Friday

April 18, 2014

April 18, 2014

Posted by **tan** on Tuesday, March 19, 2013 at 3:06am.

- calculus -
**drwls**, Tuesday, March 19, 2013 at 3:17amdR/dt = t/(t^2+1)

Assume R = 0 at t = 0

R = integral of t dt/(t^2 + 1)

= (1/2)ln(t^2 + 1)

At t = 3 minutes, R = (1/2) ln(10) = 1.15

Since R is measured in thousands, the answer is 1150.

- calculus -
**tan**, Tuesday, March 19, 2013 at 8:16ami dont understand this part:dt/(t^2 + 1)

= (1/2)ln(t^2 + 1)

- calculus -
**Steve**, Tuesday, March 19, 2013 at 11:18amIt's not dt/(t^2 + 1)

It is t dt/(t^2 + 1)

That is just (1/2) du/u if u=t^2+1. That's where the log comes from

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