Math Problem Statement

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

Solution

Given the parametric equations for the spacecraft's path:

x(t)=50(1+t41)x(t) = 50 \left( \sqrt{1 + t^4} - 1 \right) y(t)=20t32=10t3y(t) = 20 \frac{t^3}{2} = 10 t^3

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

1. Find the velocity components.

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

vx(t)v_x(t) - Velocity in the x-direction:

vx(t)=dx(t)dt=50ddt(1+t41)v_x(t) = \frac{dx(t)}{dt} = 50 \cdot \frac{d}{dt} \left( \sqrt{1 + t^4} - 1 \right) Using the chain rule: vx(t)=50121+t44t3=100t31+t4v_x(t) = 50 \cdot \frac{1}{2\sqrt{1 + t^4}} \cdot 4t^3 = \frac{100 t^3}{\sqrt{1 + t^4}}

vy(t)v_y(t) - Velocity in the y-direction:

vy(t)=dy(t)dt=10ddtt3=30t2v_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)=vx(t)2+vy(t)2v(t) = \sqrt{v_x(t)^2 + v_y(t)^2}

3. Find the direction of the velocity.

The direction of the velocity (angle θ\theta with respect to the x-axis) is: θ(t)=tan1(vy(t)vx(t))\theta(t) = \tan^{-1} \left( \frac{v_y(t)}{v_x(t)} \right)

At t=20t = 20 s:

  1. Find vx(20)v_x(20): vx(20)=1002031+204=10080001+160000=800000400.00251999.99m/sv_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 vy(20)v_y(20): vy(20)=30202=30400=12000m/sv_y(20) = 30 \cdot 20^2 = 30 \cdot 400 = 12000 \, \text{m/s}

  3. Find the magnitude of velocity at t=20t = 20: v(20)=(1999.99)2+(12000)2=4000000+144000000=14800000012165.5m/sv(20) = \sqrt{(1999.99)^2 + (12000)^2} = \sqrt{4000000 + 144000000} = \sqrt{148000000} \approx 12165.5 \, \text{m/s}

  4. Find the direction at t=20t = 20: θ(20)=tan1(120001999.99)tan1(6)80.5\theta(20) = \tan^{-1} \left( \frac{12000}{1999.99} \right) \approx \tan^{-1}(6) \approx 80.5^\circ

At t=200t = 200 s:

  1. Find vx(200)v_x(200): vx(200)=10020031+2004=10080000001+160000000000=800000000400000=2000m/sv_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 vy(200)v_y(200): vy(200)=302002=3040000=1200000m/sv_y(200) = 30 \cdot 200^2 = 30 \cdot 40000 = 1200000 \, \text{m/s}

  3. Find the magnitude of velocity at t=200t = 200: v(200)=(2000)2+(1200000)2=4000000+1440000000000=14400040000001200000.67m/sv(200) = \sqrt{(2000)^2 + (1200000)^2} = \sqrt{4000000 + 1440000000000} = \sqrt{1440004000000} \approx 1200000.67 \, \text{m/s}

  4. Find the direction at t=200t = 200: θ(200)=tan1(12000002000)=tan1(600)89.9\theta(200) = \tan^{-1} \left( \frac{1200000}{2000} \right) = \tan^{-1}(600) \approx 89.9^\circ

Summary of Results:

  • At t=20t = 20 s:

    • Magnitude of velocity: 12165.5m/s12165.5 \, \text{m/s}
    • Direction: 80.580.5^\circ
  • At t=200t = 200 s:

    • Magnitude of velocity: 1200000.67m/s1200000.67 \, \text{m/s}
    • Direction: 89.989.9^\circ

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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Parametric Equations
Velocity
Trigonometry

Formulas

Velocity components: v_x(t) = dx/dt, v_y(t) = dy/dt
Magnitude of velocity: v(t) = √(v_x^2 + v_y^2)
Direction of velocity: θ(t) = tan⁻¹(v_y/v_x)

Theorems

Pythagorean Theorem
Chain Rule for Derivatives

Suitable Grade Level

Advanced High School (Grades 11-12), College Level