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If x= a (cos 0 + 0 sin 0), y = a (sin 0 - 0 cos 0 ) then [ d² y/dx²]0=π/4= ? |
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Answer» Answer: [ d²y/dx² ] at θ = π/4 = 8√2/aπ Step-by-step EXPLANATION: X = a(cosθ + θsinθ) diff WRT θ dx/dθ = a[ - sinθ + (θcosθ +sinθ))] = a[- sinθ + +θ cosθ +sinθ] = aθcosθ y = a(sinθ - θ cosθ) diff wrt θ dy/dθ = a{cosθ - [θ (-sinθ) + cosθ]} = a{cosθ +θsinθ -cosθ} = aθsinθ ∴ dy/dx = (dy/dθ)/(dx/dθ) = aθsinθ/aθcosθ = tanθ d²y/dx² = sec²θ x dθ/dx = sec²θ /aθcosθ = 1/aθcos³θ ∴ [ d²y/dx² ] at θ = π/4 = 1/(aπ/4 xcos³π/4) = 4 x 2√2/aπ = 8√2/aπ |
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