Question

Find the radius of curvature of the curve x = a log (sec t + tan t), y = a sect​

Answer 1

Answer:

             The Radius of the Curvature is asec^2t

Step-by-step explanation:

As Given:

x=a log (sec t +tan t)

solving,

$\frac{dx}{dt}=\frac{x}{y}*sect \ tant +sec^2t\\\frac{dx}{dt}=\frac{asect(sect+tant)}{sect+tant}\\\frac{dx}{dt}=a \ sect$

As the component is given , curvature is along x-axis hence finding the y-component.

x=a sec t ,also y=a sec  giving dy/dt=a sect tant

$y_{1}=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\  =\frac{a \ sect \ tant}{a \ sect}\\y_{1}=tant$

differntiating with respect to x,

$y_{2}=sec^2t(\frac{dt}{dx})\\putting \ values \\y_{2}=sec^2t(\frac{sect}{a})$

Calculating Radius of Curvature by Cartesian Formula:

$p=\frac{(1+y^2_1)^{\frac{-3}{2}}}{y_2}\\=\frac{(1+tan^2t)^{\frac{-3}{2}}a}{sect}\\=\frac{asec^3t}{sec t}\\p=asec^2t$

Hope you find it easy :)

Answer 2

Answer:

Step-by-step explanation: