Calculating the Curvature of a Trigonometric Curve | Math Homework Help

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In summary, the conversation discussed how to calculate the curvature of a curve using the given position function. The correct answer for the curvature is found to be 2/(cos^2(t)+4sin^2(t))^(3/2). Several attempts were made to reach this answer, but the mistake of dropping +1 from the bottom was identified and corrected. The final formula for the curvature is simplified to 2/(cos^2(t)+4sin^2(t))^(3/2).
  • #1
roam
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Homework Statement



I am trying to calculate the curvature of a curve given by the position function:

[itex]\vec{r}(t)= sin(t) \vec{i} + 2 \ cos (t) \ vec{j}[/itex]

The correct answer must be:

[itex]\kappa (t) = \frac{2}{(cos^2(t) + 4 \ sin^2 (t))^{3/2}}[/itex]

I tried several times but I can't arrive at this answer. :confused:

Homework Equations



Curvature is given by:

[itex]\kappa(t) = \frac{||T'(t)||}{||r'(t)||}[/itex]

Where

[itex]T (t) = \frac{r'(t)}{||r'(t)||}[/itex]

The Attempt at a Solution



[itex]r'(t) = \left\langle cos(t), \ -2 sin(t) \right\rangle[/itex]

[itex]||r'(t)|| = \sqrt{cos^2(t) + (2 \ sin(t))^2} = \sqrt{3 \ sin^2t+1}[/itex]

Therefore

[itex]T = \frac{cos(t)}{\sqrt{3 \ sin^2 (t) +1}} \ \vec{i}, \ \frac{-2 \ sin(t)}{\sqrt{3 \ sin^2 (t) +1}} \vec{j}[/itex]

[itex]\frac{dT}{dt} = \frac{-4 \ sin(t)}{(3 \ sin^2(t) +1)^{(3/2)}} \ \vec{i} + \frac{-2 \cos (t)}{(3 \ sin^2(t) +1)^{(3/2)}} \ \vec{j}[/itex]

[itex]||T'(t)|| = \sqrt{\left( \frac{-4 \ sin(t)}{3 \ sin^2(t) + 1)^{3/2}} \right)^2 + \left( \frac{-2 \cos (t)}{(3 \ sin^2(t) + 1)^{3/2}} \right)^2}[/itex]

[itex]= \sqrt{ \frac{16 \ sin^2(t)}{(3 \ sin^2 t)^3} + \frac{4 \ cos^2(t)}{(3 \ sin^2 t)^3}}[/itex]

Putting this in the equation given

[itex]\kappa = \sqrt{ \frac{16 \ sin^2(t)}{(3 \ sin^2 t)^3} + \frac{4 \ cos^2(t)}{(3 \ sin^2 t)^3}}. \frac{1}{\sqrt{3 \ sin^2t+1}}[/itex]

But I can't see how this can be simplified to get to the correct answer. I appreciate any guidance.
 
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  • #2


You dropped +1 from the bottom, while calculating ||T'||. The top equals [itex]12sin^2(t)+4 = 4(3sin^2(t)+1)[/itex] and it all works out nicely.
 
  • #3


hamsterman said:
You dropped +1 from the bottom, while calculating ||T'||. The top equals [itex]12sin^2(t)+4 = 4(3sin^2(t)+1)[/itex] and it all works out nicely.

Thank you very much! I got it! :)
 
Last edited:

1. What is a curvature math problem?

A curvature math problem is a mathematical question that involves calculating the curvature of a curve or surface. It typically requires knowledge of calculus and geometry.

2. How is curvature calculated?

Curvature can be calculated using the formula: κ = |dT/ds| / |dR/ds|, where κ represents curvature, T represents the tangent vector, and R represents the radius of curvature.

3. What is the difference between positive and negative curvature?

Positive curvature refers to a curve that is bending outward, while negative curvature refers to a curve that is bending inward. This can also be seen in the corresponding values of κ, where positive curvature has a positive value and negative curvature has a negative value.

4. How is curvature used in real life?

Curvature is used in many fields, including physics, engineering, and computer graphics. It can help in understanding the shape and behavior of objects, designing structures, and creating realistic 3D models.

5. What are some common applications of curvature math problems?

Some common applications of curvature math problems include finding the optimal path for a vehicle to travel on, analyzing the shape of a lens or mirror in optics, and determining the stress and strain on materials in mechanical engineering.

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