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.

Question

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 slope of the path at time t = 4. Show the computations that lead to your answer.

Answer 1

r = <2 + 6t cos(t), (15 − t) (1 sin(t))>

x = 2 + 6tcost
dx/dt = 6cost - 6tsint <---- you did not have that

y = 15sint - tsint ,
dy/dt = 15cost - tcost - sint <---- does not match yours

slope of path
= dy/dx = (dy/dt)/(dx/dt) = (15cost - tcost - sint)/(6cost - 6tsint)
when t = 4, we get

dy/dx = (15cos4 - 4cos4 - sin4)/(6cos4 - 6sin4), set your calculator to RAD
= approx -10.4

check my arithmetic

Answer 2

dy/dx = (dy/dt)/(dx/dt)
Using your values, that would be
dy/dx = (−15cos(t) + t cos(t) − 1 + sin(t)) / (6 − 2sin(t))
so plug in t=4

But actually,
d/dt 2 + 6t cos(t) = 6cost - 6tsint