Set up an integral for the length of the curve.
-$$y = 7 \cot x , \frac { \pi } { 4 } \leq x \leq \frac { \pi } { 2 }$$
A) $\int _ { \pi / 4 } ^ { \pi / 2 } \sqrt { 1 + 49 \csc ^ { 4 } x } d x$
B) $\int _ { \pi / 4 } ^ { \pi / 2 } \sqrt { 1 + 7 \csc ^ { 2 } x } d x$
C) $\int _ { \pi / 4 } ^ { \pi / 2 } \sqrt { 1 - 49 \csc ^ { 2 } x } d x$
D) $\int _ { \pi / 4 } ^ { \pi / 2 } \sqrt { 1 + 49 \csc ^ { 2 } x } d x$

Question

Quizplus · Accepted Answer

The length of a curve defined by $y = f(x)$ from $x = a$ to $x = b$ is given by the integral $\int_a^b \sqrt{1 + (f'(x))^2} dx$. For $y = 7\cot(x)$, $f'(x) = -7\csc^2(x)$. Squaring this derivative gives $49\csc^4(x)$. Thus, the integral for the length of the curve is $\int _ { \pi / 4 } ^ { \pi / 2 } \sqrt { 1 + 49 \csc ^ { 4 } x } d x$.

Set Up an Integral for the Length of the Curve y=7cot⁡x,π4≤x≤π2y = 7 \cot x , \frac { \pi } { 4 } \leq x \leq \frac { \pi } { 2 }y=7cotx,4π​≤x≤2π​

Set Up an Integral for the Length of the Curve $y = 7 \cot x , \frac { \pi } { 4 } \leq x \leq \frac { \pi } { 2 }$