The depth of water, h, in metres at time t, in hours, is given by the formula: h(t)=7.8+3.5sin[pi/6(t−3)]. Its a 24 hours period and t=0 is midnight. Provide an algebraic solution to determine the time(s) of day, the water reaches the depth of 10.29 m
i got one time 4.51 but i have no idea how to get the others? i have graphed it on desmos and seen which point are the right answer but i don't know how to derive them algebraically. help plz!
All you have to do is follow the same steps I used to get the answer of 4.51
http://www.jiskha.com/display.cgi?id=1469238292
thanks and sorry i didn't know some one already posted this.
To determine the time(s) of day when the water reaches a depth of 10.29 m, we can set up the equation:
h(t) = 7.8 + 3.5sin[(π/6)(t − 3)] = 10.29
First, let's subtract 7.8 from both sides:
3.5sin[(π/6)(t − 3)] = 10.29 - 7.8
3.5sin[(π/6)(t − 3)] = 2.49
Next, divide both sides by 3.5:
sin[(π/6)(t − 3)] = 2.49/3.5
sin[(π/6)(t − 3)] ≈ 0.7114
To find the time(s) when the water reaches a depth of 10.29 m, we need to find the values of (π/6)(t - 3) that have a sine value of approximately 0.7114.
The sine function repeats every 2π, so we can rewrite the equation as:
(π/6)(t − 3) = arcsin(0.7114) + 2πn
where n is an integer representing the number of full cycles.
Now, we can solve for (t - 3):
(t − 3) = (arcsin(0.7114) + 2πn) * (6/π)
Add 3 to both sides:
t = (arcsin(0.7114) + 2πn) * (6/π) + 3
This equation gives the time(s) of day when the water reaches a depth of 10.29 m. You already found one solution, t ≈ 4.51. To find the other solution(s), you can substitute different values of n (positive and negative integers) in the equation and solve for t.
For example, when n = -1:
t = (arcsin(0.7114) + 2π(-1)) * (6/π) + 3
Calculate the right-hand side to find the corresponding value(s) for t.
Repeat this process with different values of n to find all the time(s) of day when the water reaches the depth of 10.29 m.