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inessss [21]
3 years ago
6

Find the line integral with respect to arc length ∫C(9x+5y)ds, where C is the line segment in the xy-plane with endpoints P=(2,0

) and Q=(0,7).
(a) Find a vector parametric equation r⃗ (t) for the line segment C so that points P and Q correspond to t=0 and t=1, respectively

(b) Rewrite integral using parametrization found in part a

(c) Evaluate the line integral with respect to arc length in part b
Mathematics
1 answer:
Gemiola [76]3 years ago
6 0

(a) You can parameterize <em>C</em> by the vector function

<em>r</em><em>(t)</em> = (<em>x(t)</em>, <em>y(t)</em> ) = <em>P</em> (1 - <em>t </em>) + <em>Q</em> <em>t</em> = (2 - 2<em>t</em>, 7<em>t</em> )

where 0 ≤ <em>t</em> ≤ 1.

(b) From the above parameterization, we have

<em>r</em><em>'(t)</em> = (-2, 7)   ==>   ||<em>r</em><em>'(t)</em>|| = √((-2)² + 7²) = √53

Then

d<em>s</em> = √53 d<em>t</em>

and the line integral is

\displaystyle\int_C(9x(t)+5y(t))\,\mathrm ds = \boxed{\sqrt{53}\int_0^1(17t+18)\,\mathrm dt}

(c) The remaining integral is pretty simple,

\displaystyle\sqrt{53}\int_0^1(17t+18)\,\mathrm dt = \sqrt{53}\left(\frac{17}2t^2+18t\right)\bigg|_{t=0}^{t=1} = \boxed{\frac{53^{3/2}}2}

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