Answer:
a. [tex]\mathbf{36 \sqrt{5}}[/tex]
b. [tex]\mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}[/tex]
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
[tex]\int \limits _c \ x \ ds[/tex]
where;
x = t , y = t/2
the derivative of x with respect to t is:
[tex]\dfrac{dx}{dt}= 1[/tex]
the derivative of y with respect to t is:
[tex]\dfrac{dy}{dt}= \dfrac{1}{2}[/tex]
and t varies from 0 to 12.
we all know that:
[tex]ds=\sqrt{ (\dfrac{dx}{dt})^2 + ( \dfrac{dy}{dt} )^2}} \ \ dt[/tex]
∴
[tex]\int \limits _c \ x \ ds = \int \limits ^{12}_{t=0} \ t \ \sqrt{1+(\dfrac{1}{2})^2} \ dt[/tex]
[tex]= \int \limits ^{12}_{0} \ \dfrac{\sqrt{5}}{2}(\dfrac{t^2}{2}) \ dt[/tex]
[tex]= \dfrac{\sqrt{5}}{2} \ \ [\dfrac{t^2}{2}]^{12}_0[/tex]
[tex]= \dfrac{\sqrt{5}}{4}\times 144[/tex]
= [tex]\mathbf{36 \sqrt{5}}[/tex]
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:
[tex]\dfrac{dx}{dt}= 1[/tex]
the derivative of y with respect to t is:
[tex]\dfrac{dy}{dt} = 6t[/tex]
[tex]ds = \sqrt{1+36 \ t^2} \ dt[/tex]
Hence; the integral _C x ds is:
[tex]\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt[/tex]
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
[tex]tdt = \dfrac{du}{76}[/tex]
The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145
Thus;
[tex]\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt[/tex]
[tex]\mathtt{= \int \limits ^{145}_{0} \sqrt{u} \ \dfrac{1}{72} \ du}[/tex]
[tex]= \dfrac{1}{72} \times \dfrac{2}{3} \begin {pmatrix} u^{3/2} \end {pmatrix} ^{145}_{1}[/tex]
[tex]\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}[/tex]
[tex]\mathbf{= \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}[/tex]
