Problem 42
Question
\(39-44=\) Find the derivative of the vector function. $$ \mathbf{r}(t)=a t \cos 3 t \mathbf{i}+b \sin ^{3} t \mathbf{j}+c \cos ^{3} t \mathbf{k} $$
Step-by-Step Solution
Verified Answer
The derivative is \( (a \cos 3t - 3at \sin 3t) \mathbf{i} + 3b \sin^2 t \cos t \mathbf{j} - 3c \cos^2 t \sin t \mathbf{k} \).
1Step 1: Understand the Function Components
The vector function \( \mathbf{r}(t) = a t \cos 3t \mathbf{i} + b \sin^3 t \mathbf{j} + c \cos^3 t \mathbf{k} \) consists of three different components: \( a t \cos 3t \) in the \( \mathbf{i} \) direction, \( b \sin^3 t \) in the \( \mathbf{j} \) direction, and \( c \cos^3 t \) in the \( \mathbf{k} \) direction.
2Step 2: Differentiate the i-component
The \( \mathbf{i} \) component is \( a t \cos 3t \). Using the product rule, where \( u = at \) and \( v = \cos 3t \):\[ u' = a, \quad v' = -3 \sin 3t \]The derivative of the \( \mathbf{i} \) component is:\[ \frac{d}{dt}(at \cos 3t) = a \cos 3t + at(-3 \sin 3t) = a \cos 3t - 3at \sin 3t \]
3Step 3: Differentiate the j-component
For the \( \mathbf{j} \) component \( b \sin^3 t \), apply the chain rule:\[ \frac{d}{dt}( \sin^3 t) = 3 \sin^2 t \cos t \]The derivative is:\[ \frac{d}{dt}(b \sin^3 t) = b \cdot 3 \sin^2 t \cos t = 3b \sin^2 t \cos t \]
4Step 4: Differentiate the k-component
For the \( \mathbf{k} \) component \( c \cos^3 t \), also apply the chain rule:\[ \frac{d}{dt}( \cos^3 t) = 3 \cos^2 t (-\sin t) = -3 \cos^2 t \sin t \]The derivative is:\[ \frac{d}{dt}(c \cos^3 t) = c \left(-3 \cos^2 t \sin t\right) = -3c \cos^2 t \sin t \]
5Step 5: Combine the Results
Combine the derivatives of each component to get the derivative of the vector function:\[ \frac{d\mathbf{r}}{dt} = \left(a \cos 3t - 3at \sin 3t\right) \mathbf{i} + 3b \sin^2 t \cos t \mathbf{j} - 3c \cos^2 t \sin t \mathbf{k} \]
Key Concepts
Derivative of vector functionsProduct ruleChain rule
Derivative of vector functions
In vector calculus, finding the derivative of a vector function involves the differentiation of each of its components separately. A vector function is typically expressed in terms of unit vectors like \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and has components along each unit direction that may depend on a parameter, often represented by \( t \).
For example, the vector function \( \mathbf{r}(t) = a t \cos 3t \mathbf{i} + b \sin^3 t \mathbf{j} + c \cos^3 t \mathbf{k} \) consists of three distinct parts that need to be differentiated separately.
Differentiation of vector functions is a fundamental operation in calculus, allowing us to understand how a vector field changes or behaves over time. Each component behaves like a regular function and can be differentiated using the principles of standard calculus. After finding the derivative of each component, they are combined to find the derivative of the entire vector function.
For example, the vector function \( \mathbf{r}(t) = a t \cos 3t \mathbf{i} + b \sin^3 t \mathbf{j} + c \cos^3 t \mathbf{k} \) consists of three distinct parts that need to be differentiated separately.
Differentiation of vector functions is a fundamental operation in calculus, allowing us to understand how a vector field changes or behaves over time. Each component behaves like a regular function and can be differentiated using the principles of standard calculus. After finding the derivative of each component, they are combined to find the derivative of the entire vector function.
Product rule
The product rule is a key concept needed when differentiating products of functions, such as those found in vector functions. To handle these multiplicative scenarios, the product rule comes into play and allows us to differentiate effectively.
The product rule formula is:
\[ \frac{d}{dt}(uv) = u'v + uv' \]
Where \( u \) and \( v \) are functions of \( t \), with \( u' \) and \( v' \) denoting their respective derivatives.
Consider the \( \mathbf{i} \) component of our original vector function: \( a t \cos 3t \). Here, \( u = a t \) and \( v = \cos 3t \). Applying the product rule involves differentiating each part:
\[ a \cos 3t - 3at \sin 3t \]
This rationale illustrates how the product rule decomposes a product function into simpler derivative calculations, which are then mathematically combined.
The product rule formula is:
\[ \frac{d}{dt}(uv) = u'v + uv' \]
Where \( u \) and \( v \) are functions of \( t \), with \( u' \) and \( v' \) denoting their respective derivatives.
Consider the \( \mathbf{i} \) component of our original vector function: \( a t \cos 3t \). Here, \( u = a t \) and \( v = \cos 3t \). Applying the product rule involves differentiating each part:
- \( u' = a \)
- \( v' = -3 \sin 3t \)
\[ a \cos 3t - 3at \sin 3t \]
This rationale illustrates how the product rule decomposes a product function into simpler derivative calculations, which are then mathematically combined.
Chain rule
The chain rule is crucial for differentiating composite functions, which occur frequently in vector calculus. It allows us to differentiate a function with respect to another function, a process that's often necessary when dealing with powers and compositions.
The chain rule is stated as:
\[ \frac{d}{dt}(f(g(t))) = f'(g(t)) \cdot g'(t) \]
In this formula, \( f \) is the outer function while \( g \) is the inner function.
For the \( \mathbf{j} \) component of the vector function \( b \sin^3 t \), the chain rule helps differentiate \( \sin^3 t \). The process is as follows:
\[ 3b \sin^2 t \cos t \]
Similarly, for the \( \mathbf{k} \) component, the chain rule allows differentiation of \( c \cos^3 t \), yielding:
\[ -3c \cos^2 t \sin t \]
The chain rule's flexibility in handling varied compositions makes it an invaluable tool in vector calculus.
The chain rule is stated as:
\[ \frac{d}{dt}(f(g(t))) = f'(g(t)) \cdot g'(t) \]
In this formula, \( f \) is the outer function while \( g \) is the inner function.
For the \( \mathbf{j} \) component of the vector function \( b \sin^3 t \), the chain rule helps differentiate \( \sin^3 t \). The process is as follows:
- Recognize the outer function \( f(u) = u^3 \) and inner function \( u = \sin t \).
- Differentiate: \( f'(u) = 3u^2 \) and \( u' = \cos t \).
- The chain rule then gives: \( 3 \sin^2 t \cos t \).
\[ 3b \sin^2 t \cos t \]
Similarly, for the \( \mathbf{k} \) component, the chain rule allows differentiation of \( c \cos^3 t \), yielding:
\[ -3c \cos^2 t \sin t \]
The chain rule's flexibility in handling varied compositions makes it an invaluable tool in vector calculus.
Other exercises in this chapter
Problem 42
Find equations of the normal plane and osculating plane of the curve at the given point. $$x=t, y=t^{2}, z=t^{3} ; \quad(1,1,1)$$
View solution Problem 42
(a) Find the point at which the given lines intersect: $$\begin{aligned} \mathbf{r} &=\langle 1,1,0\rangle+ t\langle 1,-1,2\rangle \\\ \mathbf{r} &=\langle 2,0,
View solution Problem 42
Let \(\mathbf{v}=5 \mathbf{j}\) and let \(\mathbf{u}\) be a vector with length 3 that starts at the origin and rotates in the \(x y\) -plane. Find the maximum a
View solution Problem 42
If \(\mathbf{r}=\langle x, y, z\rangle, \mathbf{a}=\left\langle a_{1}, a_{2}, a_{3}\right\rangle,\) and \(\mathbf{b}=\left\langle b_{1}, b_{2}, b_{3}\right\rang
View solution