Evaluate the line integral along the curve C.
-$\int _ { C } \left( \frac { x ^ { 2 } + y ^ { 2 } } { z ^ { 2 } } \right) d s , C$ is the curve $r ( t ) = ( 2 \sin 6 t ) i + ( 2 \cos 6 t ) j + 5 t k , 2 \leq t \leq 4$
A) $- \frac { 91 } { 400 }$
B) $\frac { 13 } { 25 }$
C) $\frac { 169 } { 25 }$
D) $\frac { 1 } { 25 }$

Question

Quizplus · Accepted Answer

To evaluate the line integral $\int _ { C } \left( \frac { x ^ { 2 } + y ^ { 2 } } { z ^ { 2 } } \right) d s$, we first need to express $x$, $y$, and $z$ in terms of $t$ using the given curve $r(t) = (2\sin 6t)i + (2\cos 6t)j + 5tk$. Thus, $x = 2\sin 6t$, $y = 2\cos 6t$, and $z = 5t$.The expression $x^2 + y^2$ becomes $(2\sin 6t)^2 + (2\cos 6t)^2 = 4\sin^2 6t + 4\cos^2 6t = 4(\sin^2 6t + \cos^2 6t) = 4$, using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$.The denominator $z^2$ becomes $(5t)^2 = 25t^2$.Thus, the integrand simplifies to $\frac{4}{25t^2}$.To find $ds$, we compute the magnitude of the derivative of $r(t)$: $r'(t) = (12\cos 6t)i - (12\sin 6t)j + 5k$. The magnitude of $r'(t)$ is $\sqrt{(12\cos 6t)^2 + (-12\sin 6t)^2 + 5^2} = \sqrt{144\cos^2 6t + 144\sin^2 6t + 25} = \sqrt{144 + 25} = \sqrt{169} = 13$.Therefore, $ds = 13dt$, and the line integral becomes $\int_{2}^{4} \frac{4}{25t^2} \cdot 13dt = \frac{52}{25} \int_{2}^{4} t^{-2} dt$.Integrating, we get $\frac{52}{25} \left[ -t^{-1} \right]_{2}^{4} = \frac{52}{25} \left[ -\frac{1}{4} + \frac{1}{2} \right] = \frac{52}{25} \cdot \frac{1}{4} = \frac{13}{25}$.Hence, the correct answer is $\frac{13}{25}$.

Evaluate the Line Integral Along the Curve C ∫C(x2+y2z2)ds,C\int _ { C } \left( \frac { x ^ { 2 } + y ^ { 2 } } { z ^ { 2 } } \right) d s , C∫C​(z2x2+y2​)ds,C

Evaluate the Line Integral Along the Curve C $\int _ { C } \left( \frac { x ^ { 2 } + y ^ { 2 } } { z ^ { 2 } } \right) d s , C$