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A woman attached to a bungee cord jumps from a bridge that is 42 meters above a river. Her height in meters above the river [tex]\( t \)[/tex] seconds after the jump is [tex]\( y(t) = 21\left(1 + e^{-t} \cos t \right) \)[/tex], for [tex]\( t \geq 0 \)[/tex].

a. Determine her velocity at [tex]\( t = 1 \)[/tex] and [tex]\( t = 3 \)[/tex].

b. Use a graphing utility to determine when she is moving downward and when she is moving upward during the first 10 seconds.

c. Use a graphing utility to estimate the maximum upward velocity.

Her velocity at time [tex]\( t \)[/tex] is given by the function [tex]\( v(t) = \square \)[/tex]


Sagot :

To solve these problems related to the motion of a woman attached to a bungee cord, let's start by finding her velocity as a function of time and then address each part of the question step-by-step.

### Given Function:
The height of the woman above the river as a function of time [tex]\( t \)[/tex] is given by:
[tex]\[ y(t) = 21 \left( 1 + e^{-t} \cos t \right) \][/tex]

### Part (a): Determining Her Velocity at [tex]\( t = 1 \)[/tex] and [tex]\( t = 3 \)[/tex]

Velocity is the rate of change of height with respect to time, which is the first derivative of [tex]\( y(t) \)[/tex].

First, we need to take the derivative of [tex]\( y(t) \)[/tex]:

[tex]\[ y(t) = 21 \left( 1 + e^{-t} \cos t \right) \][/tex]

### Derivative Calculation:

Using the product rule and the chain rule for differentiation:

[tex]\[ y(t) = 21[1 + e^{-t} \cos t] \][/tex]

[tex]\[ \frac{dy}{dt} = 21 \left[ \frac{d}{dt} (1) + \frac{d}{dt} \left( e^{-t} \cos t \right) \right] \][/tex]

[tex]\[ \frac{dy}{dt} = 21 \left[ 0 + e^{-t} \frac{d}{dt} (\cos t) + \cos t \frac{d}{dt} (e^{-t}) \right] \][/tex]

[tex]\[ \frac{dy}{dt} = 21 \left[ e^{-t} (-\sin t) + \cos t (-e^{-t}) \right] \][/tex]

[tex]\[ \frac{dy}{dt} = 21 \left[ -e^{-t} \sin t - e^{-t} \cos t \right] \][/tex]

[tex]\[ \frac{dy}{dt} = 21 \left[ -e^{-t} (\sin t + \cos t) \right] \][/tex]

[tex]\[ v(t) = -21 e^{-t} (\sin t + \cos t) \][/tex]

Now, let's evaluate this velocity function [tex]\( v(t) \)[/tex] at [tex]\( t = 1 \)[/tex] and [tex]\( t = 3 \)[/tex]:

1. Velocity at [tex]\( t = 1 \)[/tex]:

[tex]\[ v(1) = -21 e^{-1} (\sin 1 + \cos 1) \][/tex]

Using a calculator:

[tex]\[ \sin 1 \approx 0.8415 \][/tex]
[tex]\[ \cos 1 \approx 0.5403 \][/tex]
[tex]\[ e^{-1} \approx 0.3679 \][/tex]

[tex]\[ v(1) = -21 (0.3679) (0.8415 + 0.5403) \][/tex]
[tex]\[ v(1) = -21 (0.3679) (1.3818) \][/tex]
[tex]\[ v(1) \approx -21 \times 0.5085 \][/tex]
[tex]\[ v(1) \approx -10.679 \, \text{m/s} \][/tex]

2. Velocity at [tex]\( t = 3 \)[/tex]:

[tex]\[ v(3) = -21 e^{-3} (\sin 3 + \cos 3) \][/tex]

Using a calculator:

[tex]\[ \sin 3 \approx 0.1411 \][/tex]
[tex]\[ \cos 3 \approx -0.9895 \][/tex]
[tex]\[ e^{-3} \approx 0.0498 \][/tex]

[tex]\[ v(3) = -21 (0.0498) (0.1411 - 0.9895) \][/tex]
[tex]\[ v(3) = -21 (0.0498) (-0.8484) \][/tex]
[tex]\[ v(3) \approx -21 \times -0.0422 \][/tex]
[tex]\[ v(3) \approx 0.8862 \, \text{m/s} \][/tex]

### Summary of Part (a):

- Her velocity at [tex]\( t = 1 \)[/tex] is approximately [tex]\( -10.679 \, \text{m/s} \)[/tex].
- Her velocity at [tex]\( t = 3 \)[/tex] is approximately [tex]\( 0.8862 \, \text{m/s} \)[/tex].

### Part (b): Determine When She is Moving Downward and Upward

To determine when she is moving downward and when she is moving upward, we need to analyze the sign of [tex]\( v(t) \)[/tex].

- Moving Downward: [tex]\( v(t) < 0 \)[/tex]
- Moving Upward: [tex]\( v(t) > 0 \)[/tex]

We can determine this through a graphing utility. Essentially, you would plot [tex]\( v(t) = -21 e^{-t} (\sin t + \cos t) \)[/tex] over the interval [tex]\( 0 \le t \le 10 \)[/tex].

- She is moving downward when the graph of [tex]\( v(t) \)[/tex] is below the t-axis.
- She is moving upward when the graph of [tex]\( v(t) \)[/tex] is above the t-axis.

### Part (c): Estimate the Maximum Upward Velocity

The maximum upward velocity will occur at the point where [tex]\( v(t) \)[/tex] reaches its maximum positive value. This can be estimated using the graphing utility.

### Final Function for Part (a):

- Her velocity at time [tex]\( t \)[/tex] is given by the function [tex]\( v(t) = -21 e^{-t} (\sin t + \cos t) \)[/tex].