Find the angle made by two tangents

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Homework Help Overview

The problem involves finding the angle made by two tangents to the parametric curve defined by the equations \(x=\sin(2t)\) and \(y=\sin(2t)\cos(2t)\) at the point \((0,0)\). The discussion centers on the calculation of slopes at specific values of \(t\) and the interpretation of these slopes in relation to the angle between the tangents.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the slopes of the tangents at various values of \(t\) and questions their interpretation in terms of the angle between the tangents. Some participants suggest reconsidering the relationship between the slopes and the angle, specifically noting that the slopes of 1 and -1 correspond to a specific angle.

Discussion Status

The discussion has progressed with participants clarifying the relationship between the calculated slopes and the angle between the tangents. There is acknowledgment that the angle between lines with slopes of 1 and -1 is 90 degrees, indicating a productive direction in understanding the problem.

Contextual Notes

The original poster expresses confusion regarding the interpretation of slope values and their connection to the angle, indicating a need for clarification on the geometric implications of the slopes derived from the parametric equations.

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Homework Statement


Find the angle made by the two tangents to the curve ##x=\sin2t## and ##y=\sin(2t)\cos(2t)## at the point ##(0,0)##

Homework Equations


##\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}##
(Derivative of a parametric equation)

The Attempt at a Solution


## x = \sin(2t) = 0 ## when ## t = 0, \frac{\pi}{2}, \pi,... ##
## y=\sin(2t)\cos(2t) = 0 ## at the same values of ## t##
Taking the derivative of the parametric equations by using the formula in part two, I get ##\frac{2\cos^2(2t)-2t\sin^2t}{2\cos(2t)}## I get to plug in any value of t, so I choose ##t=\pi/2##
With that t value, I get ## \frac{2(-1^2)-2(0)}{-2}## which is equal to ##-1##
Now I'll try ##t=0##, and I get ##1##. Using ##t=\pi## I also get one. Therefore, the angle must be between -1 and 1 and be equal to ## t = 0, \frac{\pi}{2}, \pi,... ##
My answer is 0.
The correct answer is ##\frac{\pi}{2}##
Could someone please enlighten me as to my mistake? That would be highly appreciated. Thanks.
 
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Think about what you calculated...
The slope you got at one time was 1, the slope you got at another time was -1 (both at the same point).
What is the angle between a line of slope 1 and a line of slope -1?

You don't want to guess that because you got -1 and 1 as answers, the best thing to do is average them to get 0. -1 and 1 have nothing directly to do with angles, those are your slopes (dy/dx)
 
I see! The angle between them is 90 degrees. Thank you so much.
 
Brian T said:
What is the angle between a line of slope 1 and a line of slope -1?

Now I get it! The angle is 90 degrees. Thank you so much!
 
Calpalned said:
Now I get it! The angle is 90 degrees. Thank you so much!

Glad to help. and just so you can visualize the parametrization:
http://www4b.wolframalpha.com/Calculate/MSP/MSP11162061a3322c6679i6000012f58ad84fgbh939?MSPStoreType=image/gif&s=64&w=286.&h=97.&cdf=Animation
We have two lines that cross the origin, with the slopes that you figured out.
 
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