Question

Let C be the curve given parametrically by r(t) = ⟨t + 3, 4 − 2t⟩, t : 1 → 2; if f(x, y) = 2x + 4y, the value of ∫C f(x, y) ds is?

A. 0
B. 13√5 5
C. 4√5 5
D. 26√5 5
E. 22√5

Answer 1

Replace x and y with the corresponding components of r(t), where

`\mathbf r(t)=\langle x(t),y(t)\rangle=\langlet+3,4-2t\rangle`

We have

`\displaystyle\int_Cf(x,y)\,\mathrm ds=\int_1^2f(x(t),y(t))\sqrt{\left((\mathrm dx)/(\mathrm dt)\right)^2+\left((\mathrm dy)/(\mathrm dt)\right)^2}\,\mathrm dt`

`=\displaystyle\int_1^2(2(t+3)+4(4-2t))√(1^2+(-2)^2)\,\mathrm dt`

`=\displaystyle\sqrt5\int_1^2(22-6t)\,\mathrm dt`

`=\sqrt5(22t-3t^2)\bigg|_1^2=\boxed{13\sqrt5}`

I'm tempted to say the answer is B, but it doesn't seem to match up exactly. It's possible that choice contains a typo.