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inessss [21]
2 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]2 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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A cube is a three-dimensional solid object having six square faces. The measure of the length of the diagonal of the face of the cube is 12.2 cm.

<h3>What is a cube?</h3>

A cube is a three-dimensional solid object having six square faces, facets, or sides, three of which meet at each vertex. The cube is one of the five Platonic solids and the only regular hexahedron.

Given the length of an edge of the cube is √75 cm, therefore, the length of the diagonal of the face of the cube will be √2 times the length of an edge of the cube. Thus, the measure of the length of the diagonal of the face of the cube is,

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