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To find [tex]\(\frac{d}{dt}[u(t) \cdot v(t)]\)[/tex] where [tex]\(u(t) = e^{8t} \mathbf{i} + e^{-8t} \mathbf{j}\)[/tex] and [tex]\(v(t) = t \mathbf{i} - 9 t^5 \mathbf{j}\)[/tex], we need to follow these steps:
1. Compute the dot product of [tex]\(u(t)\)[/tex] and [tex]\(v(t)\)[/tex].
2. Differentiate the dot product with respect to [tex]\(t\)[/tex].
### 1. Compute the dot product [tex]\(u(t) \cdot v(t)\)[/tex]
Given [tex]\(u(t) = e^{8t} \mathbf{i} + e^{-8t} \mathbf{j}\)[/tex] and [tex]\(v(t) = t \mathbf{i} - 9 t^5 \mathbf{j}\)[/tex], we compute the dot product by multiplying the corresponding components of the vectors and adding the results.
The dot product is given by:
[tex]\[ u(t) \cdot v(t) = (e^{8t} \mathbf{i} + e^{-8t} \mathbf{j}) \cdot (t \mathbf{i} - 9 t^5 \mathbf{j}) \][/tex]
Expanding this, we have:
[tex]\[ u(t) \cdot v(t) = (e^{8t} \mathbf{i}) \cdot (t \mathbf{i}) + (e^{8t} \mathbf{i}) \cdot (-9 t^5 \mathbf{j}) + (e^{-8t} \mathbf{j}) \cdot (t \mathbf{i}) + (e^{-8t} \mathbf{j}) \cdot (-9 t^5 \mathbf{j}) \][/tex]
The perpendicular components [tex]\(\mathbf{i} \cdot \mathbf{j}\)[/tex] and [tex]\(\mathbf{j} \cdot \mathbf{i}\)[/tex] are zero:
[tex]\[ \mathbf{i} \cdot \mathbf{j} = 0 \quad \text{and} \quad \mathbf{j} \cdot \mathbf{i} = 0 \][/tex]
Thus, the terms involving [tex]\((e^{8t} \mathbf{i}) \cdot (-9 t^5 \mathbf{j})\)[/tex] and [tex]\((e^{-8t} \mathbf{j}) \cdot (t \mathbf{i})\)[/tex] drop out:
[tex]\[ u(t) \cdot v(t) = (e^{8t} \mathbf{i}) \cdot (t \mathbf{i}) + (e^{-8t} \mathbf{j}) \cdot (-9 t^5 \mathbf{j}) \][/tex]
Now compute the remaining terms:
[tex]\[ (e^{8t} \mathbf{i}) \cdot (t \mathbf{i}) = e^{8t} \cdot t = t e^{8t} \][/tex]
[tex]\[ (e^{-8t} \mathbf{j}) \cdot (-9 t^5 \mathbf{j}) = e^{-8t} \cdot (-9 t^5) = -9 t^5 e^{-8t} \][/tex]
Combining these, we get:
[tex]\[ u(t) \cdot v(t) = t e^{8t} - 9 t^5 e^{-8t} \][/tex]
### 2. Differentiate the dot product with respect to [tex]\(t\)[/tex]
Now we need to find [tex]\(\frac{d}{dt} \left( t e^{8t} - 9 t^5 e^{-8t} \right)\)[/tex].
Using the product rule [tex]\(\frac{d}{dt} (f(t) g(t)) = f'(t) g(t) + f(t) g'(t)\)[/tex]:
#### Differentiate [tex]\(t e^{8t}\)[/tex]:
Let [tex]\(f(t) = t\)[/tex] and [tex]\(g(t) = e^{8t}\)[/tex].
[tex]\[ f'(t) = 1 \][/tex]
[tex]\[ g'(t) = 8 e^{8t} \][/tex]
So,
[tex]\[ \frac{d}{dt}(t e^{8t}) = 1 \cdot e^{8t} + t \cdot 8 e^{8t} = e^{8t} + 8 t e^{8t} \][/tex]
#### Differentiate [tex]\(-9 t^5 e^{-8t}\)[/tex]:
Let [tex]\(-9 t^5 = f(t)\)[/tex] and [tex]\(e^{-8t} = g(t)\)[/tex].
[tex]\[ f'(t) = -45 t^4 \][/tex]
[tex]\[ g'(t) = -8 e^{-8t} \][/tex]
So,
[tex]\[ \frac{d}{dt}(-9 t^5 e^{-8t}) = (-45 t^4) e^{-8t} + (-9 t^5)(-8 e^{-8t}) = -45 t^4 e^{-8t} + 72 t^5 e^{-8t} \][/tex]
Combining these differentiation results, we have:
[tex]\[ \frac{d}{dt}[u(t) \cdot v(t)] = \left(e^{8t} + 8 t e^{8t}\right) + \left(-45 t^4 e^{-8t} + 72 t^5 e^{-8t}\right) \][/tex]
Simplifying:
[tex]\[ = e^{8t} + 8 t e^{8t} - 45 t^4 e^{-8t} + 72 t^5 e^{-8t} \][/tex]
Thus, the derivative [tex]\(\frac{d}{dt}[u(t) \cdot v(t)]\)[/tex] is:
[tex]\[ \boxed{e^{8t} + 8 t e^{8t} - 45 t^4 e^{-8t} + 72 t^5 e^{-8t}} \][/tex]
1. Compute the dot product of [tex]\(u(t)\)[/tex] and [tex]\(v(t)\)[/tex].
2. Differentiate the dot product with respect to [tex]\(t\)[/tex].
### 1. Compute the dot product [tex]\(u(t) \cdot v(t)\)[/tex]
Given [tex]\(u(t) = e^{8t} \mathbf{i} + e^{-8t} \mathbf{j}\)[/tex] and [tex]\(v(t) = t \mathbf{i} - 9 t^5 \mathbf{j}\)[/tex], we compute the dot product by multiplying the corresponding components of the vectors and adding the results.
The dot product is given by:
[tex]\[ u(t) \cdot v(t) = (e^{8t} \mathbf{i} + e^{-8t} \mathbf{j}) \cdot (t \mathbf{i} - 9 t^5 \mathbf{j}) \][/tex]
Expanding this, we have:
[tex]\[ u(t) \cdot v(t) = (e^{8t} \mathbf{i}) \cdot (t \mathbf{i}) + (e^{8t} \mathbf{i}) \cdot (-9 t^5 \mathbf{j}) + (e^{-8t} \mathbf{j}) \cdot (t \mathbf{i}) + (e^{-8t} \mathbf{j}) \cdot (-9 t^5 \mathbf{j}) \][/tex]
The perpendicular components [tex]\(\mathbf{i} \cdot \mathbf{j}\)[/tex] and [tex]\(\mathbf{j} \cdot \mathbf{i}\)[/tex] are zero:
[tex]\[ \mathbf{i} \cdot \mathbf{j} = 0 \quad \text{and} \quad \mathbf{j} \cdot \mathbf{i} = 0 \][/tex]
Thus, the terms involving [tex]\((e^{8t} \mathbf{i}) \cdot (-9 t^5 \mathbf{j})\)[/tex] and [tex]\((e^{-8t} \mathbf{j}) \cdot (t \mathbf{i})\)[/tex] drop out:
[tex]\[ u(t) \cdot v(t) = (e^{8t} \mathbf{i}) \cdot (t \mathbf{i}) + (e^{-8t} \mathbf{j}) \cdot (-9 t^5 \mathbf{j}) \][/tex]
Now compute the remaining terms:
[tex]\[ (e^{8t} \mathbf{i}) \cdot (t \mathbf{i}) = e^{8t} \cdot t = t e^{8t} \][/tex]
[tex]\[ (e^{-8t} \mathbf{j}) \cdot (-9 t^5 \mathbf{j}) = e^{-8t} \cdot (-9 t^5) = -9 t^5 e^{-8t} \][/tex]
Combining these, we get:
[tex]\[ u(t) \cdot v(t) = t e^{8t} - 9 t^5 e^{-8t} \][/tex]
### 2. Differentiate the dot product with respect to [tex]\(t\)[/tex]
Now we need to find [tex]\(\frac{d}{dt} \left( t e^{8t} - 9 t^5 e^{-8t} \right)\)[/tex].
Using the product rule [tex]\(\frac{d}{dt} (f(t) g(t)) = f'(t) g(t) + f(t) g'(t)\)[/tex]:
#### Differentiate [tex]\(t e^{8t}\)[/tex]:
Let [tex]\(f(t) = t\)[/tex] and [tex]\(g(t) = e^{8t}\)[/tex].
[tex]\[ f'(t) = 1 \][/tex]
[tex]\[ g'(t) = 8 e^{8t} \][/tex]
So,
[tex]\[ \frac{d}{dt}(t e^{8t}) = 1 \cdot e^{8t} + t \cdot 8 e^{8t} = e^{8t} + 8 t e^{8t} \][/tex]
#### Differentiate [tex]\(-9 t^5 e^{-8t}\)[/tex]:
Let [tex]\(-9 t^5 = f(t)\)[/tex] and [tex]\(e^{-8t} = g(t)\)[/tex].
[tex]\[ f'(t) = -45 t^4 \][/tex]
[tex]\[ g'(t) = -8 e^{-8t} \][/tex]
So,
[tex]\[ \frac{d}{dt}(-9 t^5 e^{-8t}) = (-45 t^4) e^{-8t} + (-9 t^5)(-8 e^{-8t}) = -45 t^4 e^{-8t} + 72 t^5 e^{-8t} \][/tex]
Combining these differentiation results, we have:
[tex]\[ \frac{d}{dt}[u(t) \cdot v(t)] = \left(e^{8t} + 8 t e^{8t}\right) + \left(-45 t^4 e^{-8t} + 72 t^5 e^{-8t}\right) \][/tex]
Simplifying:
[tex]\[ = e^{8t} + 8 t e^{8t} - 45 t^4 e^{-8t} + 72 t^5 e^{-8t} \][/tex]
Thus, the derivative [tex]\(\frac{d}{dt}[u(t) \cdot v(t)]\)[/tex] is:
[tex]\[ \boxed{e^{8t} + 8 t e^{8t} - 45 t^4 e^{-8t} + 72 t^5 e^{-8t}} \][/tex]
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