For a certain day,the depth of water,h, in metres in PEI, in hours is given by the formula:h(t) = 7.8sin (pi/6(t-3)), t E [0,24], assume t=0 represents midnight. Provide an algebraic solution to determine the time(s) of day, the water reaches the depth of 10.29 m( give your answer in hours and minutes)
Hmmm.
7.8sin (pi/6(t-3)) = 10.29
sin (pi/6(t-3)) = 1.319 <--- ??
I suspect that you have left out a term in the definition of h(t). Fix it and then proceed with the solution.
Sorry it's h(t)=7.8+3.5sin((pi/6(t-3))
To find the time(s) of day when the water reaches a depth of 10.29 meters, we need to solve the equation h(t) = 10.29 for t.
The given equation is h(t) = 7.8sin((π/6)(t-3)), where t is the time in hours.
Substituting the given depth, we have 10.29 = 7.8sin((π/6)(t-3)).
To solve this equation, we can isolate the sin term by dividing both sides by 7.8:
10.29/7.8 = sin((π/6)(t-3)).
Now, we need to find the inverse of sine to get the value of the argument:
sin^(-1)(10.29/7.8) = (π/6)(t-3).
Using a calculator, evaluate sin^(-1)(10.29/7.8) to get the value of the argument.
Let's say it's equal to A, so we have:
A = (π/6)(t-3).
Next, we solve for t by isolating it:
t-3 = (6A)/π.
Finally, we solve for t:
t = (6A)/π + 3.
This gives us the time in hours when the water reaches the depth of 10.29 meters.
To convert this into hours and minutes, you can use decimal approximation if needed. Multiply the fractional part (after the decimal point) by 60 to get the number of minutes.
For example, if the calculated value of t is 8.456 (approximately), the decimal part is 0.456. Multiply 0.456 by 60 to get 27.36 minutes. Therefore, the time would be 8 hours and 27 minutes.