# find the max & min values of voltage given by the expression: v=5cos4θ + 5 between and including values of θ=0 and θ=2π radians?

### 3 Answers

- 冷眼旁觀Lv 78 months agoFavorite Answer
0 ≤ θ ≤ 2π

0 ≤ 4θ ≤ 8π

v = 5 cos(4θ) + 5

dv/dθ = -20 sin(4θ)

d²v/dθ² = -80 cos(4θ)

When dv/dθ = 0:

-20 sin(4θ) = 0

4θ = 0, π, 2π, 3π, 4π, 5π, 6π, 7π, 8π

θ = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π

When θ = 0, π/2, π, 3π/2, 2π:

v = 5 cos(4θ) + 5 = 10

dv/dθ = 0

d²v/dθ² = -80 cos(4θ) = -80 < 0

When θ = π/4, 3π/4, 5π/4, 7π/4

v = 5 cos(4θ) + 5 = 0

dv/dθ = 0

d²v/dθ² = -80 cos(4θ) = 80 > 0

Answers:

Maximum voltage = 10 V when θ = 0, π/2, π, 3π/2, 2π

Minimum voltage = 0 V when θ =π/4, 3π/4, 5π/4, 7π/4

- Ian HLv 78 months ago
V = 5cos(4θ) + 5, (θ from 0 to 2π radians)

Graph looks like this

https://www.wolframalpha.com/input/?i=V+%3D+5cos%2...

Values of voltage at the 5 maxima are 10v.

Values of voltage at the 4 minima are zero v.

- billrussell42Lv 78 months ago
The 4θ term means there are 4 cycles between 0 and 2π, so the max and min are +10 and 0