Question

consider the line integral

$$ int _ { c } ( 4 x + 9 y ) d s $$

where ( c ) is the line segment in the ( x y )-plane with endpoints ( p = ( 8,0 ) ) and ( q = ( 0,4 ) ).
(a) find a parameterization ( mathbf { r } ( t ) ), ( 0 leq t leq 1 ), for the line segment ( c ) so that points ( p ) and ( q ) correspond to ( t = 0 ) and ( t = 1 ),
respectively.
( mathbf { r } ( t ) = )
(b) using the parametrization in part (a), the line integral with respect to arc length is

$$ int _ { c } ( 4 x + 9 y ) d s = int _ { a } ^ { b } d t $$

with limits of integration ( a = )
and ( b = )
(c) evaluate the integral in part (b)

$$ int _ { c } ( 4 x + 9 y ) d s = $$

submit answer
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Explanation:

Step1: Find the parameterization

The formula for the parameterization of a line segment from \((x_1,y_1)\) to \((x_2,y_2)\) is \(\mathbf{r}(t)=(x_1+(x_2 - x_1)t,y_1+(y_2 - y_1)t)\), where \(t\in[0,1]\).
Given \(P=(8,0)\) and \(Q=(0,4)\), then \(\mathbf{r}(t)=(8-8t,0 + 4t)=(8-8t,4t)\).

Step2: Calculate \(ds\)

If \(\mathbf{r}(t)=(x(t),y(t))\), then \(ds=\sqrt{(x^{\prime}(t))^{2}+(y^{\prime}(t))^{2}}dt\).
Here \(x(t)=8-8t\), \(x^{\prime}(t)=-8\); \(y(t)=4t\), \(y^{\prime}(t)=4\).
So \(ds=\sqrt{(-8)^{2}+4^{2}}dt=\sqrt{64 + 16}dt=\sqrt{80}dt = 4\sqrt{5}dt\).

Step3: Substitute \(x,y\) and \(ds\) into the line - integral

The line integral \(\int_{C}(4x + 9y)ds=\int_{a}^{b}[4x(t)+9y(t)]\sqrt{(x^{\prime}(t))^{2}+(y^{\prime}(t))^{2}}dt\).
Substitute \(x(t)=8-8t\), \(y(t)=4t\) and \(ds = 4\sqrt{5}dt\) with \(a = 0\) and \(b = 1\).

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Step4: Evaluate the integral

Use the integral formula \(\int_{0}^{1}(32 + 4t)dt=\int_{0}^{1}32dt+\int_{0}^{1}4tdt\).
\(\int_{0}^{1}32dt=32t\big|_{0}^{1}=32\) and \(\int_{0}^{1}4tdt=4\times\frac{t^{2}}{2}\big|_{0}^{1}=2\).
So \(4\sqrt{5}\int_{0}^{1}(32 + 4t)dt=4\sqrt{5}(32 + 2)=136\sqrt{5}\).

Answer:

(a) \(\mathbf{r}(t)=(8-8t,4t)\), (b) \(a = 0\), \(b = 1\), (c) \(\int_{C}(4x + 9y)ds=136\sqrt{5}\)