Find the times (to the nearest hundredth of a second) that the weight is halfway to its maximum negative position over the interval 0<_t<_0.5. Solve algebraically, and show your work and final answer in the response box. Hint: Use the amplitude to determine what y(t) must be when the weight is halfway to its maximum negative position. Graph the equation and explain how it confirms your solution(s).

0.20 and 0.36

Step-by-step explanation:

y(t) = 2 sin (4π t) + 5 cos (4π t)

We wish to convert this to:

y = A sin(ωt + φ)

We know that ω = 4π.  We also know the following:

5 = A sin φ

2 = A cos φ

Divide the first equation by the second equation:

5/2 = tan φ

φ = tan⁻¹(5/2)

Now, square the two equations and add them together.

5² + 2² = (A sin φ)² + (A cos φ)²

29 = A²

A = √29

The equation of the wave is therefore:

y = √29 sin(4π t + tan⁻¹(5/2))

The maximum negative position is -√29.  And half of that is -½√29.

-½√29 = √29 sin(4π t + tan⁻¹(5/2))

-½ = sin(4π t + tan⁻¹(5/2))

7π/6 + 2kπ or 11π/6 + 2kπ = 4π t + tan⁻¹(5/2)

7 + 12k or 11 + 12k = 24t + 6 tan⁻¹(5/2) / π

t = (7 + 12k − 6 tan⁻¹(5/2) / π) / 24 or (11 + 12k − 6 tan⁻¹(5/2) / π) / 24

Trying different integer values of k, we find there are two possible values for t between 0 and 0.5, both when k = 0.

t = (7 − 6 tan⁻¹(5/2) / π) / 24 or (11 − 6 tan⁻¹(5/2) / π) / 24

t ≈ 0.20 or 0.36

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

step-by-step explanation: