Part of a cross-country skier's path can be described with the vector function r = <2 + 6t cos(t), (15 − t) (1 sin(t))> for 0 ≤ t ≤ 15 minutes, with x and y measured in meters.
The derivatives of these functions are given by x′(t) = 6 − 2sin(t) and y′(t) = −15cos(t) + t cos(t) − 1 + sin(t).
Find the total distance in meters that the skier travels from t = 0 to t = 15 minutes.
To find the total distance that the skier travels from t = 0 to t = 15 minutes, we need to integrate the magnitude of the velocity vector over the given time interval.
The velocity vector is given by the derivatives of the position vector:
v(t) = <x'(t), y'(t)> = <6 - 2sin(t), -15cos(t) + tcos(t) - 1 + sint(t)>
The magnitude of the velocity vector is given by the square root of the sum of the squares of its components:
|v(t)| = √[(6 - 2sin(t))^2 + (-15cos(t) + tcos(t) - 1 + sint(t))^2]
Now, we integrate the magnitude of the velocity vector over the interval 0 ≤ t ≤ 15.
∫[0,15] |v(t)| dt = ∫[0,15] √[(6 - 2sin(t))^2 + (-15cos(t) + tcos(t) - 1 + sint(t))^2] dt
This integral cannot be solved analytically, but we can approximate it numerically using numerical integration methods such as the trapezoidal rule or Simpson's rule.
Alternatively, we can use a software program or online calculator that can perform numerical integration to approximate the integral and find the total distance traveled by the skier.
as usual,
s = ∫ √((dx/dt)^2 + (dy/dt)^2) dt
Using the x(t) you provided, dx/dt = 6cost - 6t sint
s = ∫[0,15] √((6cost - 6t sint)^2 + (-15cos(t) + t cos(t) - 1 + sin(t))^2) dt