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. $$ \vec{r}(t)=\left\langle t^{2}, t^{2}-t^{3}\right\rangle \text { on }[-1,1] $$

Short answer

The speed is minimized at \(t = 0\) and maximized at \(t = -1\) on the interval \([-1, 1]\).

Step-by-step solution

Differentiate the Position Function

To find the speed, we first need to determine the velocity vector by differentiating the position function \(\vec{r}(t)\) with respect to \(t\). Given \(\vec{r}(t) = \langle t^2, t^2 - t^3 \rangle\), we find \(\vec{v}(t) = \frac{d}{dt}\vec{r}(t) = \langle 2t, 2t - 3t^2 \rangle\).

Calculate the Speed

The speed of the object is the magnitude of the velocity vector \(\vec{v}(t)\). We calculate it as \(||\vec{v}(t)|| = \sqrt{(2t)^2 + (2t - 3t^2)^2}\). Simplifying, we have \(||\vec{v}(t)|| = \sqrt{4t^2 + (2t - 3t^2)^2} = \sqrt{4t^2 + (4t^2 - 12t^3 + 9t^4)}\). Thus, the speed \(s(t) = \sqrt{9t^4 - 12t^3 + 8t^2}\).

Find Critical Points

To determine where the speed is minimized/maximized, we find the critical points of \(s(t) = \sqrt{9t^4 - 12t^3 + 8t^2}\). First, find the derivative \(s'(t)\) and solve \(s'(t) = 0\). Differentiating, we have: \(s'(t) = \frac{1}{2\sqrt{9t^4 - 12t^3 + 8t^2}}(36t^3 - 36t^2 + 16t)\). Solving \(36t^3 - 36t^2 + 16t = 0\) gives the critical points. Factor to get \(t(36t^2 - 36t + 16) = 0\). Therefore, possible solutions are \(t=0\) or the roots of \(36t^2 - 36t + 16=0\).

Evaluate the Quadratic

Solve the quadratic \(36t^2 - 36t + 16 = 0\) using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 36\), \(b = -36\), \(c = 16\). The discriminant \(b^2 - 4ac = 1296 - 2304 = -1008\) which is negative, indicating there are no real roots. So the only real critical point is \(t=0\).

Test Critical Points and Endpoints

Evaluate \(s(t)\) at the critical point \(t = 0\) and the endpoints \(t = -1\) and \(t = 1\) of the interval \([-1, 1]\). For \(t = 0\), \(s(0) = \sqrt{0} = 0\). For \(t = 1\), \(s(1) = \sqrt{9 - 12 + 8} = \sqrt{5}\). For \(t = -1\), \(s(-1) = \sqrt{9 + 12 + 8} = \sqrt{29}\).

Determine Maxima and Minima

By comparing the calculated speeds, we find that the minimum speed is \(0\) at \(t=0\), and the maximum speed is \(\sqrt{29}\) at \(t=-1\).

Key concepts

Position Function

In calculus, a position function \(\vec{r}(t)\) provides the position of an object at a given time \(t\). For this specific problem, the position function is given as \(\vec{r}(t) = \langle t^2, t^2 - t^3 \rangle\). This vector function shows how both the x and y positions of the object change over time.
Understanding position functions is important because they track the exact path of a moving object. This function is derived from the object's motion and can include multiple components in space.
In practical scenarios, the position function allows us to predict where an object will be at any time, given its initial conditions.

Critical Points

Critical points in calculus refer to values of \(t\) where the derivative of a function either equals zero or is undefined. For our problem, we want the values of \(t\) where the speed is minimized or maximized. These correspond to the critical points of the speed function.

To find them, we differentiate the speed function \(s(t) = \sqrt{9t^4 - 12t^3 + 8t^2}\) to get its derivative. Set this derivative equal to zero and solve for \(t\). The process involves simplifying the derivative's equation to find the points at which the slope of \(s(t)\) changes.

• In the example, this leads to finding where \(36t^3 - 36t^2 + 16t = 0\).
• Factors or further solving will help isolate these key \(t\) values that define potential minimum or maximum speeds.

Critical points often hold significant physical meaning in terms of moving objects, as they highlight moments of transition in speed.

Velocity Vector

The velocity vector is a key derivative of the position function and is central to understanding an object's motion. Differentiating the position function \(\vec{r}(t)\) with respect to time \(t\) gives us the velocity vector \(\vec{v}(t) = \langle 2t, 2t - 3t^2 \rangle\).

The elements of this vector describe the rate of change of position along each axis. For our problem:

• \(2t\) models the rate of change for the x-component.
• \(2t - 3t^2\) accounts for the y-component's velocity.

The velocity vector tells us both speed and direction of motion for the object and is a crucial link between position and acceleration in advanced motion problems.
This calculation is an essential technique learned in calculus, often forming the foundational basis for further exploration of dynamics in physics.

Quadratic Formula

The quadratic formula is an efficient tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). In our case, it is used to find the critical points of the derived function.

The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, the primary task is to determine whether the quadratic equation has real solutions, as real roots indicate potential critical points.

• Plug in coefficients \(a = 36\), \(b = -36\), and \(c = 16\).
• Calculate the discriminant, \(b^2 - 4ac\), to check if it is positive, zero, or negative.

In this problem, the discriminant was negative (-1008), signifying no real roots. As a consequence, the only critical point evaluated is \(t = 0\), emphasizing the importance of evaluating all options.
The quadratic formula is versatile and deeply powerful in mathematics for analyzing polynomial functions.