find dy/dx and d^2/dx^2. for which values of t is the curve concave upwards?
i got (-3/4)sec^3t as d^2/dx^2. you need that to find the concavity right? but how do you do it?
Find dy/dx and d²/dx².
For which values of θ is the curve concave upwards?
x = 2·sinθ, .y = 3·cosθ, .0 < θ < 2π
i got: .d²y/dx² = -¾·sec³θ . Good!
You need that to find the concavity, right?
But how do you do it?
As the Captain pointed out, the curve is concave up when d²y/dx² is positive.
So you need to solve this inequality: .-¾sec³θ .> .0
. . I'll take baby-steps . . .
Divide by -¾: .sec³θ .< .0
Take cube roots: .secθ .< .0
When is secθ negative?
. . When is 1/cosθ negative? .In Quadrants 2 and 3.
Hence, the curve is concave up for: .π/2 < θ < 3π/2