Problem 15 - Essential Calculus Early Transcendentals

Question

The position function of a particle is given by \(\mathbf{r}(t)=\left\langle t^{2}, 5 t, t^{2}-16 t\right\rangle .\) When is the speed a minimum?

Step-by-Step Solution

Answer

The speed is minimum at \( t = 4 \).

Step 1: Find the velocity vector

The velocity vector of the particle is obtained by differentiating the position function with respect to time. We differentiate each component of \( \mathbf{r}(t) \) separately.\[ \mathbf{v}(t) = \frac{d}{dt}\left\langle t^{2}, 5t, t^{2} - 16t \right\rangle = \left\langle 2t, 5, 2t - 16 \right\rangle. \]

Step 2: Determine the speed function

Speed is the magnitude of the velocity vector. Compute the magnitude of \( \mathbf{v}(t) = \left\langle 2t, 5, 2t - 16 \right\rangle \).\[ \|\mathbf{v}(t)\| = \sqrt{(2t)^2 + 5^2 + (2t - 16)^2}. \]

Step 3: Simplify the speed expression

Expand the terms in the speed expression and simplify:\[ \|\mathbf{v}(t)\| = \sqrt{4t^2 + 25 + (2t - 16)^2} = \sqrt{4t^2 + 25 + 4t^2 - 64t + 256}. \]Combine like terms:\[ \|\mathbf{v}(t)\| = \sqrt{8t^2 - 64t + 281}. \]

Step 4: Find the critical points of the speed

To find when the speed is a minimum, we differentiate the function \( \sqrt{8t^2 - 64t + 281} \) with respect to \( t \), and then find critical points. However, directly differentiate the expression inside the square root, \( f(t) = 8t^2 - 64t + 281 \), since differentiating square roots can be complex.\[ f'(t) = 16t - 64. \]Set the derivative equal to zero to find critical values:\[ 16t - 64 = 0 \Rightarrow 16t = 64 \Rightarrow t = 4. \]

Step 5: Verify the minimum with the second derivative

Check if \( t = 4 \) is a minimum by checking the second derivative of the function.\[ f''(t) = 16. \]Since \( f''(t) = 16 > 0 \), the function has a local minimum at \( t = 4 \).

Step 6: Conclude the result

Given that the second derivative is positive at \( t = 4 \), the speed function has a minimum at this time.

Key Concepts

Velocity Vector

A velocity vector represents the rate of change of a particle's position. It's crucial in particle motion analyses because it provides both the speed and direction of the particle's travel at any time. Given a position function, such as \(\mathbf{r}(t) = \left\langle t^{2}, 5t, t^{2} - 16t \right\rangle\), we can find the velocity vector by differentiating each component with respect to time \(t\). In the provided exercise, this process gave us the velocity function: \[\mathbf{v}(t) = \left\langle 2t, 5, 2t - 16 \right\rangle.\] Each element of this vector represents the rate of change along that axis:\(2t\) represents the change in the \(x\)-direction, \(5\) the constant rate in the \(y\)-direction, and \(2t - 16\) the change in the \(z\)-direction.

Speed Function

The speed function is the magnitude of the velocity vector. It's a scalar representation of how fast the particle is moving, without direction. To find it, we compute the magnitude of the velocity vector, \(\mathbf{v}(t)\). This involves taking the square root of the sum of the squares of its components. For the velocity vector \(\left\langle 2t, 5, 2t - 16 \right\rangle\), the speed function is calculated as follows: \[\|\mathbf{v}(t)\| = \sqrt{(2t)^2 + 5^2 + (2t - 16)^2}.\] By simplifying, we find: \[\|\mathbf{v}(t)\| = \sqrt{8t^2 - 64t + 281}.\] This expression gives us the speed of the particle at any time \(t\) as a single numerical value.

Critical Points

Critical points of a function are the values of \(t\) where the function's derivative equals zero or where the derivative does not exist. These are the points where the function might achieve a local minimum or maximum. To find the critical points of the speed function, we take the derivative of the simplified function inside the square root: \[f(t) = 8t^2 - 64t + 281,\] and differentiate it: \[f'(t) = 16t - 64.\] Setting this derivative equal to zero allows us to solve for \(t\): \[16t - 64 = 0 \Rightarrow t = 4.\] Thus, the critical point occurs at \(t = 4\). This is where we further explore using the second derivative to confirm it as a minimum.

Second Derivative Test

The second derivative test helps us determine whether a critical point is a local minimum or maximum. After finding \(t = 4\) as a critical point, we need to evaluate the second derivative of \(f(t)\) to ascertain the nature of this critical point. Calculating the second derivative, we find: \[f''(t) = 16.\] Because \(f''(t) = 16\) is greater than zero, this indicates that the function \(f(t)\) is concave up at \(t = 4\), confirming a local minimum at this point. Thus, the speed function achieves its minimum value when \(t = 4\), indicating the slowest movement in the particle's trajectory.