Let L = x + y L=x+y L = x + y
cos θ = 2 y , sin θ = h x \cos \theta=\dfrac{2}{y}, \sin \theta=\dfrac{h}{x} cos θ = y 2 , sin θ = x h Then
L = x + y = h sin θ + 2 cos θ L=x+y=\dfrac{h}{\sin\theta}+\dfrac{2}{\cos \theta} L = x + y = sin θ h + cos θ 2
L = h sin θ + 2 cos θ L=\dfrac{h}{\sin\theta}+\dfrac{2}{\cos \theta} L = sin θ h + cos θ 2 Find the first derivative with respect to θ \theta θ
d L d θ = − h cos θ sin 2 θ + 2 sin θ cos 2 θ \dfrac{dL}{d\theta}=-\dfrac{h\cos \theta}{\sin^2\theta}+\dfrac{2\sin \theta}{\cos^2 \theta} d θ d L = − sin 2 θ h cos θ + cos 2 θ 2 sin θ Find the critical number(s)
d L d θ = 0 = > − h cos θ sin 2 θ + 2 sin θ cos 2 θ = 0 \dfrac{dL}{d\theta}=0=>-\dfrac{h\cos \theta}{\sin^2\theta}+\dfrac{2\sin \theta}{\cos^2 \theta}=0 d θ d L = 0 => − sin 2 θ h cos θ + cos 2 θ 2 sin θ = 0
h cos 3 θ = 2 sin 3 θ h\cos^3\theta=2\sin^3\theta h cos 3 θ = 2 sin 3 θ
h = 2 tan 3 θ h=2\tan^3\theta h = 2 tan 3 θ Given that the shortest ladder is 6 m.
Substitute
h sin θ + 2 cos θ = 6 \dfrac{h}{\sin\theta}+\dfrac{2}{\cos \theta}=6 sin θ h + cos θ 2 = 6
2 tan 3 θ cos θ + 2 sin θ = 6 sin θ cos θ 2\tan^3\theta\cos \theta+2\sin \theta=6\sin \theta\cos \theta 2 tan 3 θ cos θ + 2 sin θ = 6 sin θ cos θ 0 ° < θ < 90 ° 0\degree<\theta< 90\degree 0° < θ < 90°
sin 3 θ + sin θ cos 2 θ = 3 sin θ cos 3 θ \sin^3\theta+\sin \theta\cos^2\theta=3\sin \theta\cos^3\theta sin 3 θ + sin θ cos 2 θ = 3 sin θ cos 3 θ
1 = 3 cos 3 θ 1=3\cos^3\theta 1 = 3 cos 3 θ
cos 3 θ = 1 3 \cos^3\theta=\dfrac{1}{3} cos 3 θ = 3 1
cos θ = 1 3 3 \cos \theta=\dfrac{1}{\sqrt[3]{3}} cos θ = 3 3 1
sin θ = 1 − 1 3 2 3 = 9 3 − 1 3 3 \sin \theta=\sqrt{1-\dfrac{1}{\sqrt[3]{3^2}}}=\dfrac{\sqrt{\sqrt[3]{9}-1}}{\sqrt[3]{3}} sin θ = 1 − 3 3 2 1 = 3 3 3 9 − 1
h = 2 ( 9 3 − 1 ) 3 h=2\bigg(\sqrt{\sqrt[3]{9}-1}\bigg)^3 h = 2 ( 3 9 − 1 ) 3
h ≈ 2.245 m h\approx2.245\ m h ≈ 2.245 m
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