Derivatives and parallel lines

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Homework Help Overview

The discussion revolves around a curve defined by the equation x^2 + xy + y^2 = 27. Participants are exploring the derivatives of the curve, specifically focusing on the slope at various points, the conditions for parallel tangents at x-intercepts, and identifying points where tangents are vertical.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using implicit differentiation to find the slope of the curve and question the process of determining x-intercepts and their slopes. There is also exploration of conditions for vertical tangents and the implications of certain expressions in the derivative.

Discussion Status

Some participants have confirmed the correctness of the implicit differentiation approach, while others suggest alternative methods for finding x-intercepts and vertical tangents. Multiple interpretations of the problem are being explored, particularly regarding the conditions for parallel lines and the handling of specific cases in the derivative.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to find specific points on the curve and the implications of certain values in the derivative. There is an emphasis on not needing to calculate x-intercepts directly for this particular problem.

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Homework Statement



Consider the curve defined by x^2+xy +y^2=27

a. Write an expression for the slope of the curve at any point (x,y)
b. Determine whether the lines tangent to the curve at the x-intercepts of the curve are parallel. Show the analysis that leads to your conclusion.
c. Find the points on the curve where the lines tangent to the curve are vertical.


The Attempt at a Solution



a. I used implicit differentiation
x^2 +xy +y^2 =27
2x +y + xy' +2yy' = 0
y'(x+2y)= - (2x+y)
y' = -(2x+y)/(x+2y). I am pretty sure this is right.

b. i know that the lines must have the same slope for them to be parallel, but I am not actually sure what else to do with this. how do u find the x-intercepts and show that their slopes are equal?

c. the vertical line would be like x=k, then you plug k into the orginal curve and then solve for k? i got 6 and -6 is that right? can someone please help me with this
 
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a. Correct

b. Find x-intercepts of the curve by putting y=0, and find the x values. Then put these x and y in dy/dx.

c. Find where dx/dy is zero.
 
(a) is almost right, but you neglect the possibility that x+2y=0. You would either have to find an alternative expression for that case, or show it never happens on the curve (e.g. by substituting x=-2y into the original equation)

For (b), I suggest being really slick and not calculating where the x-intercepts are located if you can (try plugging y=0 into your dy/dx expression. Where can this go wrong?). Note a line that's tangent to a curve has a slope equal to dy/dx

For (c), if the curve is parallel to a vertical line, then dy/dx is +/- infinity. Taking a page from part (a), when does this occur?
 
Office_Shredder said:
For (b), I suggest being really slick and not calculating where the x-intercepts are located if you can (try plugging y=0 into your dy/dx expression. Where can this go wrong?). Note a line that's tangent to a curve has a slope equal to dy/dx

I didn't quite get what exactly you meant.
 
Last edited by a moderator:
He is saying, rather than set y= 0 in the original equation, to find the x-intercepts, set y= 0 in the formula for the derivative. It happens, for this particular problem, that you don't need to know x!
 

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