Calculus III 12.03 Velocity and Acceleration

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12.3 Velocity[1] and Acceleration[2]

• Describe the velocity and acceleration associated with a vector-valued function.
• Use a vector-valued function to describe projectile motion.

Velocity and Acceleration

Parametric equations, curves, vectors, and vector-valued functions can be combined into a model that describes motion along a curve. This discussion starts with a two-dimensional curve and extends the model to three-dimensions.

As an object moves along a curve in two-dimensions the coordinates \(x\) and \(y\) for its mass center are functions for time \(t\). Rather than using the letters \(f\) and \(g\) to represent these two functions the notation \(x=x(t)\) and \(y=y(t)\) are used. The position vector \( \textbf{r}(t) \) takes the form

\( \textbf{r}(t) = x(t)\textbf{i} + y(t)\textbf{j} \:\:\:\: \) Position vector

Velocity and acceleration are both vector quantities having magnitude and direction. This model elegantly describes the velocity as the first derivative, \( \textbf{r}^{\prime}(t) \), and acceleration as the second derivative, \( \textbf{r}^{\prime \prime}(t) \), for the vector-valued function :\( \textbf{r}(t) \). To find the velocity and acceleration vectors at a given time \(t\), consider a point \(Q(x(t+ \Delta t), y(t+ \Delta t))\) that is approaching the point \(P(x(t),y(t))\) along the curve \(C\) given by \( \textbf{r}(t) = x(t)\textbf{i} + y(t)\textbf{j} \), as shown in Figure 12.3.1. As \(\Delta t \rightarrow 0\), the direction for the vector \( \vec{PQ} \), denoted by \(\Delta \textbf{r} \), approaches the direction of motion at time \(t\).

\( \Delta \textbf{r} \)	\(= \textbf{r}(t+ \Delta t) - \textbf{r}(t) \)
$$ \frac{ \Delta \textbf{r} }{ \Delta t} $$	$$= \frac{ \textbf{r}(t+ \Delta t) - \textbf{r}(t) }{ \Delta t}$$
$$ \lim_{\Delta t \rightarrow 0} \frac{ \Delta \textbf{r} }{ \Delta t} $$	$$= \lim_{\Delta t \rightarrow 0} \frac{ \textbf{r}(t+ \Delta t) - \textbf{r}(t) }{ \Delta t}$$

If this limit exists, it is defined as the velocity vector or tangent vector to the curve at point \(P\). This is the same limit used to define \( \textbf{r}^{\prime}(t) \). Therefore, \( \textbf{r}^{\prime}(t) \) describes the direction at time \(t\). The vector \( \textbf{r}^{\prime}(t) \)'s magnitude

\( \| \textbf{r}^{\prime}(t) \| = \| x^{\prime}(t)\textbf{i} + y^{\prime}(t)\textbf{j} \| = \sqrt{ [x^{\prime}(t)]^{2} + [y^{\prime}(t)]^{2}} \)

gives the object's speed at time \(t\).

As \( \Delta t \rightarrow 0,\: \Delta \textbf{r}/ \Delta t \) approaches the velocity vector.
Figure 12.3.1

Definition 12.3.1 Velocity and Acceleration

If \(x\) and \(y\) are twice-differentiable functions for \(t\), and \( \textbf{r} \) is a vector-valued function given by \( \textbf{r}(t) = x(t)\textbf{i} + y(t)\textbf{j} \), then the velocity vector, acceleration vector, and speed at time \(t\) are as follows.

Velocity	\(= \textbf{v}(t)\)	\(= \textbf{r}^{\prime}(t) \)	\(= x^{\prime}(t)\textbf{i} + y^{\prime}(t)\textbf{j} \)	Two dimensions
			\(= x^{\prime}(t)\textbf{i} + y^{\prime}(t)\textbf{j} + z^{\prime}(t)\textbf{k} \:\:\:\: \)	Three dimensions
Acceleration	\(= \textbf{a}(t)\)	\(= \textbf{r}^{\prime \prime}(t) \)	\(= x^{\prime \prime}(t)\textbf{i} + y^{\prime \prime}(t)\textbf{j} \)
			\(= x^{\prime \prime}(t)\textbf{i} + y^{\prime \prime}(t)\textbf{j} + z^{\prime \prime}(t)\textbf{k}\)
Speed	\(= \| \textbf{v}(t) \| \)	\(= \| \textbf{r}^{\prime}(t) \| \)	\(= \sqrt{ [x^{\prime}(t)]^{2} + [y^{\prime}(t)]^{2}} \)
			\(= \sqrt{ [x^{\prime}(t)]^{2} + [y^{\prime}(t)]^{2} + [z^{\prime}(t)]^{2} }\)

Example 12.3.1 Velocity and Acceleration Along a Two-Dimensional Curve

The particle moves around the circle at a constant speed. Figure 12.3.2

Find the velocity vector, speed, and acceleration vector for a particle that moves along the plane curve \(C\) described by $$ \textbf{r}(t) = 2 \sin \frac{t}{2}\textbf{i} + 2 \cos \frac{t}{2}\textbf{j} \:\:\:\: \color{red}{ \text{Position vector }} $$ and sketch the graph. Solution The velocity vector is $$ \textbf{v}(t) = \textbf{r}^{\prime}(t) = \cos \frac{t}{2}\textbf{i} - \sin \frac{t}{2}\textbf{j} \:\:\:\: \color{red}{ \text{Velocity vector}} $$ The speed, at any time, is $$ \| \textbf{r}^{\prime}(t) \| = \sqrt{ \cos^{2} \frac{t}{2}\textbf{i} - \sin^{2} \frac{t}{2}\textbf{j} } = 1\:\:\:\: \color{red}{ \text{Speed}} $$ The acceleration vector is $$ \textbf{a}(t) = \textbf{r}^{\prime \prime}(t) = - \frac{1}{2} \sin \frac{t}{2}\textbf{i} - \frac{1}{2} \cos \frac{t}{2}\textbf{j} \:\:\:\: \color{red}{ \text{Acceleration vector}} $$ The parametric equations for the curve \(C\) are: $$ x=2 \sin \frac{t}{2}, \: y = 2 \cos \frac{t}{2}$$ Eliminating the parameter \(t\) yields the cartesian equation \( x^{2} + y^{2} = 4.\) Curve \(C\) is a circle with radius 2 centered at the origin, as shown in Figure 12.3.2. Because the velocity vector has a constant magnitude but a changing direction as \(t\) increases, the particle moves around the circle at a constant speed.

Example 12.3.2 Velocity and Acceleration Vectors in Two-Dimensions

At each point on the curve, the acceleration vector points to the right. Figure 12.3.3 At each point in the comet’s orbit, the acceleration vector points toward the sun. Figure 12.3.4

Sketch the object's path as it moves along the a two-dimensional curve given by \( \textbf{r}(t) = (t^{2}-4)\textbf{i} + t \textbf{j} \:\:\:\: \)Position vector and find the velocity and acceleration vectors when \(t=0\) and \(t=2\). Solution The position vector yields the parametric equations \(x=t^{2}-4\) and \(y=t\). Substituting \(y\) for \(t\) in the first equation produces a parabola described by \(x=y^{2}-4 \:\:\:\: \) Cartesian equation as shown in Figure 12.3.3. The velocity vector, at any time, is \( \textbf{v}(t) = \textbf{r}^{\prime}(t) = 2t \textbf{i} + \textbf{j} \:\:\:\: \) Velocity equation The acceleration vector, at any time, is \( \textbf{a}(t) = \textbf{r}^{\prime \prime}(t) = 2\textbf{i}. \:\:\:\: \) Acceleration equation When \(t=0\), the velocity and acceleration vectors are \( \textbf{v}(\color{red}{0}) = 2(\color{red}{0})\textbf{i}+\textbf{j} = \textbf{j}\) and \(\textbf{a}(\color{red}{0}) = 2(\color{red}{0})\textbf{i}.\) The acceleration vector is constant with a magnitude two and moves from right to left. This mean the speed is decreasing as the object moves toward the parabola's vertex at the origin and increases as it moves away from the parabola's vertex. Comets that travel on parabolic paths through our solar system do not display this motion. Their acceleration vector always points to the origin, the sun, which increases the comet’s speed as it approaches the vertex and decreases as it moves away from the vertex, as shown in Figure 12.3.4.

Example 12.3.3 Velocity and Acceleration Vectors in Three-Dimensions

Figure 12.3.5

Sketch the path for an object moving along the three-dimensional curve \(C\) given by \( \textbf{r}(t) = t\textbf{i} + t^{2}\textbf{j} +3t\textbf{k} \:\:\:\: \)Position vector and find the velocity and acceleration vectors when \(t=1\). Solution The position vector yields the parametric equations \(x=t\) and \(y=t^{3}\). Substituting \(x\) for \(t\) produces a cubic cylinder described by \(y=x^{3}. \:\:\:\: \) Cartesian equation Because \(z=3t\), the object starts at the origin, (0,0,0), and moves upward as \(t\) increases, as shown in Figure 12.3.5. Taking the first and second derivatives produces the velocity and acceleration vectors. \( \textbf{v}(t) = \textbf{r}^{\prime}(t) = \textbf{i} + 3t^{2}\textbf{j} + 3\textbf{k} \:\:\:\: \) Velocity vector \( \textbf{a}(t) = \textbf{r}^{\prime \prime}(t) = 6t\textbf{j} \:\:\:\: \) Acceleration vector When \(t=1\), the velocity and acceleration vectors are: \( \textbf{v}(\color{red}{1}) = \textbf{r}^{\prime}(\color{red}{1}) = \textbf{i} + 3\textbf{j} + 3\textbf{k} \) \( \textbf{a}(\color{red}{1}) = \textbf{r}^{\prime \prime}(\color{red}{1}) = 6\textbf{j} \)

Example 12.3.4 Finding a Position Vector by Integration

The object takes two seconds to move from point (1,2,0) to point (1,4,4) along the curve. Figure 12.3.6

An object starts from rest at the point (1,2,0) and moves with an acceleration \( \textbf{a}(t) = \textbf{j} + 2\textbf{k} \:\:\:\: \) Acceleration vector where \( \| \textbf{a}(t) \| \) is measured in feet per second per second. Find the object's location after \(t=2\). Solution From the question, the object’s initial conditions are at rest at point (1,2,0). This makes the initial velocity, \( \textbf{v}(0) = \textbf{0} \). Because the the object starts at \((x,y,z)= (1,2,0)\) the initial position vector is \( \textbf{r}(0) = x(0)\textbf{i} + y(0)\textbf{j} + z(0)\textbf{k} = 1\textbf{i} + 2\textbf{j} + 0\textbf{k} = \textbf{i} + 2\textbf{j}.\:\:\:\: \)Position vector To find the position vector, integrate twice, each time using an initial condition to solve for the integration constant. \( \textbf{v}(t) \) $$= \int \textbf{a}(t)\:dt \:\:\:\: \color{red}{\text{ Velocity vector }}$$ $$= \int ( \textbf{j} + 2\textbf{k}) \:dt$$ \(= t\textbf{j} + 2t\textbf{k} + \textbf{C} \) where \( \textbf{C}= C_{1}\textbf{i} + C_{2}\textbf{j} + C_{3}\textbf{k} \). Letting \(t=0\) and applying the initial condition \( \textbf{v}(0) = \textbf{0} \) produces \( \textbf{v}(0) = C_{1}\textbf{i} + C_{2}\textbf{j} + C_{3}\textbf{k} = \textbf{0} \rightarrow C_{1} = C_{2} = C_{3} = 0. \) The velocity at any time \(t\) is \( \textbf{v}(t) = t\textbf{j} + 2t\textbf{k}. \) Integrating once more produces \( \textbf{r}(t) \) $$= \int \textbf{v}(t)\:dt \color{red}{\text{ Velocity vector }}$$ $$= \int ( t\textbf{j} + 2t\textbf{k}) \:dt$$ $$= \frac{t^{2}}{2}\textbf{j} + t^{2}\textbf{k} + \textbf{C} $$ where \( \textbf{C}= C_{4}\textbf{i} + C_{5}\textbf{j} + C_{6}\textbf{k} \). Letting \(t=0\) and applying the initial condition \( \textbf{r}(0) = \textbf{j} + 2\textbf{k} \) produces \( \textbf{r}(0) = C_{4}\textbf{i} + C_{5}\textbf{j} + C_{6}\textbf{k} = \textbf{j} + 2\textbf{k} \rightarrow C_{4} = 1,\: C_{5} = 2,\: C_{6} = 0. \) The position vector is $$ \textbf{r}(t) = \textbf{i} + \left( \frac{t^{2}}{2} + 2 \right)\textbf{j} + t^{2}\textbf{k}. \:\:\:\: \color{red}{ \text{Position vector}} $$ The object's location after \(t=2\) is given by \( \textbf{r}(2) = \textbf{i} + 4\textbf{j} +4\textbf{k} \) as shown in Figure 12.3.6

Projectile Motion

Figure 12.3.7

A projectile in motion through the air with only gravity acting on it can be described using a vertical plane using the \(xy\)-coordinate system with the origin and terminal points on the Earth's surface, as shown in Figure 12.3.7. For a projectile with mass \(m\), the force due to gravity is \( \textbf{F} = -mg \textbf{j} \:\:\:\: \)Force due to gravity where the acceleration due to gravity is \(g=32\) feet per second per second, or 9.81 meters per second per second. By Newton’s Second Law of Motion, this same force produces an acceleration \( \textbf{a} = \textbf{a}(t) \) and satisfies the equation \(\textbf{F}=m\textbf{a} \). The projectile's acceleration is given by \( m\textbf{a} = -mg \textbf{j} \), which reduces to \( \textbf{a} = -g \textbf{j}. \:\:\:\: \)Projectile acceleration

Example 12.3.5 Find the Position Vector for a Projectile using Derivation

A projectile with mass \(m\) is launched from an initial position \( \textbf{r}_{0} \) with an initial velocity \( \textbf{v}_{0} \). Find its position vector as a function for time.
Solution Begin with the acceleration equation \( \textbf{a} = -g \textbf{j} \) and integrate twice.

\( \textbf{v}(t) \)	$$= \int \textbf{a}(t) \:dt = \int -g \textbf{j} \:dt = -gt \textbf{j} + \textbf{C}_{1} $$
\( \textbf{r}(t) \)	$$= \int \textbf{v}(t) \:dt = \int ( -gt \textbf{j} + \textbf{C}_{1}) \:dt = - \frac{1}{2}gt^{2} \textbf{j} + \textbf{C}_{1}t + \textbf{C}_{2} $$

The constant vectors \( \textbf{C}_{1} \) and \( \textbf{C}_{2} \) can be solved using the fact that \( \textbf{v}(0) = \textbf{v}_{0} \) and \( \textbf{r}(0) = \textbf{r}_{0} \) to produce

\( \textbf{C}_{1} = \textbf{v}_{0} \)

\( \textbf{C}_{2} = \textbf{r}_{0} \).

Therefore, the position vector is

$$ \textbf{r}(t) = - \frac{1}{2}gt^{2} \textbf{j} + t\textbf{v}_{0} + \textbf{r}_{0}. $$

Figure 12.3.8

In many projectile problems, the constant vectors \( \textbf{r}_{0} \) and \( \textbf{v}_{0} \) are not given explicitly. Often only the initial height \(h\) the initial speed \(v_{0} \), and the launch angle \( \theta \) relative to the Earth are given, as shown in Figure 12.3.8. From the given height, \( \textbf{r}_{0} =h \textbf{j}\) can be deduced. Because the speed gives the magnitude for the initial velocity, it follows that \(v_{0} = \| \textbf{v}_{0} \| \) which produces \( \textbf{v}_{0} \) \(= x\textbf{i} + y\textbf{j} \) \(= \left( \| \textbf{v}_{0} \| \cos \theta \right)\textbf{i} + \left( \| \textbf{v}_{0} \| \sin \theta \right)\textbf{j} \) \(= v_{0} \cos \theta \textbf{i} + v_{0} \sin \theta \textbf{j} \). The position vector can be written in the form \( \textbf{r}(t) \) $$ = - \frac{1}{2}gt^{2} \textbf{j} + t\textbf{v}_{0} + \textbf{r}_{0} $$ Position vector $$ = - \frac{1}{2}gt^{2} \textbf{j} + tv_{0} \cos \theta \textbf{i} + tv_{0} \sin \theta \textbf{j} + h\textbf{j} $$ $$= \left( v_{0} \cos \theta \right)t\textbf{i} + \left[ h + (v_{0} \sin \theta )t - \frac{1}{2}gt^{2} \right]\textbf{j}.$$

Theorem 12.3.1 Position Vector for a Projectile

Neglecting air resistance, the path for a projectile launched from an initial height \(h\) with initial speed \(v_{0} \) and an elevation angle \( \theta \) is described by the vector function

$$ \textbf{r}(t) = \left( v_{0} \cos \theta \right)t\textbf{i} + \left[ h + (v_{0} \sin \theta )t - \frac{1}{2}gt^{2} \right]\textbf{j} $$

where \(g\) is the acceleration due to gravity.

Example 12.3.6 Describing a Baseball's Path

Figure 12.3.9

A baseball is hit 3 feet above ground level at 100 feet per second and at a 45° angle with respect to the ground, as shown in Figure 12.3.9. Find the maximum height reached by the baseball. Will it clear a 10-foot-high fence located 300 feet from home plate? Solution From the question \(h=3, \: v_{0} = 100 \text{, and } \theta = 45^{\circ } \). Using Theorem 12.3.1 with \(g=32 \) feet per second per second produces \( \textbf{r}(t) \) $$= \left( 100 \cos \frac{\pi}{4} \right)t\textbf{i} + \left[3 + \left( 100 \sin \frac{\pi}{4} \right)t - 16t^{2} \right]\textbf{j}$$ \(= \left( 50 \sqrt{2}t \right)\textbf{i} + \left(3 + 50 \sqrt{2}t - 16t^{2} \right)\textbf{j} \). The velocity vector is \( \textbf{v}(t) = \textbf{r}^{\prime}(t) = 50 \sqrt{2}\textbf{i} +(50 \sqrt{2}- 32t)\textbf{j}. \) The maximum height occurs when \( y^{\prime}(t)= 50 \sqrt{2}- 32t \) is equal to 0, which implies that $$ t = \frac{25 \sqrt{2}}{16} \approx 2.21 \text{ seconds. } $$ The maximum height reached by the ball is \(y\) $$= 3 + 50 \sqrt{2} \left( \frac{25 \sqrt{2}}{16} \right) - 16 \left( \frac{25 \sqrt{2}}{16} \right)^{2} $$ $$= \frac{649}{8} $$ \( \approx 81 \: \) feet. Maximum height when \(t \approx 2.21\) seconds The ball is 300 feet from where it was hit when \( 300 = x(t) \rightarrow 300 = 50 \sqrt{2}t.\) Solving this equation for \(t\) produces \(t = 3 \sqrt{2} \approx 4.24\) seconds. At this time, the ball's height is \(y\) \(= 3 + 50 \sqrt{2}( 3 \sqrt{2}) - 16 ( 3 \sqrt{2})^{2} \) \(= 303-288\) \(= 15\) feet. Height when \(t \approx 4.24\) seconds Therefore, the ball clears the 10-foot fence for a home run.

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Parent Article: Calculus III 12 Vector-Valued Functions