Trigonometric Model: depth D(t) = -y(pi/6)t + 11.5 at t hrs

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sail0r

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I was given the equation D(t)=-6(?/6)t+11.5, where t is time in hours, and asked to solve for when the depth will be 7m

I substituted 7 in for D(t) and got the answer 1.38.

The given answers were 1:24 a.m, 10:36 a.m, 1:24 p.m, 10:36 p.m

What did I do wrong? or What did I not do at all?
 
Re: Trigonometric Model

sail0r said:
I was given the equation D(t)=-6(?/6)t+11.5, where t is time in hours, and asked to solve for when the depth will be 7m

I substituted 7 in for D(t) and got the answer 1.38.

The given answers were 1:24 a.m, 10:36 a.m, 1:24 p.m, 10:36 p.m

What did I do wrong? or What did I not do at all?
Click to expand...

Is your original equation correct? It seems to me that the two 6's would be cancelled leaving you with D(t)=-?t+11.5. Possibly you have a misread or a typo somewhere???
 
Re: Trigonometric Model

Whoops the equation is D(t)=-6cos(?/6)t+11.5
 
Re: Trigonometric Model

Just set it equal to 7 and solve for t.

\(\displaystyle -6cos(\frac{\pi}{6})t+\frac{23}{2}=7\)

Subtract 11 and divide by -6:

\(\displaystyle cos(\frac{\pi}{6})t=\frac{3}{4}\)

Finish?. It's rather straightforward. Nothing fancy needed.
 
When I solved the equation exactly like that I got an answer of 1.38. This is either incorrect or I am not interpreting it correctly.
 
That is incorrect.

\(\displaystyle t=\frac{\frac{3}{4}}{cos(\frac{\pi}{6})}\)

I assume you know \(\displaystyle cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\)

It's a good idea to leave your solution in a form like that instead of a decimal unless instructed otherwise.
 
Ok so if I am not mistaken when solved like that the calculated answer is 0.86

How do I interpret this to get the given answers of 1:24 a.m, 10:36 a.m, 1:24 p.m, 10:36 p.m?
 
I don't know, you must have been given an initial time?. Then, you would add the result to that.
 
sail0r said:
... How do I interpret this to get the given answers of 1:24 a.m, 10:36 a.m, 1:24 p.m, 10:36 p.m?
Click to expand...


Hello Sail0r:

Periodic functions require the variable to be a part of the sinusoidal function's argument. Despite two attempts to provide us with the correct definition for D(t), it appears that you still have not. (Both of the function definitions that you typed for D(t) are linear, not periodic.)

Try this one.

D(t) = -6 * cos(?/6 * t) + 11.5

When D(t) = 7, then a couple of the infinite solutions for t are t ? 10.6197 and t ? 13.3803.

These solutions correspond to 10:37 AM and 1:23 PM, respectively.

In order to restrict the solutions to the specific four that you posted, you must have been given a restricted domain.

Perhaps 0 < t < 24.

Does this information help you?

Cheers,

~ Mark :)

 
Re:

(Both of the function definitions that you typed for D(t) are linear, not periodic.)

Try this one.

D(t) = -6 * cos(?/6 * t) + 11.5
Click to expand...

DUH on my part...I should've seen right off the t was on the outside. :roll: :oops:
 
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