Question

Evaluate integral _C x ds, where C is

Answer 1

Answer:

a.    $\mathbf{36 \sqrt{5}}$

b.   $\mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}$

Step-by-step explanation:

Evaluate integral _C x ds  where C is

a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)

i . e

$\int \limits _c \ x \ ds$

where;

x = t   , y = t/2

the derivative of x with respect to t is:

$\dfrac{dx}{dt}= 1$

the derivative of y with respect to t is:

$\dfrac{dy}{dt}= \dfrac{1}{2}$

and t varies from 0 to 12.

we all know that:

$ds=\sqrt{ (\dfrac{dx}{dt})^2 + ( \dfrac{dy}{dt} )^2}} \ \ dt$

∴

$\int \limits _c \ x \ ds = \int \limits ^{12}_{t=0} \ t \ \sqrt{1+(\dfrac{1}{2})^2} \ dt$

$= \int \limits ^{12}_{0} \ \dfrac{\sqrt{5}}{2}(\dfrac{t^2}{2}) \ dt$

$= \dfrac{\sqrt{5}}{2} \ \ [\dfrac{t^2}{2}]^{12}_0$

$= \dfrac{\sqrt{5}}{4}\times 144$

= $\mathbf{36 \sqrt{5}}$

b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)

Given that:

x = t  ; y = 3t²

the derivative of  x with respect to t is:

$\dfrac{dx}{dt}= 1$

the derivative of y with respect to t is:

$\dfrac{dy}{dt} = 6t$

$ds = \sqrt{1+36 \ t^2} \ dt$

Hence; the  integral _C x ds is:

$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

Let consider u to be equal to  1 + 36t²

1 + 36t² = u

Then, the differential of t with respect to u is :

76 tdt = du

$tdt = \dfrac{du}{76}$

The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145

Thus;

$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

$\mathtt{= \int \limits ^{145}_{0} \sqrt{u} \ \dfrac{1}{72} \ du}$

$= \dfrac{1}{72} \times \dfrac{2}{3} \begin {pmatrix} u^{3/2} \end {pmatrix} ^{145}_{1}$

$\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}$

$\mathbf{= \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}$