Finding the equation of a circle with a curvature equal to the curvature of a given curve at a point

Question

"Find an equation for the circle of curvature of the curve r(t)=8t i + sin(5t) j at the point (4pi,1). (the curve parameterizes the graph of y=sin(5/8X) in the xy plane)

Ive already found v(t)=8 i+5cos(5t) j and a(t)= -25sin(5t) j

from there I found the curvature using mag(v cross a) / mag(v(t)) but when I use 4pi as x=t i get 0 for the curvature?

im very confused about this one (how does the parametrization aspect play into this?) and the only worked examples are behind chegg paywalls.

Thank you for any help you can give.

Answer 1

When you say the point (4π,1), those appear to be (x,y) coordinates. From the x expression x=8t or t=x/8.

1) Therefore the t value that corresponds to x = 4π is t = π/2

2) The equation for curvature is |v x a|/ |v • v|^3/2

curvature is |(8,5cos(5t)) x (0,-25sin(5t))/ (64 +cos²(5t))^1.5

8*(25)|sin(5t)|/(64+cos²(5t))^1.5

Now insert t=pi/2 : 8*(25)/64^1.5 or 25/64. The radius is 1/κ

Hope that helps. I checked the result with Wolfram Alpha.