Calculus 8th Eighth Edition 10.3 #22

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In summary, the problem is asking for the equations of the tangent lines at the point where the curves intersect. To solve this, we can observe where the curve intersects y=0 and plug in the values for x and y to find the values of t and s. After solving for s and t, it is clear that the x and y values for s and t are the same, giving us the equations for the tangent lines.
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calculushelp
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Homework Statement


Find the equations of the tangent lines at the point where the curves crosses itself.

x=2-piCos(t), y=2t-piSin(t)


Homework Equations





The Attempt at a Solution

 
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  • #2
What have you tried so far?
 
  • #3
well, all i need are the points and after that its easy. But the PROBLEM is the points that's is where i am stuck
 
  • #4
Points? The problem only mentions one. Anyway, there wasn't any trick to it rather than realizing that the textbook wouldn't make a calculus problem very difficult in precalculus terms. I'd observe where the curve intercepts y=0. Any time you see such a strange formulation of the x and y coordinates, you know something is bound to simplify dramatically, so start plugging in things!
 
  • #5
At the point where the graph crosses itself, you must have the point [itex]2- \pi cos(t)= 2- \pi sin(s)[/itex] for some s and t and [itex]2t- \pi sin(t)= 2s-\pi sin(s)[/itex]. Solve those equations for s and t.
 
  • #6
sorry, but still confusion.. can you take it step by step please
 
  • #7
[itex]2- \pi cos(t)= 2- \pi sin(s)[/itex] and is this right? or a typo because I thought its was suppose to be [itex]2- \pi sin(t)= 2- \pi sin(s)[/itex]
 
  • #8
calculushelp said:
[itex]2- \pi cos(t)= 2- \pi sin(s)[/itex] and is this right? or a typo because I thought its was suppose to be [itex]2- \pi sin(t)= 2- \pi sin(s)[/itex]
Sorry that was a typo. It should be [itex]2- \pi cos(t)= 2- \pi cos(s)[/itex] which says that the x values for s and t are the same. Remember that [itex]x= 2- \pi cos(t)[/itex].
The other equation is [itex]2t- \pi sin(t)= 2s- \pi sin(s)[itex] which says that the y values for s and t are the same- from [itex]y= 2t- \pi cos(t)[/itex].

It should be obvious that [itex]2- \pi cos(t)= 2- \pi cos(x)[/itex] gives cos(t)= cos(s) which does not mean t= s but rather that [itex]t= 2\pi- s[/itex].
 

Related to Calculus 8th Eighth Edition 10.3 #22

1. What is Calculus 8th Eighth Edition 10.3 #22?

Calculus 8th Eighth Edition 10.3 #22 refers to problem 22 in section 10.3 of the 8th edition of the Calculus textbook. It is a specific problem related to the topic being covered in that section.

2. Is the 8th edition of the Calculus textbook the most recent version?

No, the 8th edition of the Calculus textbook is not the most recent version. There may be newer editions available, depending on when the book was published and if any updates have been made.

3. What is the purpose of including specific problems like #22 in a textbook?

The purpose of including specific problems like #22 in a textbook is to provide students with practice and application of the concepts being taught. These problems allow students to test their understanding and improve their problem-solving skills.

4. What is the topic covered in section 10.3 of the Calculus textbook?

Section 10.3 of the Calculus textbook covers the concept of limits and continuity. This section introduces the idea of limits and how they are used to determine the behavior of a function at a specific point.

5. Is it necessary to solve all the problems in a textbook to understand the material?

No, it is not necessary to solve all the problems in a textbook to understand the material. However, solving a variety of problems can help reinforce understanding and improve problem-solving skills. It is important to prioritize and focus on the most relevant problems for your learning goals.

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