My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree1Thanks
  • 1 Post By skeeter
Reply
 
LinkBack Thread Tools Display Modes
May 16th, 2017, 08:45 AM   #1
Newbie
 
Joined: Jan 2017
From: brussels

Posts: 21
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
aaronmooy47 is offline  
 
May 16th, 2017, 09:04 AM   #2
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,731
Thanks: 707

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?
Country Boy is offline  
May 16th, 2017, 12:17 PM   #3
Newbie
 
Joined: Jan 2017
From: brussels

Posts: 21
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
aaronmooy47 is offline  
May 16th, 2017, 12:59 PM   #4
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,640
Thanks: 1319

$s = 4t+\ln(1-t)$

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 Country Boy
skeeter is offline  
May 22nd, 2017, 08:52 AM   #5
Newbie
 
Joined: Jan 2017
From: brussels

Posts: 21
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
aaronmooy47 is offline  
May 22nd, 2017, 09:15 AM   #6
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,640
Thanks: 1319

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
skeeter is offline  
May 22nd, 2017, 01:10 PM   #7
Newbie
 
Joined: Jan 2017
From: brussels

Posts: 21
Thanks: 0

Thank you that link was very helpful
aaronmooy47 is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
differentiation



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Difference between vector differentiation and parametric differentiation Mr Davis 97 Calculus 1 March 9th, 2015 04:50 AM
differentiation 3 Raptor Calculus 2 May 8th, 2014 06:12 AM
Differentiation Thepiman Calculus 1 May 8th, 2014 06:11 AM
Differentiation 2 Raptor Calculus 6 May 7th, 2014 11:33 AM
Differentiation Raptor Calculus 1 May 7th, 2014 07:47 AM





Copyright © 2017 My Math Forum. All rights reserved.