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 slope of the path at time t = 4. Show the computations that lead to your answer.
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
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
To find the slope of the path at time t = 4, we need to calculate the derivative dy/dx at that specific time value. The slope of the path is given by the ratio dy/dx.
The given vector function is r = <2 + 6t cos(t), (15 − t) (1 sin(t))>. We will use the derivative formulas x'(t) = 6 - 2sin(t) and y'(t) = -15cos(t) + tcos(t) - 1 + sin(t) to calculate the derivatives.
First, let's calculate x'(t) at t = 4:
x'(4) = 6 - 2sin(4)
= 6 - 2(-0.7568) (using a calculator to find sin(4) ≈ -0.7568)
= 6 + 1.5136
= 7.5136
Next, let's calculate y'(t) at t = 4:
y'(4) = -15cos(4) + 4cos(4) - 1 + sin(4)
= -15(0.6536) + 4(0.6536) - 1 + (-0.7568) (using a calculator to find cos(4) ≈ 0.6536 and sin(4) ≈ -0.7568)
= -9.804 + 2.6144 - 1 - 0.7568
= -9.9454
Now, we have both x'(4) and y'(4). We can calculate dy/dx by dividing y'(4) by x'(4):
dy/dx = y'(4) / x'(4)
= (-9.9454) / (7.5136)
≈ -1.3256
Therefore, the slope of the path at time t = 4 is approximately -1.3256.