Question

given the graph of y=cos θ + pi/2 from 0≤ θ ≤ 2pi,
a) for what values of θ does the instantaneous rate of change appear to equal 0? (I said pi/2 and 3pi/2)

b) for what values of θ does the instantaneous rate of change reach its maximum? its minimum? (this part I don't get)

Answer 1

If you wrote what you meant, then y' = -sinθ y'=0 at θ=0,π,2π
If you really meant y = cos(θ + π/2) = -sinθ, then y' = -cosθ, so y'=0 at θ = π/2, 3π/2 as you said.

so, using the 2nd version, -cosθ achieves its maximum at θ=π, minimum at 0,2π

You can always check the graph for confirmation.

Answer 2

The instantaneous rate of change is your first derivative which would be
dy/dθ = -sinθ
if -sinθ = 0, θ = 0, π, and 2π for your given interval

if your initial equation is y = cos (θ + π/2) , then
dy/dθ = -sin(θ+π/2)
and for -sin(θ+π/2) = 0, θ = π/2, 3π/2
That was your answer, so the brackets ARE NECESSARY

In that case for the first derivative to have a max, its derivative, namely the
2nd derivative of the original must be zero.

then y '' = -cos(θ+π/2) = 0 would be for the function with brackets.
θ = 0, π, 2π

you decide which way the original equation was

Answer 3

sorry I forgot the brackets. I did mean cos(θ + π/2)