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Sagot :
To solve the given problem, we need to find the flow of the velocity field [tex]\( \mathbf{F} = 2xy\mathbf{i} + 2y\mathbf{j} + \mathbf{k} \)[/tex] along the curve [tex]\( \mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + \mathbf{k} \)[/tex] from [tex]\( t = 0 \)[/tex] to [tex]\( t = 4 \)[/tex].
Here are the steps:
Step 1: Parametrize the curve
The given curve is already parametrized as:
[tex]\[ \mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + \mathbf{k} \][/tex]
Step 2: Compute the derivative of [tex]\( \mathbf{r}(t) \)[/tex]
[tex]\[ \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t\mathbf{i} + t^2\mathbf{j} + \mathbf{k}) = \mathbf{i} + 2t\mathbf{j} + 0\mathbf{k} \][/tex]
Thus,
[tex]\[ \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t\mathbf{j} \][/tex]
Step 3: Substitute [tex]\( \mathbf{r}(t) \)[/tex] into [tex]\( \mathbf{F} \)[/tex]
Let’s find [tex]\( \mathbf{F}( \mathbf{r}(t) ) \)[/tex]:
[tex]\[ x = t, \quad y = t^2, \quad z = 1 \][/tex]
Hence,
[tex]\[ \mathbf{F}(t, t^2, 1) = 2xt^2\mathbf{i} + 2y\mathbf{j} + \mathbf{k} = 2t(t^2)\mathbf{i} + 2(t^2)\mathbf{j} + \mathbf{k} \][/tex]
[tex]\[ \mathbf{F}(t, t^2, 1) = 2t^3\mathbf{i} + 2t^2\mathbf{j} + \mathbf{k} \][/tex]
Step 4: Compute the dot product [tex]\( \mathbf{F}( \mathbf{r}(t) ) \cdot \frac{d\mathbf{r}}{dt} \)[/tex]
[tex]\[ \mathbf{F}(t, t^2, 1) \cdot \frac{d\mathbf{r}}{dt} = (2t^3\mathbf{i} + 2t^2\mathbf{j} + \mathbf{k}) \cdot (\mathbf{i} + 2t\mathbf{j}) \][/tex]
[tex]\[ = (2t^3)(1) + (2t^2)(2t) + (1)(0) \][/tex]
[tex]\[ = 2t^3 + 4t^3 \][/tex]
[tex]\[ = 6t^3 \][/tex]
Step 5: Integrate [tex]\( 6t^3 \)[/tex] from [tex]\( t = 0 \)[/tex] to [tex]\( t = 4 \)[/tex]
The integrand is [tex]\( 6t^3 \)[/tex], so we need to compute:
[tex]\[ \int_{0}^{4} 6t^3 \, dt \][/tex]
First, find the antiderivative of [tex]\( 6t^3 \)[/tex]:
[tex]\[ \int 6t^3 \, dt = 6 \cdot \frac{t^4}{4} = \frac{3t^4}{2} \][/tex]
Now, evaluate this at the boundaries:
[tex]\[ \left[ \frac{3t^4}{2} \right]_{0}^{4} = \frac{3(4)^4}{2} - \frac{3(0)^4}{2} \][/tex]
[tex]\[ = \frac{3 \cdot 256}{2} - 0 \][/tex]
[tex]\[ = \frac{768}{2} \][/tex]
[tex]\[ = 384 \][/tex]
Therefore, the flow along the curve is:
[tex]\[ \boxed{384} \][/tex]
Here are the steps:
Step 1: Parametrize the curve
The given curve is already parametrized as:
[tex]\[ \mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + \mathbf{k} \][/tex]
Step 2: Compute the derivative of [tex]\( \mathbf{r}(t) \)[/tex]
[tex]\[ \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t\mathbf{i} + t^2\mathbf{j} + \mathbf{k}) = \mathbf{i} + 2t\mathbf{j} + 0\mathbf{k} \][/tex]
Thus,
[tex]\[ \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t\mathbf{j} \][/tex]
Step 3: Substitute [tex]\( \mathbf{r}(t) \)[/tex] into [tex]\( \mathbf{F} \)[/tex]
Let’s find [tex]\( \mathbf{F}( \mathbf{r}(t) ) \)[/tex]:
[tex]\[ x = t, \quad y = t^2, \quad z = 1 \][/tex]
Hence,
[tex]\[ \mathbf{F}(t, t^2, 1) = 2xt^2\mathbf{i} + 2y\mathbf{j} + \mathbf{k} = 2t(t^2)\mathbf{i} + 2(t^2)\mathbf{j} + \mathbf{k} \][/tex]
[tex]\[ \mathbf{F}(t, t^2, 1) = 2t^3\mathbf{i} + 2t^2\mathbf{j} + \mathbf{k} \][/tex]
Step 4: Compute the dot product [tex]\( \mathbf{F}( \mathbf{r}(t) ) \cdot \frac{d\mathbf{r}}{dt} \)[/tex]
[tex]\[ \mathbf{F}(t, t^2, 1) \cdot \frac{d\mathbf{r}}{dt} = (2t^3\mathbf{i} + 2t^2\mathbf{j} + \mathbf{k}) \cdot (\mathbf{i} + 2t\mathbf{j}) \][/tex]
[tex]\[ = (2t^3)(1) + (2t^2)(2t) + (1)(0) \][/tex]
[tex]\[ = 2t^3 + 4t^3 \][/tex]
[tex]\[ = 6t^3 \][/tex]
Step 5: Integrate [tex]\( 6t^3 \)[/tex] from [tex]\( t = 0 \)[/tex] to [tex]\( t = 4 \)[/tex]
The integrand is [tex]\( 6t^3 \)[/tex], so we need to compute:
[tex]\[ \int_{0}^{4} 6t^3 \, dt \][/tex]
First, find the antiderivative of [tex]\( 6t^3 \)[/tex]:
[tex]\[ \int 6t^3 \, dt = 6 \cdot \frac{t^4}{4} = \frac{3t^4}{2} \][/tex]
Now, evaluate this at the boundaries:
[tex]\[ \left[ \frac{3t^4}{2} \right]_{0}^{4} = \frac{3(4)^4}{2} - \frac{3(0)^4}{2} \][/tex]
[tex]\[ = \frac{3 \cdot 256}{2} - 0 \][/tex]
[tex]\[ = \frac{768}{2} \][/tex]
[tex]\[ = 384 \][/tex]
Therefore, the flow along the curve is:
[tex]\[ \boxed{384} \][/tex]
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