We have x=acosθ+aθsinθ
\(∴\frac{dx}{dθ}=-asinθ+asinθ+aθcosθ=aθcosθ\)
\(y=asinθ-aθcosθ\)
∴\(\frac{dy}{dθ}\)\(=acosθ-acosθ+aθsinθ=aθsinθ\)
∴ \(\frac{dy}{dx}\)=\(\frac{dy}{dθ}\) . \(\frac{dθ}{dx}\)=\(\frac{aθsinθ}{aθcosθ}\)=tanθ
∴ Slope of the normal at any point θ is -\(\frac{1}{tanθ}\).
The equation of the normal at a given point (x,y) is given by,
\(y-asinθ+aθcosθ=\)-\(\frac{1}{tanθ}\)\((x-acosθ-aθsinθ)\)
\(⇒ysinθ-asin^2θ+aθsinθcosθ=-xcosθ+acos^2θ+aθsinθcosθ\)
\(⇒xcosθ+ysinθ-a(sin^2θ+cos^2θ)=0\)
\(⇒xcosθ+ysinθ-a=0\)
Now, the perpendicular distance of the normal from the origin is
\(\frac{|-a|}{\sqrt{cos^2θ+sin^2θ}}\) \(=\frac{|-a|}{\sqrt1}=|-a|,\) which is independent of θ.
Hence, the perpendicular distance of the normal from the origin is constant
If \( x = a(0 - \sin \theta) \), \( y = a(1 + \cos \theta) \), find \[ \frac{dy}{dx}. \]
Find the least value of ‘a’ for which the function \( f(x) = x^2 + ax + 1 \) is increasing on the interval \( [1, 2] \).
If f (x) = 3x2+15x+5, then the approximate value of f (3.02) is