Question

A position function \(\vec{r}(t)\) of an object is given. Find the speed of the object in terms of \(t,\) and find where the speed is minimized/maximized on the indicated interval. Projectile Motion: \(\vec{r}(t)=\left\langle\left(v_{0} \cos \theta\right) t,-\frac{1}{2} g t^{2}+\left(v_{0} \sin \theta\right) t\right\rangle\) on \(\left[0, \frac{2 v_{0} \sin \theta}{g}\right]\)

Short answer

Minimum speed is at \( t = \frac{v_{0} \sin \theta}{g} \) with speed \( v_{0} \cos \theta \); maximum speed is \( v_{0} \) at endpoints.

Step-by-step solution

Understand the position vector

The position vector \( \vec{r}(t) \) is given as \( \left\langle \left(v_{0} \cos \theta\right) t, -\frac{1}{2} g t^{2}+\left(v_{0} \sin \theta\right) t \right\rangle \). This represents the trajectory of an object moving under the influence of gravity, where \( v_0 \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

Find the velocity vector

Differentiate the position vector \( \vec{r}(t) \) with respect to \( t \) to find the velocity vector \( \vec{v}(t) \).\[\vec{v}(t) = \frac{d}{dt} \left\langle \left(v_{0} \cos \theta\right) t, -\frac{1}{2} g t^{2}+\left(v_{0} \sin \theta\right) t \right\rangle = \left\langle v_{0} \cos \theta, -g t + v_{0} \sin \theta \right\rangle \]

Calculate the speed

The speed of the object is the magnitude of the velocity vector \( \vec{v}(t) \). Calculate it by finding the magnitude.\[speed(t) = \left\| \vec{v}(t) \right\| = \sqrt{(v_{0} \cos \theta)^2 + (-g t + v_{0} \sin \theta)^2}\]

Simplify the speed function

Simplify the expression inside the square root for the speed:\[speed(t) = \sqrt{(v_{0}^2 \cos^2 \theta) + (g^2 t^2 - 2 g t v_{0} \sin \theta + v_{0}^2 \sin^2 \theta)}\]Simplify further:\[speed(t) = \sqrt{v_{0}^2(\cos^2 \theta + \sin^2 \theta) - 2 g t v_{0} \sin \theta + g^2 t^2}\]Since \( \cos^2 \theta + \sin^2 \theta = 1 \), we have:\[speed(t) = \sqrt{v_{0}^2 - 2 g t v_{0} \sin \theta + g^2 t^2}\]

Find critical points of the speed function

To find where the speed is minimized or maximized, find the derivative of the speed function and set it equal to zero to find critical points.Compute \( \frac{d}{dt} \left( \sqrt{v_{0}^2 - 2 g t v_{0} \sin \theta + g^2 t^2} \right) \) and set it to zero. For simplicity, find the critical points by solving:\[- g v_{0} \sin \theta + g^2 t = 0 \]This gives:\[t = \frac{v_{0} \sin \theta}{g}\]

Evaluate endpoints and critical points

Evaluate the speed at the endpoints and the critical point to determine where the speed is minimized or maximized.- At \( t = 0 \):\[\text{Speed} = \sqrt{v_{0}^2} = v_{0}\]- At \( t = \frac{2 v_{0} \sin \theta}{g} \):\[\text{Speed} = v_{0}\]- At \( t = \frac{v_{0} \sin \theta}{g} \):\[\text{Speed} = \sqrt{v_{0}^2 - g \times \frac{v_{0}^2 \sin \theta}{g} \times \sin \theta + \left(g \times \frac{v_{0} \sin \theta}{g}\right)^2} = \sqrt{v_{0}^2 - v_{0}^2 \sin^2 \theta} = v_{0} \cos \theta \]

Conclusion

The minimum speed occurs at \( t = \frac{v_{0} \sin \theta}{g} \) with a speed of \( v_{0} \cos \theta \). The speed is maximum at \( t = 0 \) and \( t = \frac{2 v_{0} \sin \theta}{g} \), both equal to \( v_{0} \).

Key concepts

Position Vector Analysis

In the realm of calculus, position vector analysis serves as a critical foundation in understanding how objects move through space. A position vector essentially provides a way to pinpoint the location of an object at any given time, and is often represented in terms of time, \(t\). It's like a roadmap that tells us where an object is at different times as it moves. For our exercise, the position vector is given by \( \vec{r}(t) = \left\langle \left(v_{0} \cos \theta\right) t, -\frac{1}{2} g t^{2}+\left(v_{0} \sin \theta\right) t \right\rangle \). This expression describes the motion of a projectile, factoring in initial velocity \(v_0\), the angle of launch \(\theta\), and the constant pull of gravity \(g\). By understanding the position vector, we can derive other important quantities, such as velocity and speed.

Velocity Vector Differentiation

The next crucial concept is velocity vector differentiation, which involves calculating how the position of an object changes with time. This is done by differentiating the position vector with respect to time. The velocity vector essentially shows the direction and speed of the trajectory of the object at any point in time. For the given position vector, differentiating with respect to \(t\) produces the velocity vector \( \vec{v}(t) = \left\langle v_{0} \cos \theta, -g t + v_{0} \sin \theta \right\rangle \). This velocity vector tells us how fast and in what direction the object is moving at any specific point on its path. The horizontal component \(v_{0} \cos \theta\) remains constant, as no forces act horizontally, while the vertical component \(-g t + v_{0} \sin \theta\) changes due to gravity.

Critical Points in Calculus

Identifying critical points is a pivotal step in calculus, especially when determining where a function reaches its maximum or minimum values. To find these points in the context of speed, we differentiate the speed function and equate it to zero, which allows us to determine the values of \(t\) that minimize or maximize speed. Critical points often correspond to changes in direction or extreme points on a graph. For our exercise, setting the derivative of the speed equation equal to zero, \(- g v_{0} \sin \theta + g^2 t = 0\), gives the critical point \( t = \frac{v_{0} \sin \theta}{g} \). This critical point indicates the moment the projection reaches its peak speed in the vertical direction and slows to change its motion vertically, which is where the velocity is entirely horizontal, thus minimizing the speed.

Projectile Motion Calculations

Projectile motion calculations apply the concepts we've discussed to determine how an object behaves under the influence of gravity after being launched at a specific angle. By understanding both the position and velocity vectors, we can analyze how objects move through air, predict where they will land, and evaluate at what points their speed is minimized or maximized. In our exercise, the interval \( \left[0, \frac{2 v_{0} \sin \theta}{g} \right] \) represents the entire flight duration of the projectile. Evaluations of the speed at calculated points show us that at \(t = 0\) and \(t = \frac{2 v_{0} \sin \theta}{g}\), the speed is at its maximum, equating to \(v_{0}\). However, at \(t = \frac{v_{0} \sin \theta}{g}\), the speed is minimized at \(v_{0} \cos \theta\), due to the entire velocity being horizontal at this peak of the projectile's path.