Question

Part of a cross-country skier's path can be described with the vector function r = <2 + 6t + 2 cos(t), (15t)(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)sin(t).
Find the slope of the path at time t = 4. Show the computations that lead to your answer.

Answer 1

The slope of the skier's path at time t = 4 is calculated as the ratio of the derivatives of the y-component to the x-component of the velocity, resulting in a value of approximately 1.0412.

Step-by-step explanation:

To find the slope of the skier's path at time t = 4, we can use the given derivatives of the position function r. The slope is the ratio of the derivatives of the y-component to the x-component of the position vector, which is the same as the derivative of the y with respect to x. Using the provided derivatives for x'(t) and y'(t), we evaluate them at t = 4.

1. Substitute t = 4 into x'(t) to get x'(4) = 6 + 2sin(4).
2. Substitute t = 4 into y'(t) to get y'(4) = -15cos(4) + 4cos(4)sin(4).
3. The slope at t = 4 is the ratio y'(4) / x'(4).

Now, we perform the actual calculations:

1. Calculate x'(4):

x'(4) = 6 + 2sin(4)

= 6 + 2(0.7568) [using a calculator]

= 6 + 1.5136

= 7.5136

1. Calculate y'(4):

y'(4) = -15cos(4) + 4cos(4)sin(4)

= -15(−0.6536) + 4(−0.6536)sin(4) [using a calculator]

= 9.804 + 4(−0.6536)(0.7568)

= 9.804 − 1.9806

= 7.8234

1. Finally, find the slope at t = 4:

Slope = y'(4) / x'(4)

= 7.8234 / 7.5136

= 1.0412