Question

An object moves along a straight line from the point \(P(1,2,4)\) to the point \(Q(-6,8,10)\) a. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with a constant speed over the time interval [0,5] b. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with speed \(e^{t}\)

Short answer

Question: Determine a position function that describes the motion of an object moving along a straight line from point P(1, 2, 4) to point Q(-6, 8, 10) in 3 dimensions. Part a: The motion occurs with constant speed during the time interval [0, 5]. Part b: The motion occurs with speed e^t. Answer: Part a: The position function for constant speed during the time interval [0, 5] is: $$\mathbf{r}(t) = \begin{pmatrix}1-\frac{7}{5}t \\ 2+\frac{6}{5}t \\ 4+\frac{6}{5}t\end{pmatrix}$$ Part b: The position function for speed e^t is: $$\mathbf{r}(t) = \begin{pmatrix}1\\2\\4\end{pmatrix}+\frac{1}{11}\begin{pmatrix}-7\\6\\6\end{pmatrix}(e^{t}+C)$$ where C is an arbitrary constant.

Step-by-step solution

Find the vector PQ

To find the vector PQ, subtract the coordinates of point P from the coordinates of point Q:$$\mathbf{PQ} = \mathbf{Q} - \mathbf{P} = \begin{pmatrix}-6 -1 \\ 8-2\\ 10-4 \end{pmatrix} = \begin{pmatrix}-7 \\ 6 \\ 6\end{pmatrix}$$

Step 2a: Find the unit vector of PQ and the constant speed

Find the magnitude of vector PQ and divide the components of PQ by it to find the unit vector:$$|\mathbf{PQ}| = \sqrt{(-7)^{2} + 6^{2} + 6^{2}} =\sqrt{49+36+36}=\sqrt{121}=11$$Then, find the unit vector: $$\mathbf{\hat{PQ}} = \frac{\mathbf{PQ}}{|\mathbf{PQ}|} = \frac{1}{11}\begin{pmatrix}-7\\6\\6\end{pmatrix}$$The total time it takes for the object to move from P to Q is 5 seconds. So, find the constant speed by dividing the distance by the time: $$v=\frac{11}{5}$$

Step 3a: Create the position function for constant speed

To create the position function, we want to find a vector function that describes the motion from P to Q. We have the unit vector of PQ and the constant speed. Multiply the unit vector by the constant speed and integrate with respect to time. The position function is:$$\mathbf{r}(t)=\mathbf{P}+v\mathbf{\hat{PQ}}t= \begin{pmatrix}1\\2\\4\end{pmatrix}+\frac{11}{5}\frac{1}{11}\begin{pmatrix}-7\\6\\6\end{pmatrix}t= \begin{pmatrix}1-\frac{7}{5}t \\ 2+\frac{6}{5}t \\ 4+\frac{6}{5}t\end{pmatrix}$$

Step 2b: Integrate the speed function e^t

In this case, the speed function is given by $$v(t)=e^{t}$$. Hence, to find the position function, we need to integrate the speed function with respect to time.$$\int{e^{t}dt}=e^{t}+C$$

Step 3b: Create the position function for speed e^t

We need to find a position function that satisfies the speed function integrated from step 2b. Multiply the unit vector of PQ by the integral and add the point P to the result:$$\mathbf{r}(t)=\mathbf{P}+\mathbf{\hat{PQ}}\int{e^{t}dt} = \begin{pmatrix}1\\2\\4\end{pmatrix}+\frac{1}{11}\begin{pmatrix}-7\\6\\6\end{pmatrix}(e^{t}+C)$$The given function is nonspecific because the arbitrary constant C cannot be determined. Nonetheless, this position function accurately represents the motion with a speed of \(e^{t}\).

Key concepts

Constant Speed

When an object moves at a constant speed, it covers equal distances in equal intervals of time. In the context of this problem, we have a straight motion from point P to point Q over 5 seconds. To describe this motion, we first find the vector PQ and its magnitude, which helps us establish the constant speed by dividing the distance by the time period.

• Calculate the distance between P and Q using the Euclidean distance formula.
• Divide this distance by the total time to find the constant speed.

This constant speed ensures that the velocity of the object remains uniform; hence the motion can be described easily with a simple linear equation. This gives us a straightforward way to express the position at any time `t` using the velocity and direction.

Unit Vector

A unit vector is a vector that has a magnitude of 1 and indicates direction. In our example, we find the unit vector along the line from P to Q to determine the direction of motion.
To get the unit vector:

• Subtract the coordinates of P from Q to obtain the vector PQ.
• Calculate the magnitude of PQ using the square root of the sum of the squares of its components.
• Divide each component of PQ by this magnitude.

This unit vector is vital because it normalizes the vector PQ, allowing us to focus solely on direction. It remains constant and shows the direction of travel from P to Q, irrespective of speed.

Integrating Speed Function

When dealing with varying speeds, integrating the speed function allows us to find the position function. In the exercise, the speed is given as an exponential function, \(v(t) = e^{t}\). By integrating this speed function with respect to time, we find the velocity function.
The steps include:

• Finding the indefinite integral of the speed function \( \int{e^{t}dt} = e^{t} + C \), where C is an integration constant.
• Using this integral result with the unit vector we found earlier to develop the position function.

Integrating speed functions is essential because it helps in determining how position changes over time when speeds are not constant, by effectively capturing the accumulation of distance.

Vector Function

A vector function describes motion in space and can be used to track the position of an object at any time \(t\). Here, the position vector function is given in terms of initial position and the movement details we calculated before.

• The base of our vector function is point P — the starting point.
• The function adds the product of the constant speed or integrated speed with the unit vector to point P.
• In essence, the vector function integrates position, direction, and speed to provide a complete description of the object's trajectory.

By understanding the vector function, we can easily predict where the object will be at any given moment, which is practical for both constant and variable speed situations.