## Calculus 10th Edition

y = tanx $\frac{dy}{dt} = sec^2(x) \frac{dx}{dt}$ dx/dt = 3 ft/sec a) x = -$\frac{\pi}{3}$ $\frac{dy}{dt} = sec^2(-\frac{\pi}{3} )(3) = 12$ b) x = -$\frac{\pi}{4}$ $\frac{dy}{dt} = sec^2(-\frac{\pi}{4} )(3) = 6$ c) x = 0 $\frac{dy}{dt} = sec^2(0 )(3) = 3$