1. ## trig word problem

I have done the first part, and have gotten;

$\sqrt{5}cos(\theta - 1.11)$

I am stuck on part 'c'. I know the equation looks similar to my answer in part 'a'.

so if I change it to $15+ \sqrt{5}cos(\frac{\pi t}{12} - 1.11)$

Would this be valid? If so how do I find the maximum depth of the water and at what time this occurs from this expression?

Thank you.

2. Originally Posted by Tweety
I have done the first part, and have gotten;

$\sqrt{5}cos(\theta - 1.11)$

I am stuck on part 'c'. I know the equation looks similar to my answer in part 'a'.

so if I change it to $15+ \sqrt{5}cos(\frac{\pi t}{12} - 1.11)$

Would this be valid? If so how do I find the maximum depth of the water and at what time this occurs from this expression?

Thank you.
It is valid, there are many questions where you will be asked to use a previous idea to help you solve it.

Consider the equation $a+b\cos(\theta)$

$\theta$ will be a max at $a+b$ and a min at $a-b$

Does this help? How did you arrive at your answer for part a)?

3. Originally Posted by pickslides
It is valid, there are many questions where you will be asked to use a previous idea to help you solve it.

Consider the equation $a+b\cos(\theta)$

$\theta$ will be a max at $a+b$ and a min at $a-b$

Does this help? How did you arrive at your answer for part a)?
Thanks,

for part a I expanded $Rcos(\theta -\alpha)$

to $Rcos\theta cos\alpha - Rsin\theta sin\alpha$

and than equated coefficients.

$Rcos\alpha = 1 , Rsin\alpha = 2$

$tan\alpha = 2$
$\alpha = 1.11$

$R =\sqrt{ 1^2 + 2^2}$

But I am still not sure how I find out the answer? Could you show me please?

I know for this expression $\sqrt{5}cos(\theta - 1.11)$

the maximum is $\sqrt{5}$
minimum is $-\sqrt{5}$

So I am not sure how I apply all this to part 'c'.

4. No different for part c)

$15+ \sqrt{5}cos(\frac{\pi t}{12} - 1.11)$

max: $15+ \sqrt{5}$

main: $15- \sqrt{5}$