Question

Radius of curvature x=a(cost+t sint) , y=a(sint-tcost)

Answer 1

Solution:

Given,

    x = a (cost + t sint)

    y = a (sint - t cost)

Then, x' (derivative of x w.r. to t)

    = a (- sint + sint + t cost)

    = at cost

and y' (derivative of y w.r. to t)

    = a (cost - cost + t sint)

    = at sint

Also, x" (derivative of x' w.r. to t)

    = a (cost - t sint)

and y" (derivative of y' w.r. to t)

    = a (sint + t cost)

Hence, radius of curvature (ρ)

    = $\mathsf{\frac{(x'^{2}+y'^{2})^{3/2}}{x'y''-y'x''}}$

    = $\mathsf{\frac{\{a^{2}t^{2}(sin^{2}t+cos^{2}t)\}^{3/2}}{a^{2}(t\:sint\:cost+t^{2}cos^{2}t-t\:sint\:cost+t^{2}sin^{2}t)}}$

    = $\mathsf{\frac{a^{3}t^{3}}{a^{2}t^{2}(cos^{2}t+sin^{2}t)}}$

    = at, at the point 't'.  

Answer 2

Step-by-step explanation:

first u have to differentiate the x and y equations with respect to t and then divide dy/dt and dx/dt to get y1 and differentiate y1 with respect to x to get y2. and apply y1 and y2 to the formula.