Did I Solve This Exam Prep Problem Correctly?

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Homework Help Overview

The discussion revolves around a problem related to finding the tangent vector for a curve defined by a position vector. The original poster is preparing for an exam and seeks validation of their work, which is not yielding an answer in the textbook.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the correct expression for the tangent vector at a specific point on the curve and question the implications of the problem's constraints, such as the absence of calculators during the exam. There are inquiries about the graphical representation of the solution.

Discussion Status

Some participants have offered guidance on deriving the tangent vector and formulating the parametric equation. There is an ongoing exploration of the necessary components, such as direction cosines, and a note of caution regarding normalization of the tangent vector.

Contextual Notes

The problem includes a calculator sign, suggesting it may be complex to graph, and there are specific exam rules that restrict the use of calculators.

mr_coffee
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Hello everyone! I need to seee if i did this problem right...The answer isn't in the back of the book and I'm studying for an exam! here is my work and problem:
http://show.imagehosting.us/show/775021/0/nouser_775/T0_-1_775021.jpg
if that link is slow check this one:
http://img132.imageshack.us/img132/8606/lastscan3kj.jpg
Thanks.
 
Last edited by a moderator:
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mr_coffee said:
Hello everyone! I need to seee if i did this problem right...The answer isn't in the back of the book and I'm studying for an exam! here is my work and problem:
http://show.imagehosting.us/show/775021/0/nouser_775/T0_-1_775021.jpg
if that link is slow check this one:
http://img132.imageshack.us/img132/8606/lastscan3kj.jpg
Thanks.
You still didn't answer the question: What would the graph look like?

Alex
 
Last edited by a moderator:
The problem has alittle calculator sign next to it...so that probably means its hard to graph, and our professor doesn't allow calculators on the exam, so I'm guessing he won't put it on there, but did i do the first part right?
 
First, you have not found the correct expression for the tangent vector at
(1,1,1)
You have been given the POSITION VECTOR FOR THE CURVE:
[tex]\vec{r}(t)=(t^{2},t^{4},t^{3})[/tex]
What is the general expression for the TANGENT vector at an arbitrary point "t" now?
 
You know that the tangent will pass through the point (1,1,1) (given in the question) , now for formulating the parametric equation for tangent , you also need the corresponding direction-cosines , Because you know that the corresponding Velocity vector = dr/dt will be parallel to the tangent , you can try differentiating the expression , and put in value of 't' to get the direction-cosines.

BJ
 
Dr.Brain said:
You know that the tangent will pass through the point (1,1,1) (given in the question) , now for formulating the parametric equation for tangent , you also need the corresponding direction-cosines , Because you know that the corresponding Velocity vector = dr/dt will be parallel to the tangent , you can try differentiating the expression , and put in value of 't' to get the direction-cosines.

BJ
A quibble:
They won't be direction cosines unless you normalize the tangent vector..
 

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