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. $$ \vec{r}(t)=\langle 2 \cos t, 5 \sin t\rangle \text { on }[0,2 \pi] $$

Short answer

The speed is minimized at \(t = \frac{\pi}{2}, \frac{3\pi}{2}\) with speed 2, and maximized at \(t = 0, \pi, 2\pi\) with speed 5.

Step-by-step solution

Find the velocity function

The velocity function \(\vec{v}(t)\) is the derivative of the position function \(\vec{r}(t)\) with respect to \(t\). Differentiate each component to get: \[\vec{v}(t) = \left\langle \frac{d}{dt}[2 \cos t], \frac{d}{dt}[5 \sin t] \right\rangle = \left\langle -2 \sin t, 5 \cos t \right\rangle\]

Find the speed function

The speed of the object is the magnitude of the velocity vector \(\vec{v}(t)\). Compute the magnitude as follows:\[||\vec{v}(t)|| = \sqrt{(-2 \sin t)^2 + (5 \cos t)^2} = \sqrt{4 \sin^2 t + 25 \cos^2 t}\]

Simplify the speed function

Apply the Pythagorean identity \(\sin^2 t + \cos^2 t = 1\) to express the speed function more simplistically: \[||\vec{v}(t)|| = \sqrt{4 \sin^2 t + 25 (1 - \sin^2 t)} = \sqrt{4 \sin^2 t + 25 - 25 \sin^2 t} = \sqrt{25 - 21 \sin^2 t}\]

Set up the derivative to find critical points

To find the points where the speed is minimized or maximized, differentiate the speed function with respect to \(t\) and set it equal to zero. Let \(u = 21\sin^2 t\) so \[f(t) = \sqrt{25 - u}\] then \[f(t) = (25 - u)^{1/2}\]\[\frac{df}{dt} = \frac{1}{2}(25 - u)^{-1/2} \cdot (-21 \sin(2t)) = 0\]This gives \(-21 \sin(2t) = 0\), implying \(\sin(2t) = 0\).

Solve for critical values of t

\(\sin(2t) = 0\) gives \(2t = n\pi\), where \(n\) is an integer. Solve for \(t\):\[t = \frac{n\pi}{2}, \quad n \in \{0, 1, 2, 3, 4\}\]Given the interval \([0, 2\pi]\), the valid \(t\)s are \(t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\).

Evaluate and determine minimum and maximum speeds

Calculate \(||\vec{v}(t)||\) for each critical value and endpoints:- \(t = 0\) and \(t = 2\pi\) : \(||\vec{v}(t)|| = \sqrt{25} = 5\) - \(t = \frac{\pi}{2}\) and \(t = \frac{3\pi}{2}\) : \(||\vec{v}(t)|| = \sqrt{4} = 2\)- \(t = \pi\) : \(||\vec{v}(t)|| = \sqrt{25} = 5\)The speed is minimized at \(t = \frac{\pi}{2}, \frac{3\pi}{2}\) and maximized at \(t = 0, \pi, 2\pi\).

Key concepts

Velocity Function

The velocity function represents the rate of change of an object's position with respect to time. Essentially, it tells us how fast, and in which direction, an object is moving at any given time. This is calculated by taking the derivative of the position function. In our example, the position function given is \(\vec{r}(t) = \langle 2\cos t, 5\sin t \rangle\). We derive each component separately:

• \(\frac{d}{dt}[2 \cos t] = -2 \sin t\)
• \(\frac{d}{dt}[5 \sin t] = 5 \cos t\)

Thus, the velocity function is \(\vec{v}(t) = \langle -2 \sin t, 5 \cos t \rangle\). This vector tells us how the position changes as time progresses. The \(-2 \sin t\) indicates movement in the direction of the first component, and \(5 \cos t\) in the direction of the second.

Speed Function

Speed is a scalar quantity that reflects how fast an object is moving, irrespective of its direction. To determine the speed from the velocity function, we compute the magnitude of the velocity vector. The formula for magnitude is: \[||\vec{v}(t)|| = \sqrt{(-2 \sin t)^2 + (5 \cos t)^2}\]When broken down, the speed is calculated as: - \((-2 \sin t)^2 = 4 \sin^2 t\)- \((5 \cos t)^2 = 25 \cos^2 t\)Thus, the speed is:\[||\vec{v}(t)|| = \sqrt{4 \sin^2 t + 25 \cos^2 t}\]This illustrates the magnitude of movement at any point \(t\), combining both the \(x\) and \(y\) changes included in the velocity.

Critical Points

Critical points are crucial in finding where a function achieves its minimum or maximum values. For our speed function, critical points occur when the derivative of the speed function equals zero. After simplifying the speed function using the Pythagorean identity, we differentiate and set the derivative to zero: - Substitute \(u = 21\sin^2 t\) into the speed function.- Differentiate: \[\frac{df}{dt} = \frac{1}{2}(25 - u)^{-1/2} \cdot (-21 \sin(2t)) = 0\]- Solve the equation \(\sin(2t) = 0\).This results in angles where the sine function equals zero, giving points \(t = \frac{n\pi}{2}\) within the interval \([0, 2\pi]\). Evaluating these points helps identify where the speed is at its extremum.

Pythagorean Identity

The Pythagorean identity is a fundamental trigonometric identity: \(\sin^2 t + \cos^2 t = 1\). It allows us to simplify the combination of squared trigonometric functions, which is useful for calculating magnitudes and simplifying functions involving sine and cosine. In examining our speed function, we applied the identity to simplify:\[-25 \cos^2 t\] as \[-25 (1 - \sin^2 t)\].Thus, the simplified speed function becomes:\[\sqrt{4 \sin^2 t + 25 - 25 \sin^2 t} = \sqrt{25 - 21 \sin^2 t}\].This transformation assists in differentiating accurately and understanding the behavior of the speed over time, especially identifying minima and maxima using derivatives.