What went wrong in calculating the curvature for r(t)=<t^2,lnt,tlnt>?

  • Thread starter bl4ke360
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In summary, the student is asking for an explanation as to why their answer to a problem involving a parametric equation was marked incorrect. The professor's answer sheet used a different method, resulting in a different answer. The student's confusion is due to a mistake in normalizing a unit vector, which affected the calculation of T'(t). By plugging in a specific value for r'(t) instead of leaving it as a variable, the student lost important terms in the differentiation process.
  • #1
bl4ke360
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Homework Statement



r(t)=<t^2,lnt,tlnt>

Homework Equations



k= |T '(t)| / |r '(t)|

The Attempt at a Solution



2llfsyu.jpg



My professor's answer sheet solved the problem using the other method, k(t)=|r '(t) x r ''(t)| / |r '(t)|^3
and that answer ends up being 0.3, while mine is 0.4. I can't see where I made any mistakes, and why I got 9 points marked off. Can someone explain?
 
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  • #2
bl4ke360 said:

Homework Statement



r(t)=<t^2,lnt,tlnt>

Homework Equations



k= |T '(t)| / |r '(t)|

The Attempt at a Solution



2llfsyu.jpg



My professor's answer sheet solved the problem using the other method, k(t)=|r '(t) x r ''(t)| / |r '(t)|^3
and that answer ends up being 0.3, while mine is 0.4. I can't see where I made any mistakes, and why I got 9 points marked off. Can someone explain?

Yes. T(t) is supposed to be a unit vector for any value of t. You normalized it so that |T(1)|=1. But it's not a unit vector for any t. So your T'(t) comes out wrong.
 
  • #3
I think what you did wrong was when you calculated T(t), you plugged in |r'(1)| in the denominator, when you should have left it as |r'(t)| until after you differentiated T(t) to get T'(t). By doing this you lost the terms related to changes in |r'(t)|. Does this make sense?
 

Related to What went wrong in calculating the curvature for r(t)=<t^2,lnt,tlnt>?

1. What is curvature and why is it important?

Curvature is a measure of how much a curve deviates from being a straight line. It is important in understanding the shape of objects and surfaces in mathematics and physics. In physics, it is used to describe the bending of space and time in the theory of general relativity.

2. How can I find the curvature of a curve or surface?

The curvature of a curve can be found by calculating the rate of change of the curve's tangent vector. For a surface, the curvature can be found by calculating the principal curvatures at each point on the surface.

3. What is the difference between Gaussian and mean curvature?

Gaussian curvature measures the deviation of a surface from being flat, while mean curvature measures the average curvature of a surface. In other words, Gaussian curvature describes how much a surface is curved in both directions, while mean curvature describes how much it is curved in the direction of its normal vector.

4. How is curvature related to the shape of an object or surface?

Curvature is directly related to the shape of an object or surface. A high curvature means that the object or surface is highly curved, while a low curvature means it is relatively flat. Curvature also determines the behavior of light and other particles moving on or through the surface.

5. Can curvature be negative?

Yes, curvature can be negative. A negative curvature means that the surface is curved in the opposite direction of a positive curvature. For example, a saddle-shaped surface has both positive and negative curvature at different points on the surface.

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