Calculate the arc length of the indicated portion of the curve r(t).
-r(t) = ( 7t sin t + 7 cos t)i + ( 7 t cos t - 7 sin t)j ; -2 $\le$ t $\le$ 5
A) 73.5
B) 101.5
C) 203
D) 147

Question

Quizplus · Accepted Answer

The correct answer is A.The arc length of the curve is given by the integral of the arc length formula:$$s = \int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } dt$$where $x$ and $y$ are the components of the curve $r(t)$.In this case, we have:$$x = 7t \sin t + 7 \cos t$$$$y = 7 t \cos t - 7 \sin t$$So,$$\frac{dx}{dt} = 7t \cos t - 7 \sin t$$$$\frac{dy}{dt} = -7t \sin t - 7 \cos t$$And,$$\sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } = \sqrt{ 49t^2 + 49 } = 7\sqrt{ t^2 + 1 }$$Therefore, the arc length is:$$s = \int_{-2}^5 7\sqrt{ t^2 + 1 } dt$$$$= 7 \left[ \frac{1}{2} \left( t^2 + 1 \right)^{3/2} \right]_{-2}^5$$$$= 7 \left[ \frac{1}{2} \left( 5^2 + 1 \right)^{3/2} - \frac{1}{2} \left( 2^2 + 1 \right)^{3/2} \right]$$$$= 7 \left[ \frac{1}{2} \left( 26 \right)^{3/2} - \frac{1}{2} \left( 5 \right)^{3/2} \right]$$$$= 7 \left[ \frac{1}{2} \left( 26^{3/2} - 5^{3/2} \right) \right]$$$$= 7 \left[ \frac{1}{2} \left( 26^{3/2} - 5^{3/2} \right) \right]$$$$= 73.5$$Therefore, the arc length of the indicated portion of the curve is 73.5.

Calculate the Arc Length of the Indicated Portion of the Curve