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Instantaneous velocity and acceleration

  • Thread starter sb_4000
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  • #1
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Hi,
This is probably an easy question for some of you, but I do not Undrestand it..

x=At^4 + (Bt + C)t^2 + D * sin(Et) where A, B, C, D, and E are constants. and I have to find instantaneous acceleration and velocity.

I know Instantaneous velocity is
v_x = dx/dt

Instantaneous acceleration dv_x/dt

But I do not know how to solve this equation..
 
Last edited:

Answers and Replies

  • #2
arildno
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You're absolutely correct in which formulas you are to use!
Now, let's find the velocity first:
You've been given x(t), could you state precisely why you find differentiating x(t) looks hard? (Refer to specific terms you find troublesome)
 
  • #3
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do I just fond the derivative,
for example

4At^3+3Bt^2+2Ct+Dcos(Et), something like this, cause Im not sure how it should look.
 
  • #4
arildno
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A bit too fast, there!
Let's take it slower:
1) You know that if h(t)=f(t)+g(t), then we have:
[tex]\frac{dh}{dt}=\frac{df}{dt}+\frac{dg}{dt}[/tex]
that is, the derivative of a sum is the sum of the derivatives, right?
Let's use that to split up our task:
[tex]x(t)=At^{4}+Bt^{3}+Ct^{2}+D\sin(Et)[/tex]
Hence, we have:
[tex]\frac{dx}{dt}=\frac{dF}{dt}+\frac{dG}{dt}+\frac{dH}{dt}+\frac{dI}{dt}[/tex]
where I've introduced:
[tex]F(t)=At^{4},G(t)=Bt^{3},H(t)=Ct^{2},I(t)=D\sin(Et)[/tex]
Now, as you can see, you have computed correctly the following terms:
[tex]\frac{dF}{dt},\frac{dG}{dt},\frac{dH}{dt}[/tex]
But what about [tex]\frac{dI}{dt}[/tex]?
 
  • #5
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would it be sin(Et) only..
 
  • #6
arildno
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Hint:
Have you learnt the CHAIN RULE yet?
 
  • #7
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yeah I did in Cal1, but havent touched it in a while
 
  • #8
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Sin(Et)+Dcos(Et)? something like this, Ill go back and read my Calculus book.
 
  • #9
arildno
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OK, that's the rule you need to use, in order to differentiate Dsin(Et) correctly.
So, how would you proceed to do that?
 
  • #10
arildno
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Your posts 8+5 have wrong suggestions.
 
  • #11
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if I replace D, E with X, can I do that?

then I would get

xsin(xt)=sin(Xt)cos(Xt)?
 
  • #12
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would the instantaneous velocity and acceleration be different in this question.
 
  • #13
arildno
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NO!
It seems to me you find the chain rule kind of hard to understand;
that's fairly common, though. It IS rather difficult.
Let's suppose you've got two functions, f(t) and g(t).
You can now make yourself a new function, h(t), by substituting g(t) into all instances of "t" in f.
That is, we write:
h(t)=f(g(t))
Now, the chain rule says:
h'(t)=f'(g(t))g'(t).
Let's be absolutely clear about what this MEANS, in terms of how to compute the derivative of h(t) (that is, h'(t)) correctly:
1. Differentiate f(t) as usual, that is find [tex]\frac{df}{dt}[/tex]
2. Substitute into all instances of "t" in your expression for [tex]\frac{df}{dt}[/tex] g(t)
3. The expression you've now found, is f'(g(t)), that is, one of the factors in h'(t).
4. Differentiate g(t) as usual, that is, find [tex]\frac{dg}{dt}[/tex] (or, which is the same, g'(t))
5. Finally, multiply together f'(g(t)) (found in 2&3) and g'(t).
You have now found h'(t)=f'(g(t))g'(t)

To your problem:
In order to use the chain rule, you must first find good choices for f(t) and g(t)!
We have: [tex]I(t)=D\sin(Et)[/tex]
A good choice of f(t) is:
[tex]f(t)=D\sin(t)[/tex]
(The reason why this is a GOOD choice, is that we know how to differentiate sine!)
We therefore set:
g(t)=Et
We do this, because we now see that we may write:
I(t)=f(g(t))
Agreed?
Hence, let's calculate [tex]\frac{dI}{dt}[/tex] according to the scheme above!
1. [tex]\frac{df}{dt}=D\cos(t)[/tex]
Agreed?
2&3. [tex]f'(g(t))=D\cos(Et)[/tex]
4. [tex]\frac{dg}{dt}=E[/tex]
5. [tex]f'(g(t))g'(t)=DE\cos(Et)[/tex]
Hence, we have:
[tex]\frac{dI}{dt}=DE\cos(Et)[/tex]
 
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  • #14
arildno
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OOPS!
Dreadfully sorry, wrote "cosine" at some places where I meant "sine".
This is now edited.
 
  • #15
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If is the answer for the instantaneous velocity, what is the difference between that and instantaneous acceleration.
 
  • #16
arildno
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In order to find the acceleration, you must differentiate your new-found expression for the velocity.
 
  • #17
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This is what I get,

12At^2 + 6Bt + 2C - DEsin(Et)

is this correct?
 
  • #18
arildno
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Again, you're right about the 3 first terms; you haven't done the last term correctly, though (it's that damn chain rule again..:wink:)
 
  • #19
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-DE^2sin(Et)?
 
  • #20
arildno
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Precisely!
 
  • #21
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Thanks alot, I appreciate the help.
 
  • #22
arildno
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You're welcome!
I'd recommend you "slavishly" follow the scheme in order to use the chain rule correctly.
It IS rather difficult, but you need to get the rule "into your fingers", so to speak.
 

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