APEX Calculus

1. Taking derivatives, we find

   \begin{equation*} \vvt = \vrp(t) =\la 2t-1,2t+1\ra \text{ and } \vat = \vrp'(t) = \la 2,2\ra\text{.} \end{equation*}

   Note that acceleration is constant.

2. \(\vec v(-1) = \la -3,-1\ra\text{,}\) \(\vec a(-1) = \la 2,2\ra\text{;}\) \(\vec v(1) = \la 1,3\ra\text{,}\) \(\vec a(1) = \la 2,2\ra\text{.}\) These are plotted with \(\vrt\) in Figure 11.3.3.(a). We can think of acceleration as “pulling” the velocity vector in a certain direction. At \(t=-1\text{,}\) the velocity vector points down and to the left; at \(t=1\text{,}\) the velocity vector has been pulled in the \(\la 2,2\ra\) direction and is now pointing up and to the right. In Figure 11.3.3.(b) we plot more velocity/acceleration vectors, making more clear the effect acceleration has on velocity.

   

   

   Since \(\vat\) is constant in this example, as \(t\) grows large \(\vvt\) becomes almost parallel to \(\vat\text{.}\) For instance, when \(t=10\text{,}\) \(\vec v(10) = \la 19,21\ra\text{,}\) which is nearly parallel to \(\la 2,2\ra\text{.}\)

3. The object's speed is given by

   \begin{equation*} \norm{\vvt} = \sqrt{(2t-1)^2+(2t+1)^2} =\sqrt{8t^2+2}\text{.} \end{equation*}

   To find the minimal speed, we could apply calculus techniques (such as set the derivative equal to 0 and solve for \(t\text{,}\) etc.) but we can find it by inspection. Inside the square root we have a quadratic which is minimized when \(t=0\text{.}\) Thus the speed is minimized at \(t=0\text{,}\) with a speed of \(\sqrt{2}\) ^ft⁄_s. The graph in Figure 11.3.3.(b) also implies speed is minimized here. The filled dots on the graph are located at integer values of \(t\) between \(-3\) and 3. Dots that are far apart imply the object traveled a far distance in 1 second, indicating high speed; dots that are close together imply the object did not travel far in 1 second, indicating a low speed. The dots are closest together near \(t=0\text{,}\) implying the speed is minimized near that value.

We begin by computing the velocity and acceleration function for each object:

We immediately see that Object 1 has constant acceleration, whereas Object 2 does not.

At \(t=-1\text{,}\) we have \(\vec v_1(-1) = \la 1,-2\ra\) and \(\vec v_2(-1) = \la 3,-6\ra\text{;}\) the velocity of Object 2 is three times that of Object 1 and so it follows that the speed of Object 2 is three times that of Object 1 (\(3\sqrt{5}\) ft/s compared to \(\sqrt{5}\) ft/s.)

At \(t=0\text{,}\) the velocity of Object 1 is \(\vec v(1) = \la 1,0\ra\) and the velocity of Object 2 is \(\vec 0\text{!}\) This tells us that Object 2 comes to a complete stop at \(t=0\text{.}\)

In Figure 11.3.5, we see the velocity and acceleration vectors for Object 1 plotted for \(t=-1, -1/2, 0, 1/2\) and \(t=1\text{.}\) Note again how the constant acceleration vector seems to “pull” the velocity vector from pointing down, right to up, right. We could plot the analogous picture for Object 2, but the velocity and acceleration vectors are rather large (\(\vec a_2(-1) = \la -6,30\ra\text{!}\))

Instead, we simply plot the locations of Object 1 and 2 on intervals of \(1/10^{\text{ th } }\) of a second, shown in Figure 11.3.6.(a) and Figure 11.3.6.(b) . Note how the \(x\)-values of Object 1 increase at a steady rate. This is because the \(x\)-component of \(\vec a(t)\) is 0; there is no acceleration in the \(x\)-component. The dots are not evenly spaced; the object is moving faster near \(t=-1\) and \(t=1\) than near \(t=0\text{.}\)

In Figure 11.3.6.(b), we see the points plotted for Object 2. Note the large change in position from \(t=-1\) to \(t=-0.8\text{;}\) the object starts moving very quickly. However, it slows considerably at it approaches the origin, and comes to a complete stop at \(t=0\text{.}\) While it looks like there are 3 points near the origin, there are in reality 5 points there.

Since the objects begin and end at the same location, they have the same displacement. Since they begin and end at the same time, with the same displacement, they have the same average rate of change (i.e., they have the same average velocity). Since they follow the same path, they have the same distance traveled. Even though these three measurements are the same, the objects obviously travel the path in very different ways.

1. The ball whirls in a circle. Since the string is 2ft long, the radius of the circle is 2. The position function \(\vrt = \la 2\cos(t) , 2\sin(t) \ra\) describes a circle with radius 2, centered at the origin, but makes a full revolution every \(2\pi\) seconds, not two revolutions per second. We modify the period of the trigonometric functions to be 1/2 by multiplying \(t\) by \(4\pi\text{.}\) The final position function is thus

   \begin{equation*} \vrt = \la 2\cos(4\pi t), 2\sin(4\pi t)\ra\text{.} \end{equation*}

   (Plot this for \(0\leq t\leq 1/2\) to verify that one revolution is made in 1/2 a second.)

2. To find \(\vat\text{,}\) we take the derivative of \(\vrt\) twice.

   \begin{align*} \vvt = \vrp(t) \amp = \la -8\pi \sin(4\pi t), 8\pi \cos(4\pi t)\ra\\ \vat =\vrp'(t) \amp = \la -32\pi^2 \cos(4\pi t), -32\pi^2 \sin(4\pi t) \ra\\ \amp = -32\pi^2\la \cos(4\pi t), \sin(4\pi t)\ra\text{.} \end{align*}

   Note how \(\vat\) is parallel to \(\vrt\text{,}\) but has a different magnitude and points in the opposite direction. Why is this?

   Recall the classic physics equation, “Force \(=\) mass × acceleration.” A force acting on a mass induces acceleration (i.e., the mass moves); acceleration acting on a mass induces a force (gravity gives our mass a weight). Thus force and acceleration are closely related. A moving ball “wants” to travel in a straight line. Why does the ball in our example move in a circle? It is attached to the boy's hand by a string. The string applies a force to the ball, affecting its motion: the string accelerates the ball. This is not acceleration in the sense of “it travels faster;” rather, this acceleration is changing the velocity of the ball. In what direction is this force/acceleration being applied? In the direction of the string, towards the boy's hand.

   The magnitude of the acceleration is related to the speed at which the ball is traveling. A ball whirling quickly is rapidly changing direction/velocity. When velocity is changing rapidly, the acceleration must be “large.”

3. When the boy releases the string, the string no longer applies a force to the ball, meaning acceleration is \(\vec 0\) and the ball can now move in a straight line in the direction of \(\vec v(t)\text{.}\)

   Let \(t=t_0\) be the time when the boy lets go of the string. The ball will be at \(\vec r(t_0)\text{,}\) traveling in the direction of \(\vec v(t_0)\text{.}\) We want to find \(t_0\) so that this line contains the point \((0,10)\) (since the tree is 10 ft directly in front of the boy).

   

   There are many ways to find this time value. We choose one that is relatively simple computationally. As shown in Figure 11.3.8, the vector from the release point to the tree is \(\la 0,10\ra - \vec r(t_0)\text{.}\) This line segment is tangent to the circle, which means it is also perpendicular to \(\vec r(t_0)\) itself, so their dot product is 0.

   \begin{align*} \vec r(t_0) \cdot \big(\la 0,10\ra - \vec r(t_0)\big) \amp =0\\ \la 2\cos(4\pi t_0), 2\sin(4\pi t_0)\ra \cdot \la -2\cos(4\pi t_0),10-2\sin(4\pi t_0)\ra \amp =0\\ -4\cos^2(4\pi t_0) + 20\sin(4\pi t_0)-4\sin^2(4\pi t_0) \amp = 0\\ 20\sin(4\pi t_0) - 4 \amp =0\\ \sin(4\pi t_0) \amp =1/5\\ 4\pi t_0 \amp = \sin^{-1}(1/5)\\ 4\pi t_0 \amp \approx 0.2 + 2\pi n,\\ \end{align*}

   where \(n\) is an integer. Solving for \(t_0\) we have:

   \begin{align*} t_0 \amp \approx 0.016 + n/2 \end{align*}

   This is a wonderful formula. Every 1/2 second after \(t=0.016\,\text{s}\) the boy can release the string (since the ball makes 2 revolutions per second, he has two chances each second to release the ball).

With \(\vrt = \la \cos(t) ,\sin(t) , t\ra\text{,}\) we have:

The speed of the object is \(\norm{\vvt} = \sqrt{(-\sin(t) )^2+\cos^2(t) +1} = \sqrt{2}\) ^m⁄_min; it moves at a constant speed. Note that the object does not accelerate in the \(z\)-direction, but rather moves up at a constant rate of 1 ^m⁄_min.