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Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\langle\cos 3 t, \sin 4 t, \cos 6 t\rangle$$

Short Answer

Expert verified
Question: Find the second and third derivatives of the vector function \(\mathbf{r}(t) = \langle\cos 3t, \sin 4t, \cos 6t\rangle\). Answer: The second derivative of the function is \(\mathbf{r}^{\prime \prime}(t) = \langle -9\cos 3t, -16\sin 4t, -36\cos 6t \rangle\), and the third derivative is \(\mathbf{r}^{\prime \prime \prime}(t) = \langle 27\sin 3t, -64\cos 4t, 216\sin 6t \rangle\).

Step by step solution

01

Differentiate the component functions r(t) with respect to t

Differentiate the function component-wise: $$\mathbf{r}(t) = \langle\cos 3t, \sin 4t, \cos 6t\rangle$$ Calculate the derivatives of the component functions: $$\frac{d}{dt}\cos 3t = -3\sin 3t$$ $$\frac{d}{dt}\sin 4t = 4\cos 4t$$ $$\frac{d}{dt}\cos 6t = -6\sin 6t$$
02

Combine the derivatives into the first derivative 饾憻鈥(t)

Combine the derivatives of the component functions into the first derivative: $$\mathbf{r}^{\prime}(t) = \langle -3\sin 3t, 4\cos 4t, -6\sin 6t \rangle$$
03

Differentiate the component functions of 饾憻鈥(t) with respect to t

Differentiate the function component-wise: $$\mathbf{r}^{\prime}(t) = \langle -3\sin 3t, 4\cos 4t, -6\sin 6t \rangle$$ Calculate the derivatives of the component functions: $$\frac{d}{dt}(-3\sin 3t) = -9\cos 3t$$ $$\frac{d}{dt}(4\cos 4t) = -16\sin 4t$$ $$\frac{d}{dt}(-6\sin 6t) = -36\cos 6t$$
04

Combine the derivatives into the second derivative 饾憻鈥测(t)

Combine the derivatives of the component functions into the second derivative: $$\mathbf{r}^{\prime \prime}(t) = \langle -9\cos 3t, -16\sin 4t, -36\cos 6t \rangle$$
05

Differentiate the component functions of 饾憻鈥测(t) with respect to t

Differentiate the function component-wise: $$\mathbf{r}^{\prime \prime}(t) = \langle -9\cos 3t, -16\sin 4t, -36\cos 6t \rangle$$ Calculate the derivatives of the component functions: $$\frac{d}{dt}(-9\cos 3t) = 27\sin 3t$$ $$\frac{d}{dt}(-16\sin 4t) = -64\cos 4t$$ $$\frac{d}{dt}(-36\cos 6t) = 216\sin 6t$$
06

Combine the derivatives into the third derivative 饾憻鈥测测(t)

Combine the derivatives of the component functions into the third derivative: $$\mathbf{r}^{\prime \prime \prime}(t) = \langle 27\sin 3t, -64\cos 4t, 216\sin 6t \rangle$$ In conclusion, the second derivative of the function is \(\mathbf{r}^{\prime \prime}(t) = \langle -9\cos 3t, -16\sin 4t, -36\cos 6t \rangle\), and the third derivative is \(\mathbf{r}^{\prime \prime \prime}(t) = \langle 27\sin 3t, -64\cos 4t, 216\sin 6t \rangle\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives of Vector Functions
Vector functions are functions that have more than one component, usually defined in terms of a parameter such as time, denoted by \( t \). Each component of a vector function corresponds to a coordinate dimension, often in two-dimensional or three-dimensional space.
When we talk about the derivative of a vector function, we appreciate that we are essentially finding the rate of change of that vector with respect to \( t \).
This involves differentiating each component of the vector function separately.
The key is to treat each component as a normal function of \( t \) and apply traditional differentiation rules.
  • Understand that the derivative gives a new vector which indicates how the original vector is changing at any point \( t \).
  • The magnitude and direction of this vector can tell us a lot about the motion described by the vector function.
The example given involves finding second and third derivatives, where the process becomes iterative, involving further differentiation of the resultant derivative vectors.
Component-wise Differentiation
In vector calculus, component-wise differentiation is a method used to differentiate each part of a vector function separately. This is crucial because each part of the vector operates independently according to standard calculus rules.
To perform component-wise differentiation, you:
  • Identify each component of the vector function. For example, given \( \mathbf{r}(t) = \langle\cos 3t, \sin 4t, \cos 6t\rangle \), there are three components to differentiate: \( \cos 3t \), \( \sin 4t \), and \( \cos 6t \).
  • Apply basic differentiation rules to each component individually. Recall derivatives such as \( \frac{d}{dt}\cos\theta = -\sin\theta\) and \( \frac{d}{dt}\sin\theta = \cos\theta\).
This is repeated with any subsequent derivatives until the required derivative order is achieved.
Component-wise differentiation simplifies the process by handling each dimension of the vector function on its own terms.
Second and Third Derivatives
Once the first derivative of a vector function is obtained, additional derivatives can provide deeper insights into the function's behavior, especially in physical applications such as acceleration and jerk in motion.
The second derivative is often termed "acceleration" as it provides the derivative of the velocity vector or the rate of change of velocity.
The steps for obtaining higher-order derivatives, like the second and third, follow the same principle of differentiating each component again.
  • In the given problem, the first derivative \( \mathbf{r}^{\prime}(t) \) was found, leading to the differentiation of \(-3\sin 3t\), \(4\cos 4t\), and \(-6\sin 6t\) to obtain \( \mathbf{r}^{\prime \prime}(t) \).
  • Further differentiating \( \mathbf{r}^{\prime \prime}(t) \) gives the third derivative, \( \mathbf{r}^{\prime \prime \prime}(t) \), adding another layer of depth to the function's analysis.
Understanding successive derivatives enriches our modeling capability, particularly in dynamics and other fields requiring precise measurement of change rates.

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Most popular questions from this chapter

Consider the curve \(\mathbf{r}(t)=(\cos t, \sin t, c \sin t),\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. It can be shown that the curve lies in a plane. Prove that the curve is an ellipse in that plane.

Zero curvature Prove that the curve $$ \mathbf{r}(t)=\left\langle a+b t^{p}, c+d t^{p}, e+f t^{p}\right\rangle $$ where \(a, b, c, d, e,\) and \(f\) are real numbers and \(p\) is a positive integer, has zero curvature. Give an explanation.

Alternative derivation of the curvature Derive the computational formula for curvature using the following steps. a. Use the tangential and normal components of the acceleration to show that \(\left.\mathbf{v} \times \mathbf{a}=\kappa|\mathbf{v}|^{3} \mathbf{B} . \text { (Note that } \mathbf{T} \times \mathbf{T}=\mathbf{0} .\right)\) b. Solve the equation in part (a) for \(\kappa\) and conclude that \(\kappa=\frac{|\mathbf{v} \times \mathbf{a}|}{\left|\mathbf{v}^{3}\right|},\) as shown in the text.

Maximum curvature Consider the "superparabolas" \(f_{n}(x)=x^{2 n},\) where \(n\) is a positive integer. a. Find the curvature function of \(f_{n},\) for \(n=1,2,\) and 3 b. Plot \(f_{n}\) and their curvature functions, for \(n=1,2,\) and 3 and check for consistency. c. At what points does the maximum curvature occur, for \(n=1,2,3 ?\) d. Let the maximum curvature for \(f_{n}\) occur at \(x=\pm z_{n} .\) Using either analytical methods or a calculator determine \(\lim _{n \rightarrow \infty} z_{n}\) Interpret your result.

A baseball leaves the hand of a pitcher 6 vertical feet above home plate and \(60 \mathrm{ft}\) from home plate. Assume that the coordinate axes are oriented as shown in the figure. a. In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of \(\langle 130,0,-3\rangle \mathrm{ft} / \mathrm{s}\) (about \(90 \mathrm{mi} / \mathrm{hr}\) ). How far above the ground is the ball when it crosses home plate and how long does it take for the pitch to arrive? b. What vertical velocity component should the pitcher use so that the pitch crosses home plate exactly \(3 \mathrm{ft}\) above the ground? c. A simple model to describe the curve of a baseball assumes that the spin of the ball produces a constant sideways acceleration (in the \(y\) -direction) of \(c \mathrm{ft} / \mathrm{s}^{2}\). Assume a pitcher throws a curve ball with \(c=8 \mathrm{ft} / \mathrm{s}^{2}\) (one-fourth the acceleration of gravity). How far does the ball move in the \(y\) -direction by the time it reaches home plate, assuming an initial velocity of (130,0,-3) ft/s? d. In part (c), does the ball curve more in the first half of its trip to the plate or in the second half? How does this fact affect the batter? e. Suppose the pitcher releases the ball from an initial position of (0,-3,6) with initial velocity \((130,0,-3) .\) What value of the spin parameter \(c\) is needed to put the ball over home plate passing through the point (60,0,3)\(?\)

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