A spacecraft moves along a path described by the parametric equations shown below for the first 200 s after launch.​ Here, x and y are measured in​ meters, and t is measured in seconds. Find the magnitude and direction of the velocity of the spacecraft 20.0 s and 200 s after launch.
xequals50 left parenthesis StartRoot 1 plus t Superscript 4 EndRoot minus 1 right parenthesis​,  yequals20 t Superscript 3 divided by 2

Given the parametric equations for the spacecraft's path:

\[
x(t) = 50 \left( \sqrt{1 + t^4} - 1 \right)
\]
\[
y(t) = 20 \frac{t^3}{2} = 10 t^3
\]

To find the magnitude and direction of the velocity at \( t = 20 \) s and \( t = 200 \) s, we follow these steps:

### 1. Find the velocity components.

The velocity components \( v_x(t) \) and \( v_y(t) \) are given by the derivatives of the position functions \( x(t) \) and \( y(t) \) with respect to time \( t \):

#### \( v_x(t) \) - Velocity in the x-direction:
\[
v_x(t) = \frac{dx(t)}{dt} = 50 \cdot \frac{d}{dt} \left( \sqrt{1 + t^4} - 1 \right)
\]
Using the chain rule:
\[
v_x(t) = 50 \cdot \frac{1}{2\sqrt{1 + t^4}} \cdot 4t^3 = \frac{100 t^3}{\sqrt{1 + t^4}}
\]

#### \( v_y(t) \) - Velocity in the y-direction:
\[
v_y(t) = \frac{dy(t)}{dt} = 10 \cdot \frac{d}{dt} t^3 = 30 t^2
\]

### 2. Find the magnitude of the velocity vector.
The magnitude of the velocity is given by the Pythagorean theorem:
\[
v(t) = \sqrt{v_x(t)^2 + v_y(t)^2}
\]

### 3. Find the direction of the velocity.
The direction of the velocity (angle \( \theta \) with respect to the x-axis) is:
\[
\theta(t) = \tan^{-1} \left( \frac{v_y(t)}{v_x(t)} \right)
\]

### At \( t = 20 \) s:

1. **Find \( v_x(20) \):**
\[
v_x(20) = \frac{100 \cdot 20^3}{\sqrt{1 + 20^4}} = \frac{100 \cdot 8000}{\sqrt{1 + 160000}} = \frac{800000}{400.0025} \approx 1999.99 \, \text{m/s}
\]

2. **Find \( v_y(20) \):**
\[
v_y(20) = 30 \cdot 20^2 = 30 \cdot 400 = 12000 \, \text{m/s}
\]

3. **Find the magnitude of velocity at \( t = 20 \):**
\[
v(20) = \sqrt{(1999.99)^2 + (12000)^2} = \sqrt{4000000 + 144000000} = \sqrt{148000000} \approx 12165.5 \, \text{m/s}
\]

4. **Find the direction at \( t = 20 \):**
\[
\theta(20) = \tan^{-1} \left( \frac{12000}{1999.99} \right) \approx \tan^{-1}(6) \approx 80.5^\circ
\]

### At \( t = 200 \) s:

1. **Find \( v_x(200) \):**
\[
v_x(200) = \frac{100 \cdot 200^3}{\sqrt{1 + 200^4}} = \frac{100 \cdot 8000000}{\sqrt{1 + 160000000000}} = \frac{800000000}{400000} = 2000 \, \text{m/s}
\]

2. **Find \( v_y(200) \):**
\[
v_y(200) = 30 \cdot 200^2 = 30 \cdot 40000 = 1200000 \, \text{m/s}
\]

3. **Find the magnitude of velocity at \( t = 200 \):**
\[
v(200) = \sqrt{(2000)^2 + (1200000)^2} = \sqrt{4000000 + 1440000000000} = \sqrt{1440004000000} \approx 1200000.67 \, \text{m/s}
\]

4. **Find the direction at \( t = 200 \):**
\[
\theta(200) = \tan^{-1} \left( \frac{1200000}{2000} \right) = \tan^{-1}(600) \approx 89.9^\circ
\]

### Summary of Results:

- At \( t = 20 \) s:
  - Magnitude of velocity: \( 12165.5 \, \text{m/s} \)
  - Direction: \( 80.5^\circ \)
  
- At \( t = 200 \) s:
  - Magnitude of velocity: \( 1200000.67 \, \text{m/s} \)
  - Direction: \( 89.9^\circ \)

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Do you want more details or clarifications on any part of the process?

Here are 5 related questions you might find interesting:

1. How do we calculate acceleration from parametric equations?
2. Can we find the displacement of the spacecraft after 200 seconds?
3. How do velocity and acceleration differ in parametric motion?
4. How would these results change if the functions had different forms (e.g., quadratic or linear)?
5. What is the role of time in determining the curvature of the spacecraft's path?

**Tip:** The derivative of a parametric equation gives the rate of change of the respective variable, making it a useful tool for finding velocity and acceleration.

A spacecraft moves along a path described by the parametric equations for x and y for the first 200 seconds after launch. Find the magnitude and direction of the velocity at 20.0 s and 200 s after launch. The equations are x(t) = 50(√(1 + t^4) - 1) and y(t) = 20t^3/2.

Learn how to calculate the velocity magnitude and direction of a spacecraft moving along parametric equations for x(t) = 50(√(1 + t^4) - 1) and y(t) = 20t^3/2 at 20 seconds and 200 seconds. This problem covers key calculus and trigonometry concepts, including the Pythagorean Theorem and chain rule. Suitable for advanced high school and college students.

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