Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the velocity \(\mathbf{v}\), acceleration \(\mathbf{a}\), and speed \(s\) at the indicated time \(t=t_{1}\). $$ \mathbf{r}(t)=\tan t \mathbf{i}+3 e^{t} \mathbf{j}+\cos 4 t \mathbf{k} ; t_{1}=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

At \( t = \frac{\pi}{4} \), \( \mathbf{v} = 2\mathbf{i} + 3e^{\frac{\pi}{4}}\mathbf{j} \), \( \mathbf{a} = 2\mathbf{i} + 3e^{\frac{\pi}{4}}\mathbf{j} + 16\mathbf{k} \), and speed is \( \sqrt{4 + 9e^{\frac{\pi}{2}}} \).

Step 1: Find the Velocity Vector

To find the velocity vector \( \mathbf{v}(t) \), differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \).\[ \mathbf{v}(t) = \frac{d}{dt}(\tan t \, \mathbf{i} + 3e^{t} \, \mathbf{j} + \cos 4t \, \mathbf{k}) \]The derivatives are: - \( \frac{d}{dt}(\tan t) = \sec^2 t \)- \( \frac{d}{dt}(3e^t) = 3e^t \) - \( \frac{d}{dt}(\cos 4t) = -4\sin 4t \)So, \[ \mathbf{v}(t) = \sec^2 t \, \mathbf{i} + 3e^t \, \mathbf{j} - 4\sin 4t \, \mathbf{k} \]

Step 2: Evaluate Velocity at Given Time

Substitute \( t = \frac{\pi}{4} \) into the velocity vector \( \mathbf{v}(t) \):- \( \sec^2 \left(\frac{\pi}{4}\right) = 2 \)- \( 3e^{\frac{\pi}{4}} \)- \( -4\sin \left(4\cdot\frac{\pi}{4}\right) = 0 \) since \( \sin \pi = 0 \)Thus, \[ \mathbf{v}\left(\frac{\pi}{4}\right) = 2\mathbf{i} + 3e^{\frac{\pi}{4}}\mathbf{j} \]

Step 3: Find the Acceleration Vector

Differentiate the velocity vector \( \mathbf{v}(t) \) to find the acceleration vector \( \mathbf{a}(t) \):\[ \mathbf{a}(t) = \frac{d}{dt}(\sec^2 t \, \mathbf{i} + 3e^t \, \mathbf{j} - 4\sin 4t \, \mathbf{k}) \]The derivatives are: - \( \frac{d}{dt}(\sec^2 t) = 2\sec^2 t \tan t \)- \( \frac{d}{dt}(3e^t) = 3e^t \) - \( \frac{d}{dt}(-4\sin 4t) = -16\cos 4t \)Thus, \[ \mathbf{a}(t) = 2\sec^2 t \tan t \, \mathbf{i} + 3e^t \, \mathbf{j} - 16\cos 4t \, \mathbf{k} \]

Step 4: Evaluate Acceleration at Given Time

Substitute \( t = \frac{\pi}{4} \) into the acceleration vector \( \mathbf{a}(t) \):- \( 2\sec^2 \left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) = 2 \)- \( 3e^{\frac{\pi}{4}} \)- \( -16\cos \left(\pi\right) = 16 \), because \( \cos \pi = -1 \)Thus, \[ \mathbf{a}\left(\frac{\pi}{4}\right) = 2\mathbf{i} + 3e^{\frac{\pi}{4}}\mathbf{j} + 16\mathbf{k} \]

Step 5: Calculate Speed

The speed \( s(t) \) is the magnitude of the velocity vector: \[ s(t) = \| \mathbf{v}(t) \| = \sqrt{(\sec^2 t)^2 + (3e^t)^2 + (-4\sin 4t)^2} \]Substitute \( t = \frac{\pi}{4} \):\[ s\left(\frac{\pi}{4}\right) = \sqrt{(2)^2 + (3e^{\frac{\pi}{4}})^2 + 0} = \sqrt{4 + 9e^{\frac{\pi}{2}}} \]

Key Concepts

Velocity

In vector calculus, velocity is the rate at which an object changes its position. For a position vector function \( \mathbf{r}(t) \), the velocity vector \( \mathbf{v}(t) \) is found using differentiation. By differentiating \( \mathbf{r}(t) = \tan t \, \mathbf{i} + 3e^{t} \, \mathbf{j} + \cos 4t \, \mathbf{k} \) with respect to time \( t \), we obtain the velocity vector:

• \( \frac{d}{dt}(\tan t) = \sec^2 t \)
• \( \frac{d}{dt}(3e^t) = 3e^t \)
• \( \frac{d}{dt}(\cos 4t) = -4\sin 4t \)

Thus, the velocity vector is \( \mathbf{v}(t) = \sec^2 t \, \mathbf{i} + 3e^t \, \mathbf{j} - 4\sin 4t \, \mathbf{k} \). To find the velocity at a specific time \( t_1 = \frac{\pi}{4} \), substitute \( t_1 \) into \( \mathbf{v}(t) \), resulting in \( \mathbf{v}\left(\frac{\pi}{4}\right) = 2\mathbf{i} + 3e^{\frac{\pi}{4}}\mathbf{j} \). This calculation shows that at time \( \frac{\pi}{4} \), the components of velocity change accordingly.

Acceleration

Velocity's rate of change gives us the acceleration vector, another core concept in vector calculus. To find it, differentiate the velocity vector \( \mathbf{v}(t) \) obtained previously. This gives us the acceleration vector \( \mathbf{a}(t) \):

• \( \frac{d}{dt}(\sec^2 t) = 2\sec^2 t \tan t \)
• \( \frac{d}{dt}(3e^t) = 3e^t \)
• \( \frac{d}{dt}(-4\sin 4t) = -16\cos 4t \)

Therefore, \( \mathbf{a}(t) = 2\sec^2 t \tan t \, \mathbf{i} + 3e^t \, \mathbf{j} - 16\cos 4t \, \mathbf{k} \). To find the acceleration at \( t_1 = \frac{\pi}{4} \), substitute \( t_1 \) into \( \mathbf{a}(t) \), resulting in \( \mathbf{a}\left(\frac{\pi}{4}\right) = 2\mathbf{i} + 3e^{\frac{\pi}{4}}\mathbf{j} + 16\mathbf{k} \). This shows the acceleration's effect on all three vector components at this specific time.

Magnitude of a Vector

The magnitude of a vector (commonly the vector's length or size) is crucial for calculating speed. To find it, use the velocity vector derived earlier. Its magnitude is computed with:\[ s(t) = \| \mathbf{v}(t) \| = \sqrt{(\sec^2 t)^2 + (3e^t)^2 + (-4\sin 4t)^2} \]At \( t_1 = \frac{\pi}{4} \), the velocity vector's magnitude or speed \( s(t) \) becomes:\[ s\left(\frac{\pi}{4}\right) = \sqrt{(2)^2 + (3e^{\frac{\pi}{4}})^2 + 0} = \sqrt{4 + 9e^{\frac{\pi}{2}}} \]This denotes the object's speed at \( t_1 \), showing how the combined effects of different velocity components contribute to overall speed.

Differentiation

Differentiation is a fundamental tool in vector calculus, especially for finding velocity and acceleration. By computing the derivatives of position and velocity vector functions, we unveil changes over time. Starting with the position vector \( \mathbf{r}(t) = \tan t \, \mathbf{i} + 3e^{t} \, \mathbf{j} + \cos 4t \, \mathbf{k} \), we differentiate to reveal the velocity vector \( \mathbf{v}(t) = \sec^2 t \, \mathbf{i} + 3e^t \, \mathbf{j} - 4\sin 4t \, \mathbf{k} \). Further differentiation leads us to the acceleration vector \( \mathbf{a}(t) = 2\sec^2 t \tan t \, \mathbf{i} + 3e^t \, \mathbf{j} - 16\cos 4t \, \mathbf{k} \). Differentiation unveils crucial information about how position shifts to velocity, and how velocity morphs into acceleration over time. Its application is pivotal in understanding dynamic systems and movement within mathematical frameworks.