Taking the Second Derivative w/ the Quotient Rule: What if Numerator = 0?

AI Thread Summary
The discussion focuses on calculating the second derivative using the quotient rule, specifically addressing a case where the numerator becomes zero. The original function s(t) is clarified as s(t) = (t^2 - 2)/(t + 1). The first derivative v(t) is derived as v(t) = (t^2 + 2t + 2)/(t + 1)^2, and the second derivative a(t) is found to be a(t) = -2/(t + 1)^3. It is emphasized that acceleration should be calculated from the derivative of v(t) rather than substituting a fixed t into v(t). The conversation concludes with a resolution of the initial confusion regarding the differentiation process.
mathmann
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Just wondering how you take the second derivative when using the quotient rule. After using the quotient rule to get my first derivative, I tried again and the numerator ended up as 0.
 
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Can you show us your exact working?
 
s(t) = t^2 - 2/t + 1, is the object speeing up at 4s?
v(t) = 1.04, a(t) the numerator ended up as a 0. Perhaps I made a calculating error but I went over it a couple times.
 
You mean s(t) = (t^2 - 2)/(t + 1) right?

Then v(t) = (t^2 + 2t + 2)/(t+1)^2

and a(t) = - 2/ (t+1)^3
 
When you evaluate v(t) at some fixed t you get the velocity at that point in time. You are not supposed to differentiate that particular velocity to get the acceleration at that time. You need to work out a(t) first by differentiating the function v(t) before substituting in your fixed t.
 
I understand now.. thanks for the help.
 
No problem :smile:
 

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