Problem 1 - Calculus Early Transcendentals

Question

The table gives coordinates of a particle moving through space along a smooth curve. (a) Find the average velocities over the time intervals \([0,1]\) \([0.5,1],[1,2],\) and \([1,1.5] .\) (b) Estimate the velocity and speed of the particle at \(t=1\) $$\begin{array}{|c|c|c|c|}\hline t & {x} & {y} & {z} \\ \hline 0 & {2.7} & {9.8} & {3.7} \\ {0.5} & {3.5} & {7.2} & {3.3} \\ {1.0} & {4.5} & {6.0} & {3.0} \\ {1.5} & {5.9} & {6.4} & {2.8} \\ {2.0} & {7.3} & {7.8} & {2.7} \\\ \hline\end{array}$$

Step-by-Step Solution

Answer

Average velocities are (1.8, -3.8, -0.7), (2.0, -2.4, -0.6), (2.8, 1.8, -0.3), (2.8, 0.8, -0.4); estimated velocity at t=1 is (2.4, -0.8, -0.5) with speed ≈ 2.58.

Step 1: Average velocity over [0,1]

To find the average velocity over the interval \([0, 1]\), we use the coordinates at \(t=0\) and \(t=1\). The formula for average velocity \( \mathbf{v}_{avg} \) over an interval \([t_1, t_2]\) is:\[ \mathbf{v}_{avg} = \left(\frac{x_2-x_1}{t_2-t_1}, \frac{y_2-y_1}{t_2-t_1}, \frac{z_2-z_1}{t_2-t_1}\right) \]Substituting the given coordinates, we have:- \(x\)-component: \( \frac{4.5 - 2.7}{1-0} = 1.8 \)- \(y\)-component: \( \frac{6.0 - 9.8}{1-0} = -3.8 \)- \(z\)-component: \( \frac{3.0 - 3.7}{1-0} = -0.7 \)Thus, the average velocity over \([0, 1]\) is \((1.8, -3.8, -0.7)\).

Step 2: Average velocity over [0.5,1]

For the interval \([0.5, 1]\), use the coordinates at \(t=0.5\) and \(t=1\):- \(x\)-component: \( \frac{4.5 - 3.5}{1-0.5} = 2.0 \)- \(y\)-component: \( \frac{6.0 - 7.2}{1-0.5} = -2.4 \)- \(z\)-component: \( \frac{3.0 - 3.3}{1-0.5} = -0.6 \)The average velocity over \([0.5, 1]\) is \((2.0, -2.4, -0.6)\).

Step 3: Average velocity over [1,2]

For the interval \([1, 2]\), use the coordinates at \(t=1\) and \(t=2\):- \(x\)-component: \( \frac{7.3 - 4.5}{2-1} = 2.8 \)- \(y\)-component: \( \frac{7.8 - 6.0}{2-1} = 1.8 \)- \(z\)-component: \( \frac{2.7 - 3.0}{2-1} = -0.3 \)The average velocity over \([1, 2]\) is \((2.8, 1.8, -0.3)\).

Step 4: Average velocity over [1,1.5]

For the interval \([1, 1.5]\), use the coordinates at \(t=1\) and \(t=1.5\):- \(x\)-component: \( \frac{5.9 - 4.5}{1.5-1} = 2.8 \)- \(y\)-component: \( \frac{6.4 - 6.0}{1.5-1} = 0.8 \)- \(z\)-component: \( \frac{2.8 - 3.0}{1.5-1} = -0.4 \)The average velocity over \([1, 1.5]\) is \((2.8, 0.8, -0.4)\).

Step 5: Estimate Instantaneous Velocity at t=1

To estimate the instantaneous velocity at \(t=1\), we examine the average velocities around \(t=1\). Consider the intervals \([0.5, 1]\) and \([1, 1.5]\) which both include \(t=1\). The average velocity between these two gives a good approximation:\[- \text{Average of } (2.0, -2.4, -0.6) \text{ and } (2.8, 0.8, -0.4) = \left(\frac{2.0+2.8}{2}, \frac{-2.4+0.8}{2}, \frac{-0.6-0.4}{2} \right) \]This results in:- \(x\)-component: \( 2.4 \)- \(y\)-component: \( -0.8 \)- \(z\)-component: \( -0.5 \)So, the estimated velocity is \((2.4, -0.8, -0.5)\).

Step 6: Calculate Speed at t=1

The speed of the particle is the magnitude of the velocity vector. For the estimated velocity \((2.4, -0.8, -0.5)\) at \(t=1\), calculate:\[ \text{Speed} = \sqrt{2.4^2 + (-0.8)^2 + (-0.5)^2} \]This evaluates to:- \(2.4^2 = 5.76\)- \((-0.8)^2 = 0.64\)- \((-0.5)^2 = 0.25\)Summing these gives \(5.76 + 0.64 + 0.25 = 6.65\).Thus, \( \text{Speed} = \sqrt{6.65} \approx 2.58\).

Key Concepts

Average Velocity

Average velocity is a way to determine how quickly something is moving on average over a specific time frame. It's not just a single speed, but a vector that points in the direction of motion. To calculate it, you take the change in position and divide by the change in time.

For example, when you drive your car from one city to another and check the average speed at the end of your journey, you're essentially finding the average velocity for that trip. In vector calculus, this is expressed as:

• \(\mathbf{v}_{\text{avg}} = \left( \frac{x_2-x_1}{t_2-t_1}, \frac{y_2-y_1}{t_2-t_1}, \frac{z_2-z_1}{t_2-t_1} \right)\)

This formula helps break down the velocity into each of the spatial directions, providing a comprehensive picture of motion along the x, y, and z axes. Whether a particle moves in a straight line or follows a more complex path, average velocity tells us what path it effectively covered per unit of time.

Instantaneous Velocity

While average velocity gives us an overall speed during a certain period, instantaneous velocity focuses on the speed at a specific moment. Imagine hitting a speed bump in your car—your speed before, during, and after the bump can vary greatly. Instantaneous velocity captures each of these moments.

In mathematical terms, it can be thought of as taking the derivative of the position vector with respect to time. But, when you don't have an exact formula and rely on estimated data, you can use average velocities around the point of interest to find a close approximation: - Take average velocities immediately before and after the moment. - Calculate the midpoint between them to estimate the instantaneous velocity at that moment. Using this method ensures that we account for changes in speed, direction, or acceleration that affect how an object moves at any given time.

Speed Calculation

Speed and velocity are related, but not identical. Speed is essentially the "step-sibling" of velocity. It's a scalar quantity, meaning it has size but no direction—unlike velocity, which has both. Speed tells you how fast an object is moving but not where it's headed.

• To find speed from velocity, calculate the magnitude of the velocity vector.
• This involves using the Pythagorean theorem in three dimensions.

The formula to convert a velocity vector \( (v_x, v_y, v_z) \) to speed is:\[ \text{Speed} = \sqrt{v_x^2 + v_y^2 + v_z^2}\]This equation shows how all components of motion contribute to the overall speed. Understanding this allows for comprehensive comparisons of travel efficiency and distance over time.

Magnitude of Vectors

When talking about vectors, magnitude is a key concept. It gives the "size" or length of the vector without any direction. In physical terms, it's like knowing how far something has moved, regardless of where it ended up.

• The magnitude is always a positive number or zero.
• For any vector \( \mathbf{v} = (x, y, z) \), magnitude is calculated as:\[|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}\]

Think of this as a three-dimensional ruler that measures the straight-line distance from the start to the end of a vector. Even in more complicated paths, knowing vector magnitudes helps us understand the "stretch" of a path or movement in three-dimensional space.