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

ρ = 0.0657

Explanation:

We can apply the equation

ρ = 1/║r"(t)║

we have to get r"(t):

r'(t) = 3.78*t i + 3.51*t² j

r"(t) = 3.78 i + 7.02*t j

⇒   ║r"(t)║ = √(3.78²+(7.02*t)²)

If t = 2.1 s we have

║r"(2.1)║ = √(3.78²+(7.02*2.1)²) = 15.219

⇒   ρ = 1 / 15.219 = 0.0657 m

answer:

a

explanation: