## University Calculus: Early Transcendentals (3rd Edition)

a) $\frac{dx}{dt} = -2.5\frac{ft}{sec}$ b) $\frac{d\theta}{dt}=\frac{-3}{20}$rad/sec
a) Given that $s=10, x=8, \frac{ds}{dt}=-2,\frac{dy}{dt}=0,$ and $\tan\theta=\frac{x}{6}$ By the Pythagorean theorem: $s^2=x^2+y^2$ Differentiating both sides: $s\frac{ds}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$ $8\frac{dx}{dt}+0=10(-2)$ $\frac{dx}{dt} = -2.5\frac{ft}{sec}$ b) $cos\theta=\frac{6}{r}$ Differentiating both sides: $-sin\theta\frac{d\theta}{dt}=\frac{-6}{r^2}\frac{dr}{dt}$ $\frac{d\theta}{dt}=\frac{6}{r^2\sin\theta}\frac{dr}{dt}=\frac{6*-2}{10^2\frac{8}{10}}=\frac{-3}{20}$rad/sec