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.
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.
Hello, tom ato!
My answer is: .$\displaystyle -{\color{red}2}\pi $
$\displaystyle 3\sec^{-1}(\text{-}2) \:=\: -C$
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: .$\displaystyle \sec^{-1}(\text{-}2) \;=\;\frac{2\pi}{3}$