Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To find the radius of curvature [tex]\( R \)[/tex] of the rectangular hyperbola given by the equation
[tex]\[ r^2 = a^2 \sec(2\theta), \][/tex]
we will follow a series of steps involving differentiation and the use of a specific formula for radius of curvature in polar coordinates.
### Step-by-Step Solution:
1. Given Equation:
[tex]\[ r^2 = a^2 \sec(2\theta) \][/tex]
2. Solving for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{a^2}{\cos(2\theta)}} = \frac{a}{\sqrt{\cos(2\theta)}} \][/tex]
3. First Derivative [tex]\( \frac{dr}{d\theta} \)[/tex]:
Differentiate [tex]\( r \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ r' = \frac{d}{d\theta} \left( \frac{a}{\sqrt{\cos(2\theta)}} \right) \][/tex]
Using the chain rule:
[tex]\[ r' = \frac{a \cdot \left(- \frac{\sin(2\theta)}{\sqrt{\cos(2\theta)}^3} \cdot 2\right)}{2} = \frac{a \cdot (-2\sin(2\theta))}{2\cos(2\theta)^{3/2}} = \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \][/tex]
4. Second Derivative [tex]\( \frac{d^2r}{d\theta^2} \)[/tex]:
Differentiate [tex]\( r' \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ r'' = \frac{d}{d\theta} \left( \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \right) \][/tex]
Using the product and chain rule, we get:
[tex]\[ r'' = a \left[ \frac{(2\cos(2\theta) \cdot 3\sin(2\theta))}{\cos(2\theta)^3} + \frac{2(-\cos(2\theta)^2\sin(2\theta))}{\cos(2\theta)^3} \right] = a \left( \frac{3\sin(2\theta)^2}{\cos(2\theta)^{5/2}} - \frac{2\cos(2\theta)}{\cos(2\theta)^{3/2}} \right) \][/tex]
Simplifying:
[tex]\[ r'' = \frac{3a\sin(2\theta)^2}{\cos(2\theta)^{5/2}} + \frac{2a}{\cos(2\theta)^{3/2}} \][/tex]
5. Radius of Curvature [tex]\( R \)[/tex]:
The radius of curvature formula in polar coordinates is:
[tex]\[ R = \frac{(1 + (r')^2)^{3/2}}{|r''|} \][/tex]
Substituting [tex]\( r' \)[/tex] and [tex]\( r'' \)[/tex]:
[tex]\[ R = \frac{\left(1 + \left( \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \right)^2\right)^{3/2}}{\left| \frac{3a \sin(2\theta)^2}{\cos(2\theta)^{5/2}} + \frac{2a}{\cos(2\theta)^{3/2}} \right| } \][/tex]
Simplifying the expression inside the numerator:
[tex]\[ R = \frac{\left(1 + \frac{a^2 \sin^2 (2\theta)}{\cos^3 (2\theta)} \right)^{3/2}}{\left| \frac{3a \sin^2(2\theta)}{\cos^{5/2}(2\theta)} + \frac{2a}{\cos^{3/2}(2\theta)} \right| } \][/tex]
Thus, the final formula for the radius of curvature [tex]\( R \)[/tex] is:
[tex]\[ R = \frac{\left(a^2 \frac{\sin^2 (2\theta)}{\cos^3 (2\theta)} + 1 \right)^{3/2}}{\left| \frac{3\sqrt{a^2 \sec(2 \theta)} \sin^2 (2\theta)}{\cos^{2} (2\theta)} + \frac{2\sqrt{a^2 \sec(2 \theta)}}{\cos (2 \theta)} \right|} \][/tex]
In conclusion, after simplifying, the radius of curvature of the given rectangular hyperbola is:
[tex]\[ R = \frac{\left(\frac{a^2 \sin^2 (2\theta)}{\cos^3 (2\theta)} + 1 \right)^{3/2}}{\left| \frac{3\sqrt{a^2 \sec(2 \theta)} \sin^2 (2\theta)}{\cos^{2} (2\theta)} + \frac{2\sqrt{a^2 \sec(2 \theta)}}{\cos (2 \theta)} \right| }. \][/tex]
[tex]\[ r^2 = a^2 \sec(2\theta), \][/tex]
we will follow a series of steps involving differentiation and the use of a specific formula for radius of curvature in polar coordinates.
### Step-by-Step Solution:
1. Given Equation:
[tex]\[ r^2 = a^2 \sec(2\theta) \][/tex]
2. Solving for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{a^2}{\cos(2\theta)}} = \frac{a}{\sqrt{\cos(2\theta)}} \][/tex]
3. First Derivative [tex]\( \frac{dr}{d\theta} \)[/tex]:
Differentiate [tex]\( r \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ r' = \frac{d}{d\theta} \left( \frac{a}{\sqrt{\cos(2\theta)}} \right) \][/tex]
Using the chain rule:
[tex]\[ r' = \frac{a \cdot \left(- \frac{\sin(2\theta)}{\sqrt{\cos(2\theta)}^3} \cdot 2\right)}{2} = \frac{a \cdot (-2\sin(2\theta))}{2\cos(2\theta)^{3/2}} = \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \][/tex]
4. Second Derivative [tex]\( \frac{d^2r}{d\theta^2} \)[/tex]:
Differentiate [tex]\( r' \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ r'' = \frac{d}{d\theta} \left( \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \right) \][/tex]
Using the product and chain rule, we get:
[tex]\[ r'' = a \left[ \frac{(2\cos(2\theta) \cdot 3\sin(2\theta))}{\cos(2\theta)^3} + \frac{2(-\cos(2\theta)^2\sin(2\theta))}{\cos(2\theta)^3} \right] = a \left( \frac{3\sin(2\theta)^2}{\cos(2\theta)^{5/2}} - \frac{2\cos(2\theta)}{\cos(2\theta)^{3/2}} \right) \][/tex]
Simplifying:
[tex]\[ r'' = \frac{3a\sin(2\theta)^2}{\cos(2\theta)^{5/2}} + \frac{2a}{\cos(2\theta)^{3/2}} \][/tex]
5. Radius of Curvature [tex]\( R \)[/tex]:
The radius of curvature formula in polar coordinates is:
[tex]\[ R = \frac{(1 + (r')^2)^{3/2}}{|r''|} \][/tex]
Substituting [tex]\( r' \)[/tex] and [tex]\( r'' \)[/tex]:
[tex]\[ R = \frac{\left(1 + \left( \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \right)^2\right)^{3/2}}{\left| \frac{3a \sin(2\theta)^2}{\cos(2\theta)^{5/2}} + \frac{2a}{\cos(2\theta)^{3/2}} \right| } \][/tex]
Simplifying the expression inside the numerator:
[tex]\[ R = \frac{\left(1 + \frac{a^2 \sin^2 (2\theta)}{\cos^3 (2\theta)} \right)^{3/2}}{\left| \frac{3a \sin^2(2\theta)}{\cos^{5/2}(2\theta)} + \frac{2a}{\cos^{3/2}(2\theta)} \right| } \][/tex]
Thus, the final formula for the radius of curvature [tex]\( R \)[/tex] is:
[tex]\[ R = \frac{\left(a^2 \frac{\sin^2 (2\theta)}{\cos^3 (2\theta)} + 1 \right)^{3/2}}{\left| \frac{3\sqrt{a^2 \sec(2 \theta)} \sin^2 (2\theta)}{\cos^{2} (2\theta)} + \frac{2\sqrt{a^2 \sec(2 \theta)}}{\cos (2 \theta)} \right|} \][/tex]
In conclusion, after simplifying, the radius of curvature of the given rectangular hyperbola is:
[tex]\[ R = \frac{\left(\frac{a^2 \sin^2 (2\theta)}{\cos^3 (2\theta)} + 1 \right)^{3/2}}{\left| \frac{3\sqrt{a^2 \sec(2 \theta)} \sin^2 (2\theta)}{\cos^{2} (2\theta)} + \frac{2\sqrt{a^2 \sec(2 \theta)}}{\cos (2 \theta)} \right| }. \][/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
