Rate of decreasing shadow problem solved with derivative and similar triangles

Changing Angle Solution
Known values given: x = 30 y = 90 $\frac{dx}{dt}$ = -24 tanθ = $\frac{x}{y}$ = $\frac{30}{90}$ = ⅓ Solve for $\frac{dθ}{dt}$ tanθ = $\frac{x}{90}$ Take derivative sec²θ $\frac{dθ}{dt}$ = $\frac{1}{90}$ $\frac{dx}{dt}$ Substitute sec²θ = 1 + tan²θ (1 + tan²θ) $\frac{dθ}{dt}$ = $\frac{1}{90}$ $\frac{dx}{dt}$ Substitute tanθ = ⅓ $\frac{dx}{dt}$ = -24 (1 + (⅓)²) $\frac{dθ}{dt}$ = $\frac{1}{90}$ * -24 (1 + $\frac{1}{9}$ ) $\frac{dθ}{dt}$ = $\frac{-24}{90}$ $\frac{10}{9}$ $\frac{dθ}{dt}$ = $\frac{-4}{15}$ $\frac{dθ}{dt}$ = $\frac{-4}{15}$ * $\frac{9}{10}$ = $\frac{-6}{25}$ Angle is changing $\frac{6}{25}$ radians (13.751 degrees) per second.
157

Changing Angle
Solution

Known values given:
x = 30 y = 90 $\frac{dx}{dt}$ = -24
tanθ = $\frac{x}{y}$ = $\frac{30}{90}$ = ⅓
Drawing of water filling a cone with r and h coordinates

Solve for $\frac{dθ}{dt}$
tanθ = $\frac{x}{90}$

Take derivative
sec²θ $\frac{dθ}{dt}$ = $\frac{1}{90}$ $\frac{dx}{dt}$

Substitute sec²θ = 1 + tan²θ
(1 + tan²θ) $\frac{dθ}{dt}$ = $\frac{1}{90}$ $\frac{dx}{dt}$

Substitute tanθ = ⅓ $\frac{dx}{dt}$ = -24
(1 + (⅓)²) $\frac{dθ}{dt}$ = $\frac{1}{90}$ * -24

(1 + $\frac{1}{9}$ ) $\frac{dθ}{dt}$ = $\frac{-24}{90}$

$\frac{10}{9}$ $\frac{dθ}{dt}$ = $\frac{-4}{15}$

$\frac{dθ}{dt}$ = $\frac{-4}{15}$ * $\frac{9}{10}$ = $\frac{-6}{25}$

Angle is changing

\frac{6}{25}

radians
(13.751 degrees) per second.

157

Changing AngleSolution

Changing Angle
Solution