Answer :
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].
### 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].
