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 May 16th, 2017, 08:45 AM #1 Newbie   Joined: Jan 2017 From: brussels Posts: 28 Thanks: 0 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
 May 16th, 2017, 09:04 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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? Thanks from topsquark
 May 16th, 2017, 12:17 PM #3 Newbie   Joined: Jan 2017 From: brussels Posts: 28 Thanks: 0 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.
 May 16th, 2017, 12:59 PM #4 Math Team     Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521 $s = 4t+\ln(1-t)$ 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 ... $\dfrac{ds}{dt} = 4 - \dfrac{1}{1-t} = 4 - (1-t)^{-1}$ $\dfrac{d^2s}{dt^2} = (1-t)^{-2} \cdot (-1) = -\dfrac{1}{(1-t)^2}$ Thanks from topsquark and Country Boy
 May 22nd, 2017, 08:52 AM #5 Newbie   Joined: Jan 2017 From: brussels Posts: 28 Thanks: 0 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.
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

 May 22nd, 2017, 01:10 PM #7 Newbie   Joined: Jan 2017 From: brussels Posts: 28 Thanks: 0 Thank you - that link was very helpful. Last edited by skipjack; February 8th, 2019 at 03:20 AM.
 February 7th, 2019, 08:25 PM #8 Newbie   Joined: Oct 2017 From: Texas Posts: 9 Thanks: 0 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.
February 8th, 2019, 03:35 AM   #9
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Quote:
 Originally Posted by skeeter ... 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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