Perpendicular tangent lines in calculus?

Question

link to picture: https://imgur.com/1uIrNQn

a. Find the slope of x cubed plus y cubed minus 72 xy equals 0 at the points left parenthesis 16 comma 32 right parenthesis and left parenthesis 32 comma 16 right parenthesis.

b. At what point other than the origin does the curve have a horizontal tangent​ line?

c. Find the coordinates of the point other than the origin where the curve has a vertical tangent line.

Jeff Aaron · Accepted Answer

a.
x^3 + y^3 - 72xy = 0
3x^2 + 3y^2y' - 72y - 72xy' = 0 
y^2y' - 24xy' = 24y - x^2
y'(y^2 - 24x) = 24y - x^2
y' = (24y - x^2) / (y^2 - 24x)
At (16, 32), we have:
y' = (24*32 - 16^2) / (32^2 - 24*16)
y' = (768 - 256) / (1024 - 384)
y' = 512/640
y' = 4/5
y' = 0.8
At (32, 16) , we have:
y' = (24*16 - 32^2) / (16^2 - 24*32)
y' = (384 - 1024) / (256 - 768)
y' = (-640) / (-512)
y' = 5/4
y' = 1.25

b.
y' = (24y - x^2) / (y^2 - 24x) = 0
24y - x^2 = 0
24y = x^2
y = (1/24)x^2
x^3 + ((1/24)x^2)^3 - 72x*(1/24)x^2 = 0
x^3 + (1/13824)x^6 - 3x^3 = 0
(1/13824)x^6 - 2x^3 = 0
x^6 - 27648x^3 = 0
x^3(x^3 - 27648) = 0
x^3 = 0 or x^3 - 27648 = 0
x^3 = 0 or x^3 = 27648
If x is a real number:
x = cbrt(0) or x = cbrt(27648)
x = 0 or x = 24*cbrt(2)
If x = 0, y = (1/24)*0^2 = 0, but that's the origin, so ignore that.
If x = 24*cbrt(2), y = (1/24)*(27648)^(2/3) = 24*cbrt(4)
Answer: (24*cbrt(2), 24*cbrt(4))

c.
y' = (24y - x^2) / (y^2 - 24x) = undefined, i.e. denominator is zero
y^2 - 24x = 0
24x = y^2
x = (1/24)y^2
((1/24)y^2)^3 + y^3 - 72*(1/24)y^2*y = 0
(1/13824)y^6 + y^3 - 3y^3 = 0
(1/13824)y^6 - 2y^3 = 0
(1/13824)y^6 - 2y^3 = 0
y^6 - 27648y^3 = 0
y^3(y^3 - 27648) = 0
y = 0 or y = 24*cbrt(2)
Answer: (24*cbrt(4), 24*cbrt(2))

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Perpendicular tangent lines in calculus?

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