Rate of decreasing shadow problem solved with derivative and similar triangles

Lighthouse Beacon Solution
Known values given: y = 1 z = 2 $\frac{dθ}{dt}$ = 10π radians x = $\sqrt{z² - y²}$ = $\sqrt{4 - 1}$ = $\sqrt{3}$ Solve for $\frac{dθ}{dt}$ : tan θ = $\frac{x}{y}$ = $\frac{x}{1}$ = x x = tan θ Take the derivative $\frac{dx}{dt}$ = sec²θ $\frac{dθ}{dt}$ Substitute: $\frac{dθ}{dt}$ = 10π and sec²θ = 1 + tan²θ $\frac{dx}{dt}$ = (1 + tan²θ) 10π radians Substitute: tan θ = x = $\sqrt{3}$ $\frac{dx}{dt}$ = ( 1 + ( $\sqrt{3}$ )²) 10π radians $\frac{dx}{dt}$ = (1 + 3) 10π radians $\frac{dx}{dt}$ = 40π radians Speed of the image across the shore is 40π miles per minute.
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Lighthouse Beacon
Solution

Known values given:
y = 1 z = 2 $\frac{dθ}{dt}$ = 10π radians
x = $\sqrt{z² - y²}$ = $\sqrt{4 - 1}$ = $\sqrt{3}$
Drawing lighthoue by shore with x, y and z coordinates

Solve for $\frac{dθ}{dt}$ :
tan θ = $\frac{x}{y}$ = $\frac{x}{1}$ = x
x = tan θ

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

Substitute: $\frac{dθ}{dt}$ = 10π and
sec²θ = 1 + tan²θ
$\frac{dx}{dt}$ = (1 + tan²θ) 10π radians

Substitute: tan θ = x = $\sqrt{3}$
$\frac{dx}{dt}$ = ( 1 + ( $\sqrt{3}$ )²) 10π radians
$\frac{dx}{dt}$ = (1 + 3) 10π radians
$\frac{dx}{dt}$ = 40π radians

Speed of the image across the
shore is 40π miles per minute.

156

Lighthouse BeaconSolution

Lighthouse Beacon
Solution