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Part 1
Given the following vector field and orientated curve​ C, evaluate ModifyingBelow Integral from nothing to nothing With Upper C Bold Upper F times Upper T font size decreased by 5 ds.
Bold Upper F equals StartFraction left angle x comma y right angle Over left parenthesis x squared plus y squared right parenthesis Superscript 3 divided by 2 EndFraction
on the curve Bold r left parenthesis t right parenthesis equals left angle 3 t squared comma 5 t squared right angle​, for 1 less than or equals t less than or equals 2
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Part 1
The value of the line integral of F over C is

enter your response here.
​(Type an exact​ answer, using radicals as​ needed.)

To evaluate the line integral of the vector field \(\mathbf{F}\) over the oriented curve \(C\), we start with the given vector field and curve parameterization:

\[
\mathbf{F} = \frac{\langle x, y \rangle}{(x^2 + y^2)^{3/2}}
\]

\[
\mathbf{r}(t) = \langle 3t^2, 5t^2 \rangle \quad \text{for} \quad 1 \leq t \leq 2
\]

The line integral of \(\mathbf{F}\) along \(C\) is given by:

\[
\int_C \mathbf{F} \cdot d\mathbf{r}
\]

We can express \(d\mathbf{r}\) in terms of \(t\):

\[
\mathbf{r}(t) = \langle 3t^2, 5t^2 \rangle
\]

Differentiating \(\mathbf{r}(t)\) with respect to \(t\):

\[
\mathbf{r}'(t) = \frac{d}{dt} \langle 3t^2, 5t^2 \rangle = \langle 6t, 10t \rangle
\]

So,

\[
d\mathbf{r} = \mathbf{r}'(t) \, dt = \langle 6t, 10t \rangle \, dt
\]

The line integral becomes:

\[
\int_{1}^{2} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt
\]

We need to evaluate \(\mathbf{F}(\mathbf{r}(t))\):

\[
x = 3t^2, \quad y = 5t^2
\]

\[
x^2 + y^2 = (3t^2)^2 + (5t^2)^2 = 9t^4 + 25t^4 = 34t^4
\]

\[
(x^2 + y^2)^{3/2} = (34t^4)^{3/2} = 34^{3/2} t^6
\]

\[
\mathbf{F}(\mathbf{r}(t)) = \frac{\langle 3t^2, 5t^2 \rangle}{34^{3/2} t^6} = \frac{1}{34^{3/2} t^6} \langle 3t^2, 5t^2 \rangle = \langle \frac{3t^2}{34^{3/2} t^6}, \frac{5t^2}{34^{3/2} t^6} \rangle = \langle \frac{3}{34^{3/2} t^4}, \frac{5}{34^{3/2} t^4} \rangle
\]

Now we need to compute the dot product \(\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\):

\[
\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \left\langle \frac{3}{34^{3/2} t^4}, \frac{5}{34^{3/2} t^4} \right\rangle \cdot \langle 6t, 10t \rangle
\]

\[
= \frac{3}{34^{3/2} t^4} \cdot 6t + \frac{5}{34^{3/2} t^4} \cdot 10t
\]

\[
= \frac{18t}{34^{3/2} t^4} + \frac{50t}{34^{3/2} t^4}
\]

\[
= \frac{18t + 50t}{34^{3/2} t^4} = \frac{68t}{34^{3/2} t^4}
\]

\[
= \frac{68t}{34^{3/2} t^4} = \frac{2t}{34^{1/2} t^3} = \frac{2}{34^{1/2} t^2}
\]

The integral is:

\[
\int_{1}^{2} \frac{2}{34^{1/2} t^2} \, dt
\]

\[
= \frac{2}{34^{1/2}} \int_{1}^{2} \frac{1}{t^2} \, dt
\]

\[
= \frac{2}{\sqrt{34}} \int_{1}^{2} t^{-2} \, dt
\]

\[
= \frac{2}{\sqrt{34}} \left[ -\frac{1}{t} \right]_{1}^{2}
\]

\[
= \frac{2}{\sqrt{34}} \left( -\frac{1}{2} + 1 \right)
\]

\[
= \frac{2}{\sqrt{34}} \left( \frac{1}{2} \right)
\]

\[
= \frac{2}{\sqrt{34}} \cdot \frac{1}{2}
\]

\[
= \frac{1}{\sqrt{34}}
\]

Therefore, the value of the line integral of \(\mathbf{F}\) over \(C\) is:

\[
\boxed{\frac{1}{\sqrt{34}}}
\]

Would you like any further details or have any questions?

Here are 8 related questions that expand on the previous information:

1. How do you parameterize a curve for line integrals?
2. What are the typical steps to compute a line integral of a vector field?
3. How do you interpret the physical meaning of a line integral in vector fields?
4. Can you provide an example of a line integral in a different coordinate system, like polar coordinates?
5. How does the orientation of the curve affect the line integral?
6. What is the relationship between line integrals and work done by a force field?
7. How would the problem change if the curve were defined for \(0 \leq t \leq 1\) instead?
8. What are the common applications of line integrals in engineering and physics?

**Tip:** Always check if the vector field is conservative, as this can simplify the calculation of line integrals.

Learn how to evaluate the line integral of a vector field over an oriented curve using the given parameterization and vector field formula. Detailed step-by-step solution provided.

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