Respuesta :
Answer:
The arc length is [tex]\dfrac{21}{16}[/tex]
Step-by-step explanation:
Given that,
The given curve between the specified points is
[tex]x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}[/tex]
The points from [tex](\dfrac{9}{16},1)[/tex] to [tex](\dfrac{9}{8},2)[/tex]
We need to calculate the value of [tex]\dfrac{dx}{dy}[/tex]
Using given equation
[tex]x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}[/tex]
On differentiating w.r.to y
[tex]\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})[/tex]
[tex]\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})[/tex]
[tex]\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})[/tex]
[tex]\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}[/tex]
We need to calculate the arc length
Using formula of arc length
[tex]L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}[/tex]
Put the value into the formula
[tex]L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}[/tex]
[tex]L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}[/tex]
[tex]L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}[/tex]
[tex]L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}[/tex]
[tex]L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}[/tex]
[tex]L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}[/tex]
[tex]L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}[/tex]
[tex]L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}[/tex]
Put the limits
[tex]L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})[/tex]
[tex]L=\dfrac{21}{16}[/tex]
Hence, The arc length is [tex]\dfrac{21}{16}[/tex]
The arc length of the given curve is approximately 0.806 units.
Procedure - Determination of the arc length for a segment of a curve
By calculus we know that the arc length ([tex]s[/tex]) for a function [tex]x = f(y)[/tex] is defined by the following expression:
[tex]s = \int\limits^{y_{2}}_{y_{1}} {\sqrt{1+\left(\frac{dx}{dy} \right)^{2}}} \, dx[/tex] (1)
Where:
- [tex]y_{1}[/tex], [tex]y_{2}[/tex] - Lower and upper bounds of the integral.
- [tex]\frac{dx}{dy}[/tex] - First derivative of the function.
Let be [tex]x = \frac{y^{4}}{16}+\frac{y^{2}}{2}[/tex], if we know that [tex]\frac{dx}{dy} = \frac{y^{3}}{4}+y[/tex], [tex]y_{1} = \frac{9}{16}[/tex] and [tex]y_{2} = \frac{9}{8}[/tex], then the arc length of the curve segment is:
[tex]s = \int\limits^{\frac{9}{8} }_{\frac{9}{16} } {\sqrt{1+y^{2}\cdot \left(\frac{y^{2}}{4}+1 \right)^{2}}} \, dy[/tex]
[tex]s = \frac{1}{4} \int\limits^{9/8}_{\frac{9}{16} } {\sqrt{16+16\cdot y^{2}+8\cdot y^{4}+y^{6}}} \, dy[/tex]
This integral can be solved analytically but it may required a lot of time and effort to be done, a more efficient approach consists in using numerical methods with the help of a graphic tool (i.e. Wolfram Alpha).
From the graphic we have that the arc length of the given curve is approximately 0.806 units. [tex]\blacksquare[/tex]
To learn more on arc lengths, we kindly invite to check this verified question: https://brainly.com/question/304765
