# Maximum depths of the river

1. Aug 2, 2005

### bagpiper

D = 9 + 3cos[pi(t)/14]
where D meters is the depth of the water above the river bed at time t hours after 12 noon on 7 March 2005, Monday. This means that the depth of water only depends on time.
a) Write down the minimum and maximum depths of the river
b) Find the day and time when the water first reaches its minimum.

2. Aug 2, 2005

### VietDao29

You know that:
$$-1 \leq \cos \alpha \leq 1$$
So when the D is minimum is when $$\cos \frac{\pi t}{14}$$ is minimum and vice versa, when the D is maximum is when $$\cos \frac{\pi t}{14}$$ is maximum.
Can you go from here?
Viet Dao,

3. Aug 2, 2005

### bagpiper

Yup. i got there. So minimum is 6m, maximum is 12m. But i dont know how to manipulate it formula to solve the second part.

4. Aug 2, 2005

### EnumaElish

Why is pi written as a function of time? Isn't pi a constant? I'll assume it's a typographic error or it means pi * t, i.e. multiplication.

1. Let's say you have D = f(t). If you set f'(t) = 0 that'll give you the extrema (minima and maxima) solutions, say tmin and tmax. Then take each of these solutions and look at f''(t) at this point. If f''(t) < 0 then it's a maximum; if f''(t) > 0 then it's a minimum. If f''(t) = 0 then it's an inflection point, I guess.
2. Now having found all your minima, calculate the time difference between the first such minima occuring (assuming there are more than one, if there is only one minimum then just take its time of occurance, tmin) and "today" (7 March 2005): tmin - t7Mar05.