Pythagorean Theorem
L 2 = x 2 + 15 0 2 L^2=x^2+150^2 L 2 = x 2 + 15 0 2 Differentiate both sides with respect to t t t
2 L ( d L d t ) = 2 x ( d x d t ) 2L(\dfrac{dL}{dt})=2x(\dfrac{dx}{dt}) 2 L ( d t d L ) = 2 x ( d t d x ) Solve for d L d t \dfrac{dL}{dt} d t d L
d L d t = ( x L ) d x d t \dfrac{dL}{dt}=(\dfrac{x}{L})\dfrac{dx}{dt} d t d L = ( L x ) d t d x Substitute L = x 2 + 15 0 2 L=\sqrt{x^2+150^2} L = x 2 + 15 0 2
d L d t = ( x x 2 + 15 0 2 ) d x d t \dfrac{dL}{dt}=(\dfrac{x}{\sqrt{x^2+150^2}})\dfrac{dx}{dt} d t d L = ( x 2 + 15 0 2 x ) d t d x Given d x d t = 20 f t / s e c \dfrac{dx}{dt}=20 ft/sec d t d x = 20 f t / sec
If x = 250 f t x=250\ ft x = 250 f t
d L d t = ( 250 25 0 2 + 15 0 2 ) ( 20 ) \dfrac{dL}{dt}=(\dfrac{250}{\sqrt{250^2+150^2}})(20) d t d L = ( 25 0 2 + 15 0 2 250 ) ( 20 )
d L d t = 100 34 f t / s e c ≈ 17.15 f t / s \dfrac{dL}{dt}=\dfrac{100}{\sqrt{34}}\ ft/sec\approx17.15\ ft/s d t d L = 34 100 f t / sec ≈ 17.15 f t / s
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