Consider the following position functions \( r \) and \( R \) for two objects. a. Find the interval \( [c, d] \) over which the \( R \) trajectory is the same as the \( r \) trajectory over \( [a, b] \) b. Find the velocity of both objects. c. Graph the speed of the two objects over the intervals \( [a, b] \) and \( [c, d] \) respectively. For graphing, lot \( c=1 \) and \( d=20 \). \( r(t)=\left\langle 4+t^{2}, 4-4 t^{4}, 3+3 t^{4}\right\rangle,[a, b]=[0,5] \) \( R(t)=\left\langle 4+\ln t, 4-4 \ln ^{2} t, 3+3 \ln ^{2} 1\right\rangle \), on \( [c, d] \) a. The interval is \( \square \) (Simplify your answers.)

To find the interval \([c, d]\) over which the trajectory of \(R\) is the same as that of \(r\) over \([a, b]\), we need to analyze the position functions \(r(t)\) and \(R(t)\). ### Given Functions: 1. \( r(t) = \langle 4 + t^2, 4 - 4t^4, 3 + 3t^4 \rangle \) 2. \( R(t) = \langle 4 + \ln t, 4 - 4 \ln^2 t, 3 + 3 \ln^2 t \rangle \) ### Step 1: Find the interval \([c, d]\) We need to find \(t\) values such that \(R(t) = r(t)\). This means we need to solve the following equations: 1. \( 4 + t^2 = 4 + \ln t \) 2. \( 4 - 4t^4 = 4 - 4 \ln^2 t \) 3. \( 3 + 3t^4 = 3 + 3 \ln^2 t \) #### Solving the first equation: From the first equation: \[ t^2 = \ln t \] #### Solving the second equation: From the second equation: \[ -4t^4 = -4 \ln^2 t \implies t^4 = \ln^2 t \] #### Solving the third equation: From the third equation: \[ 3t^4 = 3 \ln^2 t \implies t^4 = \ln^2 t \] ### Step 2: Equating the equations From the first equation \(t^2 = \ln t\) and the second equation \(t^4 = \ln^2 t\), we can substitute \(t^2\) from the first into the second: \[ t^4 = (t^2)^2 = (\ln t)^2 \] This means we need to solve: \[ t^2 = \ln t \] and \[ t^4 = \ln^2 t \] ### Step 3: Finding the values of \(t\) Let \(x = t^2\). Then we have: \[ x = \ln(\sqrt{x}) = \frac{1}{2} \ln x \] This leads to: \[ 2x = \ln x \] To solve \(2x = \ln x\), we can use numerical methods or graphing. The solutions can be approximated, but we can also check specific values. ### Step 4: Numerical Solutions Using numerical methods or graphing, we find that: - \(t \approx 1\) is a solution. - \(t \approx 2.718\) (which is \(e\)) is another solution. ### Step 5: Interval \([c, d]\) Given \(c = 1\) and \(d = 20\), we can conclude that the interval over which the trajectories are the same is: \[ [c, d] = [1, e] \] ### Final Answer for Part a: The interval is \([1, e]\) (approximately \([1, 2.718]\)). ### Part b: Find the velocity of both objects The velocity \(v(t)\) is the derivative of the position function \(r(t)\) and \(R(t)\). 1. **Velocity of \(r(t)\)**: \[ v_r(t) = \frac{d}{dt} r(t) = \left\langle \frac{d}{dt}(4 + t^2), \frac{d}{dt}(4 - 4t^4), \frac{d}{dt}(3 + 3t^4) \right\rangle = \langle 2t, -16t^3, 12t^3 \rangle \] 2. **Velocity of \(R(t)\)**: \[ v_R(t) = \frac{d}{dt} R(t) = \left\langle \frac{d}{dt}(4 + \ln t), \frac{d}{dt}(4 - 4 \ln^2 t), \frac{d}{dt}(3 + 3 \ln^2 t) \right\rangle \] Using the chain rule: \[ v_R(t) = \left\langle \frac{1}{t}, -8 \ln t \cdot \frac{1}{t}, 6 \ln t \cdot \frac{1}{t} \right\rangle = \left\langle \frac{1}{t}, -\frac{8 \ln t}{t}, \frac{6 \ln t}{t} \right\rangle \] ### Part c: Graphing the speed The speed is the magnitude of the velocity vector. 1. **Speed of \(r(t)\)**: \[ \text{Speed}_r(t) = \sqrt{(2t)^2 + (-16t^3)^2 + (12t^3)^2} = \sqrt{4t^2 + 256t^6 + 144t^6} = \sqrt{4t^2 + 400t^6} \] 2. **Speed of \(R(t)\)**: \[ \text{Speed}_R(t) = \sqrt{\left(\frac{1}{t}\right)^2 + \left(-\frac{8 \ln t}{t}\right)^2 + \left(\frac{6 \ln t}{t}\right)^2} = \sqrt{\frac{1}{t^2} + \frac{64 \ln^2 t}{t^2} + \frac{36 \ln^2 t}{t^2}} = \sqrt{\frac{1 + 100 \ln^2 t}{t^2}} = \frac{\sqrt{1 + 100 \ln^2 t}}{t} \] ### Summary: - The interval is \([1, e]\). - The velocities are \(v_r(t) = \langle 2t, -16t^3, 12t^3 \rangle\) and \(v_R(t) = \left\langle \frac{1}{t}, -\frac{8 \ln t}{t}, \frac{6 \ln t}{t} \right\rangle\). - The speeds are \(\sqrt{4t^2 + 400t^6}\) for \(r(t)\) and \(\frac{\sqrt{1 + 100 \ln^2 t}}{t}\) for \(R(t)\). You can graph these speeds over the specified intervals using graphing software or tools.