dv/dt = 3/[t((t^2)-1)] t>0, v(2) = 0

dv = integral 3/[t((t^2)-1)] dt

v = 3sect^-1 + C

3sec(-2)^-1 = -C

How do I evaluate inverse secant?! Is there any simple rule to go by just looking at the unit circle? My teacher says the answer is c = -pi.

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- Apr 23rd 2009, 06:22 AMtom atoEvaluating inverse trig function
dv/dt = 3/[t((t^2)-1)] t>0, v(2) = 0

dv = integral 3/[t((t^2)-1)] dt

v = 3sect^-1 + C

3sec(-2)^-1 = -C

How do I evaluate inverse secant?! Is there any simple rule to go by just looking at the unit circle? My teacher says the answer is c = -pi. - Apr 23rd 2009, 07:36 AMSoroban
Hello, tom ato!

My answer is: .$\displaystyle -{\color{red}2}\pi $

Quote:

$\displaystyle 3\sec^{-1}(\text{-}2) \:=\: -C$

How do I evaluate inverse secant?

We are expected to reason like this . . .

Let $\displaystyle \theta \:=\:\sec^{-1}(\text{-}2)$

. . Then $\displaystyle \theta$ is the angle whose secant is -2.

Hence, $\displaystyle \theta$ is the angle whose*cosine*is $\displaystyle \text{-}\tfrac{1}{2}$

The "first" such angle is: .$\displaystyle \theta \:=\:\frac{2\pi}{3}$

Therefore: .$\displaystyle \sec^{-1}(\text{-}2) \;=\;\frac{2\pi}{3}$

- Apr 23rd 2009, 09:07 AMtom ato
Thanks. And I just realized that the answer should be negative pi because I'm supposed to take the inverse secant of 2 (I had put -2)