Asked by Jennifer
A ladder 20 ft long rests against a vertical wall. Let \theta be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, how fast does x change with respect to \theta when \theta = pi / 3
Answers
Answered by
Reiny
x -- as defined
y = 20cosØ
x^2 + y^2 = 400
x^2 + 400cos^2 Ø = 400
2x dx/dØ - 2sinØcosØ = 0
dx/dØ = sinØcosØ/x
when x = π/3 , sinπ/3 =√3/2, cosπ/3 = 1/2
x^2 + 400(3/4) = 400
x = 10
dx/dØ = (√3/2)(1/2)/10 = √3/40
y = 20cosØ
x^2 + y^2 = 400
x^2 + 400cos^2 Ø = 400
2x dx/dØ - 2sinØcosØ = 0
dx/dØ = sinØcosØ/x
when x = π/3 , sinπ/3 =√3/2, cosπ/3 = 1/2
x^2 + 400(3/4) = 400
x = 10
dx/dØ = (√3/2)(1/2)/10 = √3/40
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