Part of a cross-country skier's path can be described with the vector function r = <2 + 6t + 2 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) + tcos(t) − 1 + sin(t).
1. Find the slope of the path at time t = 4. Show the computations that lead to your answer.
2. Find the time when the skier's horizontal position is x = 60.
3. Find the acceleration vector of the skier when the skier's horizontal position is x = 60.
4. Find the speed of the skier when he is at his maximum height and find his speed in meters/min.
5. Find the total distance in meters that the skier travels from t = 0 to t = 15 minutes.
x = 2 + 6t + 2 cos(t)
y = (15 − t)(1 − sin(t))
1.
dx/dt = 6 - 2sint
dy/dt = (15-t)(-cost) + (-1)(1-sint)
= -15cost + tcost -1 + sint
so when t = 4, dx/dt = 6 - 2sin4
and dy/dt = -15cos4 + 4cos4 - 1 + sin4 = sin4 - 1 - 11cos4
slope when t = 4
= (dy/dt) / (dx/dt) = dy/dx = (6-2sin4)/(sin4 - 1 - 11cos4)
= appr 1.38
2.
we want x = 60
2 + 6t + 2 cos(t) = 60
6t + 2cost = 58
3t + cost = 29
cost = -3t + 29
Had to use technology to solve for t
according to the graph by "Desmos",
t = 9.335
3.
for acceleration, take the derivatives of both dx/dt and dy/dt ,
divide them and sub in t = 9.335 . Time for you to do some work here.
4.
maximum height is maximum of y
you know y and the derivative of y
set the derivative equal to zero and find the t value
5. find your vector magnitude when t = 0
find the vector magnitude when t = 15
take over