Answer:
To evaluate the line integral \(\int_C \mathbf{F} \cdot \mathbf{T} ds\), where \(\mathbf{F} = \langle 4x, 2y \rangle\) and \(C\) is given by \(r(t) = \langle t, 4t^3 \rangle\) for \(0 \leq t \leq 4\), we need to follow these steps:
1. Find the unit tangent vector \(\mathbf{T}\).
2. Parameterize the curve \(C\) in terms of \(t\).
3. Compute \(\mathbf{F} \cdot \mathbf{T}\).
4. Find \(ds\) (differential arc length).
5. Set up and evaluate the line integral.
Let's go through each step:
1. Find the unit tangent vector \(\mathbf{T}\):
The unit tangent vector \(\mathbf{T}\) is given by the derivative of \(\mathbf{r}(t)\) with respect to \(t\) divided by its magnitude:
\[
\mathbf{T} = \frac{d\mathbf{r}}{dt} / |\frac{d\mathbf{r}}{dt}|
\]
So, let's calculate \(\frac{d\mathbf{r}}{dt}\):
\[
\frac{d\mathbf{r}}{dt} = \langle 1, 12t^2 \rangle
\]
Now, we need to find the magnitude of this vector:
\[
|\frac{d\mathbf{r}}{dt}| = \sqrt{1^2 + (12t^2)^2} = \sqrt{1 + 144t^4}
\]
Now, we can find the unit tangent vector \(\mathbf{T}\):
\[
\mathbf{T} = \frac{1}{\sqrt{1 + 144t^4}} \langle 1, 12t^2 \rangle
\]
2. Parameterize the curve \(C\) in terms of \(t\):
We already have the parameterization of \(C\) given as \(r(t) = \langle t, 4t^3 \rangle\) for \(0 \leq t \leq 4\).
3. Compute \(\mathbf{F} \cdot \mathbf{T}\):
\[
\mathbf{F} \cdot \mathbf{T} = \langle 4x, 2y \rangle \cdot \frac{1}{\sqrt{1 + 144t^4}} \langle 1, 12t^2 \rangle = \frac{4x + 24t^2y}{\sqrt{1 + 144t^4}}
\]
4. Find \(ds\) (differential arc length):
\(ds\) is given by:
\[
ds = |\frac{d\mathbf{r}}{dt}|dt = \sqrt{1 + 144t^4} dt
\]
5. Set up and evaluate the line integral:
The line integral \(\int_C \mathbf{F} \cdot \mathbf{T} ds\) becomes:
\[
\int_{0}^{4} \frac{4x + 24t^2y}{\sqrt{1 + 144t^4}} \sqrt{1 + 144t^4} dt
\]
Simplifying the integral:
\[
\int_{0}^{4} (4x + 24t^2y) dt
\]
Now, plug in the parameterization \(r(t) = \langle t, 4t^3 \rangle\):
\[
\int_{0}^{4} (4t + 24t^5) dt
\]
Integrate with respect to \(t\):
\[
\int_{0}^{4} (4t + 24t^5) dt = \left[2t^2 + 4t^6\right]_{0}^{4} = (2(4)^2 + 4(4)^6) - (2(0)^2 + 4(0)^6) = 32 + 16384 = 16416
\]
So, the value of the line integral \(\int_C \mathbf{F} \cdot \mathbf{T} ds\) is 16416.
Explanation: