The slope is -tanθ, but at what point will the tangent have slope 1 or -1?
x = a cos³θ ----> dx/dθ = −3a cos²θ sinθ
y = a sin³θ ----> dy/dθ = 3a sin²θ cosθ
dy/dx = (dy/dθ) / (dx/dθ)
= (3a sin²θ cosθ) / (−3a cos²θ sinθ)
= −sinθ/cosθ
= −tanθ
Slope = ±1
dy/dx = ±1
−tanθ = ±1
tanθ = ±1
θ = π/4 + kπ/2, for any integer k
When tanθ = 1 (slope = −1)
cosθ = 1/√2 ----> x = a (1/√2)³ = a/(2√2)
sinθ = 1/√2 ----> y = a (1/√2)³ = a/(2√2)
or
cosθ = −1/√2 ----> x = a (−1/√2)³ = −a/(2√2)
sinθ = −1/√2 ----> y = a (−1/√2)³ = −a/(2√2)
When tanθ = −1 (slope = 1)
Slope = 1 at points (−a/(2√2), a/(2√2)) and (a/(2√2), −a/(2√2))
Slope = −1 at points (a/(2√2), a/(2√2)) and (−a/(2√2), −a/(2√2))
The value of -tan(theta) will be 1 when theta is 45 or 225 or 405 degrees, etc; and -1 when theta is 135 or 315 or 495 degrees, etc.
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Answers & Comments
x = a cos³θ ----> dx/dθ = −3a cos²θ sinθ
y = a sin³θ ----> dy/dθ = 3a sin²θ cosθ
dy/dx = (dy/dθ) / (dx/dθ)
= (3a sin²θ cosθ) / (−3a cos²θ sinθ)
= −sinθ/cosθ
= −tanθ
Slope = ±1
dy/dx = ±1
−tanθ = ±1
tanθ = ±1
θ = π/4 + kπ/2, for any integer k
When tanθ = 1 (slope = −1)
cosθ = 1/√2 ----> x = a (1/√2)³ = a/(2√2)
sinθ = 1/√2 ----> y = a (1/√2)³ = a/(2√2)
or
cosθ = −1/√2 ----> x = a (−1/√2)³ = −a/(2√2)
sinθ = −1/√2 ----> y = a (−1/√2)³ = −a/(2√2)
When tanθ = −1 (slope = 1)
cosθ = −1/√2 ----> x = a (−1/√2)³ = −a/(2√2)
sinθ = 1/√2 ----> y = a (1/√2)³ = a/(2√2)
or
cosθ = 1/√2 ----> x = a (1/√2)³ = a/(2√2)
sinθ = −1/√2 ----> y = a (−1/√2)³ = −a/(2√2)
Slope = 1 at points (−a/(2√2), a/(2√2)) and (a/(2√2), −a/(2√2))
Slope = −1 at points (a/(2√2), a/(2√2)) and (−a/(2√2), −a/(2√2))
The value of -tan(theta) will be 1 when theta is 45 or 225 or 405 degrees, etc; and -1 when theta is 135 or 315 or 495 degrees, etc.