Unit Vectors for Tangent Line at (\pi/6, 1)

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Homework Help Overview

The problem involves finding unit vectors that are parallel and perpendicular to the tangent line of the curve y=2sinx at the point (\pi/6, 1). The discussion centers around understanding the relationship between the derivative and the components of the unit vectors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the derivative of the function and its implications for the slope of the tangent line. There are attempts to relate the slope to the components of unit vectors in the x and y directions. Some participants express confusion about the next steps in the process.

Discussion Status

The discussion includes various interpretations of how to represent the tangent line's slope in terms of unit vectors. Some participants offer guidance on visualizing the problem with a right-angle triangle, while others question the accuracy of proposed answers. There is no explicit consensus on the correct unit vectors at this stage.

Contextual Notes

Participants are navigating the challenge of translating the slope into vector components, with some uncertainty about the representation in terms of i and j. There is also a mention of simplifying expressions, which may indicate constraints in the problem-solving approach.

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Homework Statement


Find the unit vectors that are parallel to the tangent line to the curve y=2sinx at the point ([tex]\pi[/tex]/6, 1). Thereafter, find the unit vectors that are perpendicular to the tangent line.


Homework Equations



The Attempt at a Solution


I took the derivative of y=2sinx and got y'=2cosx. Then subbed in [tex]\pi[/tex]/6 and got slope=[tex]\sqrt{3}[/tex]. After this, I was totally confused about what to do next since I don't know how to put the function with respect to i, j. Thanks in advance.
 
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First thing you should note is that dy/dx tells you how the unit vector x-y components are related to each other. Draw the tangent line at pi/6, and a really small right-angle triangle with the vertical length denoted dy and horizontal length dx. So, see what to do next?
 
you have slop which is rise over run . so rise will be j-hat and run will be i-hat.you don't have to find k-hat.just simplify square root of 3.
 
Oh yes, I think i got it now. Just to confirm is it (i+root3j) and (-i-root3j)? Thank you so much guys.
 
Last edited:
Don't see how you got that answer. Might want to re-check your working.
 

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