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olchik [2.2K]
3 years ago
12

Evaluate integral _C x ds, where C is

Mathematics
1 answer:
borishaifa [10]3 years ago
5 0

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]}}

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