SOLUTION 13: Consider the given right triangle with legs $x$ and $y$ and angle $\theta$ radians, and assume that $x, y,$ and $\theta$ are functions of time $t$.

Using trigonometry we can say that

$$\tan \theta = \displaystyle{ opposite \over adjacent } = { y \over x } \ \ \ \longrightarrow$$ $$\theta = \arctan \Big({ y \over x } \Big)$$

GIVEN: $\ \ \ \displaystyle{ dx \over dt } = -3 \ cm/min.$ (It is negative since $x$ is decreasing.) and $\displaystyle{ dy \over dt } = 4 \ cm/min.$

FIND: $\ \ \ \displaystyle{ d \theta \over dt }$ when $\ x =5 \ cm.$ and $\ y=2 \ cm.$

Recall that

$$D \{ \arctan f(x) \} = \displaystyle{ 1 \over 1 + (f(x))^2 } \cdot f'(x)$$ Now differentiate the theta equation with respect to time $t$ getting

$$D \{ \theta \} = D \{ \arctan \Big({ y \over x } \Big) \} \ \ \ \ \longrightarrow$$ $$\displaystyle{ d \theta \over dt } = \displaystyle{ 1 \over 1 + (y/x)^2 } \cdot D\Big\{ {y \over x} \Big\} \ \ \ \ \longrightarrow$$

$\Big($ Now use the quotient rule.$\Big)$

$$\displaystyle{ d \theta \over dt } = \displaystyle{ 1 \over 1 + (y/x)^2 } \cdot { x \displaystyle{ dy \over dt } - y \displaystyle{ dx \over dt } \over x^2 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ d \theta \over dt } = \displaystyle{ 1 \over 1 + { y^2 \over x^2 } } \cdot { x \displaystyle{ dy \over dt } - y \displaystyle{ dx \over dt } \over x^2 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ d \theta \over dt } = { x \displaystyle{ dy \over dt } - y \displaystyle{ dx \over dt } \over x^2 +y^2 } \ \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = -3, \displaystyle{ dy \over dt } = 4, x=5,$ and $y=2. \Big)$

$$\displaystyle{ d \theta \over dt } = { (5)(4) - (2) (-3) \over 5^2 + 2^2 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ d \theta \over dt } = { 26 \over 29 } \ radians/min.$$