Evaluate the line integral over the given curve $C$.
$$\int _ { C } ( 5 x + 4 y ) d s ; C : \mathbf { r } ( t ) = ( t - 5 ) \mathbf { i } + t \mathbf { j } , 0 \leq t \leq 3$$
A) $- \frac { 9 } { 2 }$
B) $- \frac { 69 } { 2 } \sqrt { 2 }$
C) $\frac { 231 } { 2 } \sqrt { 2 }$
D) $\frac { 81 } { 2 }$

Question

Quizplus · Accepted Answer

To evaluate the line integral $\int _ { C } ( 5 x + 4 y ) d s$, we first express $x$ and $y$ in terms of $t$ using the given curve $C: \mathbf { r } ( t ) = ( t - 5 ) \mathbf { i } + t \mathbf { j }$, where $x = t - 5$ and $y = t$. The differential arc length $ds$ can be found by computing the magnitude of the derivative of $\mathbf{r}(t)$ with respect to $t$, which gives $ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt = \sqrt{1^2 + 1^2}dt = \sqrt{2}dt$. Substituting $x$, $y$, and $ds$ into the integral and evaluating it from $t = 0$ to $t = 3$ gives the result $- \frac { 69 } { 2 } \sqrt { 2 }$.

Evaluate the Line Integral Over the Given Curve CCC ∫C(5x+4y)ds;C:r(t)=(t−5)i+tj,0≤t≤3\int _ { C } ( 5 x + 4 y ) d s ; C : \mathbf { r } ( t ) = ( t - 5 ) \mathbf { i } + t \mathbf { j } , 0 \leq t \leq 3∫C​(5x+4y)ds;C:r(t)=(t−5)i+tj,0≤t≤3

Evaluate the Line Integral Over the Given Curve $C$ $\int _ { C } ( 5 x + 4 y ) d s ; C : \mathbf { r } ( t ) = ( t - 5 ) \mathbf { i } + t \mathbf { j } , 0 \leq t \leq 3$