find dy/dx and d^2/dx^2. for which values of t is the curve concave upwards?

x=2sint

y=3cost

0<t<2pi

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?

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- Apr 4th 2007, 07:46 PMjephconcavity
find dy/dx and d^2/dx^2. for which values of t is the curve concave upwards?

x=2sint

y=3cost

0<t<2pi

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? - Apr 5th 2007, 06:24 AMCaptainBlack
- Apr 5th 2007, 08:43 AMSoroban
Hello, jeph!

Quote:

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

- Apr 5th 2007, 12:23 PMjeph
thanks guys! i guess that - sign got me a little confused...the < and > signs only switch when you are dividing or multiplying a negative right? these small things always slip out of my mind.....