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May 16th, 2017, 08:45 AM   #1
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Differentiation

At any time t seconds the distance (s) metres of a particle moving in a straight line is given by s = 4t + ln (1-t)

Express an equation for the first differential then rewrite as a function of a function, now determine an equation for acceleration d^2s / dt^2 and its value after 1.5 seconds
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May 16th, 2017, 09:04 AM   #2
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If you are asked to do a problem like this, asking you to find the first and second derivatives, you must be taking a Calculus class. Surely, you have learned how to differentiate, haven't you? The derivative of 4t + ln (1-t) is the derivative of 4t plus the derivative of ln(1- t). I would expect you to know the derivative of 4t. To find the derivative of ln(1- t), use the chain rule. Let u= 1- t so that ln(1- t)= ln(u). Then d(ln(1- x))/dx= (d(ln(u))/du)(du/dx). Do you know the derivative of ln(u) with respect to u? Do you know the derivative of u= 1- x with respect to x?
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May 16th, 2017, 12:17 PM   #3
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Yes I am, but I missed a large part of the calculus unit due to illness, so now I'm looking for a bit of guidance. I think the part I struggle with most is understanding the equations and what it's asking for. Thanks for your help.

Last edited by skipjack; February 8th, 2019 at 03:15 AM.
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May 16th, 2017, 12:59 PM   #4
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$s = 4t+\ln(1-t)$

note the domain of this function is $t < 1$, so finding the value of the derivative at $t = 1.5$ seconds is not possible for real values of $\ln(1-t)$.

recheck the problem statement ... could the position function possibly be $s = 4t+\ln|1-t|$ ?

fyi ...

$\dfrac{ds}{dt} = 4 - \dfrac{1}{1-t} = 4 - (1-t)^{-1}$

$\dfrac{d^2s}{dt^2} = (1-t)^{-2} \cdot (-1) = -\dfrac{1}{(1-t)^2}$
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May 22nd, 2017, 08:52 AM   #5
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Can you show me the first differential in sequence order rather than all in one? I have only done basic differentiation and don't really understand the chain rule.

Last edited by skipjack; February 8th, 2019 at 03:38 AM.
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May 22nd, 2017, 09:15 AM   #6
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Quote:
I have only done basic differentiation and don't really understand the chain rule
then you should learn the chain rule before continuing ...

Chain Rule Introduction
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May 22nd, 2017, 01:10 PM   #7
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Thank you - that link was very helpful.

Last edited by skipjack; February 8th, 2019 at 03:20 AM.
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February 7th, 2019, 08:25 PM   #8
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I think you should look at the basics first

Hello

The differentiation done above is absolutely correct.

I think if you're having trouble doing differentiation, you should first remember all of its formulas.

Keep the formulas handy, and memorized...then everything will turn easier.

Here is a good list of Differentiation Formulas

Last edited by skipjack; February 8th, 2019 at 03:22 AM.
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February 8th, 2019, 03:35 AM   #9
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Quote:
Originally Posted by skeeter View Post
... could the position function possibly be $s = 4t+\ln|1-t|$ ?
That wouldn't help. The question still wouldn't make sense.
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