Solve the problem.
-Find a curve through the point $( - 4,1 )$ whose length integral, $1 \leq y \leq 2$, is $L = \int _ { 1 } ^ { 2 } \sqrt { 1 + \frac { 4 } { y ^ { 3 } } } \mathrm { dy }$.
A) $x = \frac { 2 } { \sqrt { y } }$
B) $x = \frac { - 4 } { \sqrt { y } }$
C) $x = - 4 y ^ { 5 / 2 }$
D) $x = - 4 \sqrt { y }$

Question

Quizplus · Accepted Answer

The length of a curve from $y = a$ to $y = b$ for the curve $x = f(y)$ is given by $L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy$. Comparing this with the given integral, we see that $\left(\frac{dx}{dy}\right)^2 = \frac{4}{y^3}$, which implies $\frac{dx}{dy} = \frac{2}{y^{3/2}}$. Integrating $\frac{dx}{dy}$ with respect to $y$ gives $x = \frac{-4}{\sqrt{y}} + C$. Using the point $(-4, 1)$ to solve for $C$, we find $C = 0$, so the correct curve is $x = \frac{-4}{\sqrt{y}}$.

Solve the Problem (−4,1)( - 4,1 )(−4,1) Whose Length Integral 1≤y≤21 \leq y \leq 21≤y≤2

Solve the Problem $( - 4,1 )$ Whose Length Integral $1 \leq y \leq 2$