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 (1t) 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: 3,109 Thanks: 855 
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 (1t) 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?

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

May 16th, 2017, 12:59 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 
$s = 4t+\ln(1t)$ 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 $\ln(1t)$. recheck the problem statement ... could the position function possibly be $s = 4t+\ln1t$ ? fyi ... $\dfrac{ds}{dt} = 4  \dfrac{1}{1t} = 4  (1t)^{1}$ $\dfrac{d^2s}{dt^2} = (1t)^{2} \cdot (1) = \dfrac{1}{(1t)^2}$ 
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

May 22nd, 2017, 09:15 AM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387  Quote:
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


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