Derivatives and parallel lines

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The discussion revolves around finding the slope of the curve defined by the equation x^2 + xy + y^2 = 27 using implicit differentiation, resulting in the expression y' = -(2x+y)/(x+2y). To determine if the tangent lines at the x-intercepts are parallel, participants suggest calculating the slopes at these intercepts by setting y=0 in the derivative. For vertical tangents, it is noted that dy/dx approaches infinity, which occurs when x + 2y = 0. The conversation emphasizes the importance of analyzing the derivative rather than directly finding x-intercepts to simplify the process.
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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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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