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- Thread starter whiteti
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D
Deleted member 4993
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The position of a particle at time t seconds is s(t)=e^(-t^3) meters. What is its acceleration at 1 seconds?
How do I even set this up into a solvable form?
What is the definition of acceleration and velocity?
D
Deleted member 4993
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Velocity is change in distance/ change in time
Acceleration is change in velocity over change in time?
Correct..
Now using those definitions - express velocity and acceleration as derivatives of 's'.
D
Deleted member 4993
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I'm not sure how to do that
You know that \(\displaystyle \frac{df(x)}{dx}\) expresses
rate of change of f(x) with respect to change in 'x'.
Knowing that,
How would you express
rate of change over time of any function(of time)
as derivative of that function (with respect to time)?
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HallsofIvy
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Well, then where did you get this problem. Whether in a Calculus course or "Calculus based Physics course", you should have already learned that "velocity is the derivative of the distance function" and "acceleration is the derivative of the velocity function". Do you know how to find the derivative of \(\displaystyle e^{-t^3}\)? Do you know the derivative of \(\displaystyle e^x\)? Do you know the derivative of \(\displaystyle x^3\)? Do you know the chain rule?
D
Deleted member 4993
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so then to find velocity its v=e^-3/1
then I plug that into acceleration formula and get my answer?
Can you please show your work - leading to that result?
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Well, then where did you get this problem. Whether in a Calculus course or "Calculus based Physics course", you should have already learned that "velocity is the derivative of the distance function" and "acceleration is the derivative of the velocity function". Do you know how to find the derivative of \(\displaystyle e^{-t^3}\)? Do you know the derivative of \(\displaystyle e^x\)? Do you know the derivative of \(\displaystyle x^3\)? Do you know the chain rule?
It is calculuc 1 in university, so no physics. I do not know how to find the derivative of e^-t^3.
Can please show your work - leading to that result?
There is no work, and I forgot the t.
HallsofIvy
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Please answer my other questions! Do you know how to find the derivative of \(\displaystyle e^x\)? Do you know how to find the derivative of \(\displaystyle x^3\)? Do you know the chain rule?
e^x is e^x, x^3 is 3x^2Please answer my other questions! Do you know how to find the derivative of \(\displaystyle e^x\)? Do you know how to find the derivative of \(\displaystyle x^3\)? Do you know the chain rule?
The chain rule is used for compositions..
HallsofIvy
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Excellent. so let \(\displaystyle u(x)= x^3\) so your function is \(\displaystyle f(u)= e^u\) and use the chain rule.
Excellent. so let \(\displaystyle u(x)= x^3\) so your function is \(\displaystyle f(u)= e^u\) and use the chain rule.
But its a power on a power, do I bring them down or something?
sorry, I'm not good at calculus at all
HallsofIvy
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Do you not know the "chain rule"? \(\displaystyle \frac{df}{dx}= \frac{df}{du}\frac{du}{dx}\)
Here, as I said, if you take \(\displaystyle u(x)= x^3\) then \(\displaystyle f(u)= e^u\).
So \(\displaystyle \frac{df}{du}= e^u\) and \(\displaystyle \frac{du}{dx}= 3x^2\)
By the chain rule, \(\displaystyle \frac{df}{dx}= e^u(3x^2)= 3x^2e^{x^3}\)
Here, as I said, if you take \(\displaystyle u(x)= x^3\) then \(\displaystyle f(u)= e^u\).
So \(\displaystyle \frac{df}{du}= e^u\) and \(\displaystyle \frac{du}{dx}= 3x^2\)
By the chain rule, \(\displaystyle \frac{df}{dx}= e^u(3x^2)= 3x^2e^{x^3}\)