Answer:
a.
b.
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
where;
x = t , y = t/2
the derivative of x with respect to t is:
the derivative of y with respect to t is:
and t varies from 0 to 12.
we all know that:
∴
=
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:
the derivative of y with respect to t is:
Hence; the integral _C x ds is:
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
The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145
Thus;