Find the slope of the normal to the curvex= 1 −asinθ,y=bcos2θat.

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Vinay Kumar Aggarwal

x = 1 - a sinθdifferentiate x with respect to θ,dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)y = bcos²θdifferentiate y with respect to θ,dy/dθ = b. d(cos²θ)/dθ= b. 2cosθ. (-sinθ)= -2bsinθ.cosθ --------(2) dividing...
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x = 1 - a sinθdifferentiate x with respect to θ,dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)y = bcos²θdifferentiate y with respect to θ,dy/dθ = b. d(cos²θ)/dθ= b. 2cosθ. (-sinθ)= -2bsinθ.cosθ --------(2) dividing equations (2) by (1), dividing equations (2) by (1),dy/dx = -2bsinθ.cosθ/-acosθ = 2b sinθ/aat θ = π/2 , dy/dx = 2bsinπ/2/a = 2b/aso, slope of normal = -1/slope of tangent= -1/(2b/a) = -a/2b read less
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2b/a
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2b/a
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Slope of normal and slope of tangent are perpendicular to each other.e.g., slope of normal x = 1 - a sinθdifferentiate x with respect to θ,dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)y = bcos²θdifferentiate y with respect to θ,dy/dθ = b. d(cos²θ)/dθ=...
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Slope of normal and slope of tangent are perpendicular to each other.e.g., slope of normal x = 1 - a sinθdifferentiate x with respect to θ,dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)y = bcos²θdifferentiate y with respect to θ,dy/dθ = b. d(cos²θ)/dθ= b. 2cosθ. (-sinθ)= -2bsinθ.cosθ --------(2) dividing equations (2) by (1), dividing equations (2) by (1),dy/dx = -2bsinθ.cosθ/-acosθ = 2b/a sinθat θ = π/2 , dy/dx = 2b/a sinπ/2 = 2b/aso, slope of normal = -1/slope of tangent= -1/(2b/a) = -a/2b read less
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It is given that x = 1 − a sin θ and y = b cos2θ. Therefore, the slope of the tangent at is given by, Hence, the slope of the normal at
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It is given that x = 1 − a sin θ and y = b cos2θ. Therefore, the slope of the tangent at is given by, Hence, the slope of the normal at read less
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