A particle moving in the x-y plane has a position vector given by r = 1.89t2i + 1.17t3j, where r is in inches and t is in seconds. Calculate the radius of curvature? of the path for the position of the particle when t = 2.1 sec.

r=89.970 m

Explanation:

We need to calculate radius of curvature, using first and second derivatives of the functions x (t) and y (t).

From the given equation, we can get, that:

x (t) = 1.89t^2 and y (t) = 1.17t^3

Then, the first derivative:

x' (t) = 2*1.89*t=3.78t and y' (t) = 3*1.17*t^2=3.51t^2

The second derivatives are:

x'' (t) = 3.78 and y'' (t) = 3.51*2*t=7.02t

Equation for the radius of curvature, can be found as:

r = (x'^2+y'^2) ^ (3/2) / (x''y'-x'y'')

Note, that the denominator should be taken by its absolute value.

For the given time, we should calculate numerical values for the derivatives. For t=2.1 s:

x' (2.1) = 7.938 m/s; x'' (2.1) = 3.78 m/s^2; y' (2.1) = 15.4791 m/s and y'' (2.1) = 14.742 m/s^2

Using equation of the curvature radius and the values, we can get:

r=89.970 m