, but my reasoning is that if i could convert SINE to COSINE or vice versa, then the terms can be collected and from there you can find t ( the document explains further)

No doubt i'm coming across a bit thick but its one of the sticky questions i cant grasp. Thank-you for your patience.

 

[Dr.Peterson]

You're heading in the right direction, but your algebra is very clumsy. (Often students don't really master the details of algebra until they are forced to, by having to do a lot of it in a higher course like trig or calc, so I'm not very surprised!)

You claim to have subtracted [MATH]\cos(4\pi t)[/MATH] from both sides; but that isn't what you did!

What you need to have done is to subtract [MATH]\frac{1}{2}\cos(4\pi t)[/MATH] from both sides. That is the TERM you want to get rid of. You can't just subtract part of a term and leave the rest (the 1/2) behind as a factor in the other term.

The way I explain this aspect of basic algebra is that the last operation being done on the right-hand side is the addition of the two terms; you want to undo that operation, which means subtracting one entire term from both sides.

Once you get this right, you will be able to write the equation in the form tan(...) = ..., and you will be almost finished. Just be sure to do the work in very small steps, and check each step to make sure you did what you meant to do.

 

[DCE123]

To be honest, I have interpreted it as two terms. I treated 1/2*cos(4/pi*t) as you would lets says A*B(4*t).

appreciate your help anyway, at least I know I'm heading in the right direction as you say.

I do agree with your comment, my algebra isn't up to scratch, but what's frustrating is that the first assignments were algebra & trig, which I managed fine. It seems the progression and difficulty from them to calculus is a rather steep curve.

Thank-you again

 

[Dr.Peterson]

Are you aware of the definition of "term"? A term of an expression is specifically one of the things that are ADDED. Your A and B are not called terms, but factors. And the reason for this definition is that terms are what have to be dealt with first in solving an equation.

Furthermore, if I had an expression A*B+C and subtracted B from it, I would not get A+C or AC, but literally A*B + C - B, which simplifies only to (A-1)*B + C. In carrying out steps of algebra, it's important to actually do what you claim to be doing; that is a common error of beginning algebra students.