16. The equation models the height h in centimeters after t seconds of a weight attached to the end of a spring that has been stretched and then released.
h= 7 cos (π/3 t)
a. Solve the equation for t. (I'm pretty sure i've got this one solved)
T= 3 arccos(h/7)/π
b. Find the times at which the weight is first at a height of 1 cm, of 3 cm, and of 5 cm above the rest position. Round your answers to the nearest hundredth.
c. Find the times at which the weight is at a height of 1 cm, of 3 cm, and of 5 cm below the rest position for the second time. Round your answers to the nearest hundredth.
Im so sorry please excuse my reply. I read your reply again a few more times and i completely understand now. Thankyou so very much!
Im still confused. Do i plug in the centimeters somewhere? And what do i do to solve for the "above" and "below". I literally want to pull my hair out over this question so I would really appreciate if you could help me with that.
To solve the equation for part b and c, we need to substitute the given heights into the equation and solve for t. Let's start with part b:
b. Find the times at which the weight is first at a height of 1 cm, 3 cm, and 5 cm above the rest position.
Substitute h = 1 into the equation and solve for t:
1 = 7 cos(π/3 t)
Divide both sides by 7:
1/7 = cos(π/3 t)
To isolate t, take the inverse cosine (arccos) of both sides:
arccos(1/7) = π/3 t
Now solve for t by multiplying both sides by 3 and dividing by π:
t = 3 arccos(1/7)/π
Similarly, substitute h = 3 and h = 5 into the equation and solve for t:
For h = 3:
3 = 7 cos(π/3 t)
3/7 = cos(π/3 t)
arccos(3/7) = π/3 t
t = 3 arccos(3/7)/π
For h = 5:
5 = 7 cos(π/3 t)
5/7 = cos(π/3 t)
arccos(5/7) = π/3 t
t = 3 arccos(5/7)/π
These are the times at which the weight is first at a height of 1 cm, 3 cm, and 5 cm above the rest position. Round the answers to the nearest hundredth.
Now let's move on to part c, where the weight is at a height of 1 cm, 3 cm, and 5 cm below the rest position for the second time.
Substitute h = -1 into the equation and solve for t:
-1 = 7 cos(π/3 t)
-1/7 = cos(π/3 t)
arccos(-1/7) = π/3 t
t = 3 arccos(-1/7)/π
Similarly, substitute h = -3 and h = -5 into the equation and solve for t:
For h = -3:
-3 = 7 cos(π/3 t)
-3/7 = cos(π/3 t)
arccos(-3/7) = π/3 t
t = 3 arccos(-3/7)/π
For h = -5:
-5 = 7 cos(π/3 t)
-5/7 = cos(π/3 t)
arccos(-5/7) = π/3 t
t = 3 arccos(-5/7)/π
These are the times at which the weight is at a height of 1 cm, 3 cm, and 5 cm below the rest position for the second time. Round the answers to the nearest hundredth.
Remember to use a calculator in radian mode when evaluating the inverse cosine function.
cos (π t / 3 ) = h/7
(π/3)t =arc cos h/7
t = (3/π) arc cos (h/7) agree
but note changed the notation because for example π/3 t might mean pi / (3t)
so for example if h = 1
arc cos (1/7) = 1.42744876 radians
t = (3/3.14)(1.43) seconds
that should get you started