The length of the curve $x ( t ) = 4 - 3 \cos t , y ( t ) = 3 + 3 \sin t$ with $t \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right]$ is

A) $\frac { \pi } { 2 }$
B) $2 \pi$
C) $3 \pi$
D) $\frac { \pi } { 4 }$
E) $\frac { 5 \pi } { 2 }$

Question

The length of the curve $x ( t ) = 4 - 3 \cos t , y ( t ) = 3 + 3 \sin t$ with $t \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right]$ is

A) $\frac { \pi } { 2 }$
B) $2 \pi$
C) $3 \pi$
D) $\frac { \pi } { 4 }$
E) $\frac { 5 \pi } { 2 }$

Quizplus · Accepted Answer

The length of a curve defined parametrically by $x(t)$ and $y(t)$ for $t$ in the interval $[a, b]$ is given by the integral $\int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$. For the given functions $x(t) = 4 - 3\cos(t)$ and $y(t) = 3 + 3\sin(t)$, we first find the derivatives: $\frac{dx}{dt} = 3\sin(t)$ and $\frac{dy}{dt} = 3\cos(t)$. Plugging these into the formula gives $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{(3\sin(t))^2 + (3\cos(t))^2} dt = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{9\sin^2(t) + 9\cos^2(t)} dt = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 dt = 3\left[\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right] = 3\pi$.

The Length of the Curve x(t)=4−3cos⁡t,y(t)=3+3sin⁡tx ( t ) = 4 - 3 \cos t , y ( t ) = 3 + 3 \sin tx(t)=4−3cost,y(t)=3+3sint

The Length of the Curve $x ( t ) = 4 - 3 \cos t , y ( t ) = 3 + 3 \sin t$