Question

Evaluate the line integral, where c is the given curve. C (x/y) ds, c: x = t3, y = t4, 1 ≤ t ≤ 4.

Answer 1

We have

x = t³   ===>   dx/dt = 3t²

y = t⁴   ===>   dy/dt = 4t³

Then with the given parameteriztion, the line integral along C of x/y is

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^3}{t^4} \sqrt{(3t^2)^2 + (4t^3)^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac1t \sqrt{9t^4 + 16t^6} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{\sqrt{t^4}}t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^2}t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \int_1^4 t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 32t \sqrt{9 + 16t^2} \, dt$

$\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 \sqrt{9 + 16t^2} \, d\left(9+16t^2\right)$

$\displaystyle \int_C \frac xy \, ds = \frac1{32} \cdot \frac23 \left(9+16t^2\right)^{\frac32}\bigg|_1^4$

$\displaystyle \int_C \frac xy \, ds = \frac1{48} \left(265^{\frac32} - 25^{\frac32}\right) = \boxed{\frac{265\sqrt{265}-125}{48}}$