Problem 77
Question
Find \(f^{\prime}(2),\) where \(f(t)=\mathbf{u}(t) \cdot \mathbf{v}(t), \mathbf{u}(2)=\langle 1,2,-1\rangle\) $$\mathbf{u}^{\prime}(2)=\langle 3,0,4\rangle,\( and \)\mathbf{v}(t)=\left\langle t, t^{2}, t^{3}\right\rangle$$
Step-by-Step Solution
Verified Answer
The derivative \( f'(2) \) is 35.
1Step 1: Differentiate Using the Product Rule
The function given is a dot product of two vector functions, i.e., \( f(t) = \mathbf{u}(t) \cdot \mathbf{v}(t) \). The derivative of a dot product \( f'(t) \) is given by \( \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t) \). We will apply this rule to find \( f'(2) \).
2Step 2: Calculate \( \mathbf{v}'(t) \)
Differentiate the vector function \( \mathbf{v}(t) = \langle t, t^2, t^3 \rangle \) with respect to \( t \). This gives us \( \mathbf{v}'(t) = \langle 1, 2t, 3t^2 \rangle \).
3Step 3: Find \( \mathbf{v}(2) \) and \( \mathbf{v}'(2) \)
Substitute \( t = 2 \) into \( \mathbf{v}(t) \):\( \mathbf{v}(2) = \langle 2, 4, 8 \rangle \).Substitute \( t = 2 \) into \( \mathbf{v}'(t) \):\( \mathbf{v}'(2) = \langle 1, 4, 12 \rangle \).
4Step 4: Compute \( \mathbf{u}'(2) \cdot \mathbf{v}(2) \)
\( \mathbf{u}'(2) = \langle 3, 0, 4 \rangle \) and \( \mathbf{v}(2) = \langle 2, 4, 8 \rangle \).Calculate the dot product:\( \mathbf{u}'(2) \cdot \mathbf{v}(2) = 3 \times 2 + 0 \times 4 + 4 \times 8 = 6 + 0 + 32 = 38 \).
5Step 5: Compute \( \mathbf{u}(2) \cdot \mathbf{v}'(2) \)
\( \mathbf{u}(2) = \langle 1, 2, -1 \rangle \) and \( \mathbf{v}'(2) = \langle 1, 4, 12 \rangle \).Calculate the dot product:\( \mathbf{u}(2) \cdot \mathbf{v}'(2) = 1 \times 1 + 2 \times 4 + (-1) \times 12 = 1 + 8 - 12 = -3 \).
6Step 6: Add the Results to Find \( f'(2) \)
Now add the results from Steps 4 and 5 to find \( f'(2) \):\( f'(2) = 38 + (-3) = 35 \).
Key Concepts
Product RuleVector DifferentiationDot Product
Product Rule
The Product Rule is a fundamental concept in calculus, specifically used in differentiation. When you have a function that is the product of two differentiable functions, the derivative of that product is not simply the product of their derivatives. Instead, the product rule states that for functions \( u(t) \) and \( v(t) \), the derivative \( f'(t) \) of \( f(t) = u(t) \cdot v(t) \) is given by the formula: \[ f'(t) = u'(t)v(t) + u(t)v'(t) \].
This formula tells us to take the derivative of the first function and multiply it by the second, then add the product of the first function and the derivative of the second function.
In our problem, we apply the product rule to differentiate the dot product of two vector functions \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \), which means we have to find both \( \mathbf{u}'(t) \) and \( \mathbf{v}'(t) \) and compute their respective dot products with the original vector functions.
This formula tells us to take the derivative of the first function and multiply it by the second, then add the product of the first function and the derivative of the second function.
In our problem, we apply the product rule to differentiate the dot product of two vector functions \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \), which means we have to find both \( \mathbf{u}'(t) \) and \( \mathbf{v}'(t) \) and compute their respective dot products with the original vector functions.
Vector Differentiation
Vector differentiation involves applying the principles of calculus to vector functions, treating each component of the vector independently. When differentiating a vector function like \( \mathbf{v}(t) = \langle t, t^2, t^3 \rangle \), we find the derivative of each component separately.
For our function, \[ \mathbf{v}'(t) = \langle 1, 2t, 3t^2 \rangle \],
which means we apply basic differentiation rules to each part of the vector:
For our function, \[ \mathbf{v}'(t) = \langle 1, 2t, 3t^2 \rangle \],
which means we apply basic differentiation rules to each part of the vector:
- \( t \) becomes \( 1 \),
- \( t^2 \) becomes \( 2t \),
- \( t^3 \) becomes \( 3t^2 \).
Dot Product
The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation combines components of vectors using multiplication and then sums the results.
The formula for the dot product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is:
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \].
In our problem, we compute the dot product of differentiated vectors and original vectors to find parts of the derivative,
The formula for the dot product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is:
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \].
In our problem, we compute the dot product of differentiated vectors and original vectors to find parts of the derivative,
- \( \mathbf{u}'(2) \cdot \mathbf{v}(2) = 38 \)
- \( \mathbf{u}(2) \cdot \mathbf{v}'(2) = -3 \)
Other exercises in this chapter
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Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are vector functions that possess limits as \(t \rightarrow a\) and let \(c\) be a constant. Prove the following prope
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Find an expression for $$\frac{d}{d t}[\mathbf{u}(t) \cdot(\mathbf{v}(t) \times \mathbf{w}(t))]$$
View solution