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A smoke alarm emits alod oscillating signal when  activated.the signal comes frome A and B. The volum from first is giving by  Va =7cos 12t and from the second by Vb =6sin 12t. for what percentage of the total volum(Va+Vb) greater than 9 units

 Mar 25, 2015

Best Answer 

 #3
avatar+890 
+10

I really like Chris's answer to this question, using the graph plotter makes life so much easier.

Here's the 'on paper' method that we had to use before these sophisticated plotters were available.

 

Let

$$7\cos12t+6\sin12t\equiv R\cos(12t-\alpha)=R\cos12t\cos\alpha+R\sin12t\sin\alpha,$$

then

$$R\cos\alpha=7 \text{ and } R\sin\alpha=6.$$

Squaring and adding,       $$R^{2}=7^{2}+6^{2}=85, \text{ so }R=\sqrt{85}.$$

Dividing,                         $$\tan\alpha = 6/7, \text{ so }\alpha = 0.7086 \text{ rad}.$$

 

We have then                 $$7\cos12t+6\sin12t=\sqrt{85}\cos(12t-0.7086)>9.$$

 

The maximum for the cosine occurs when 12t - 0.7086 = 0, that is, when t = 0.0591 (4dp, but I kept greater accuracy for all decimal values during the calculation).

For equality with the 9,           $$12t - 0.7086 = \cos^{-1}(9/\sqrt{85}),\quad t=0.0773.$$

 

It follows that the wave is greater than 9 for the time span 2(0.0773 - 0.0591) = 0.0364.

 

Since the period of the wave is 2pi/12, the proportion of the time that the wave is greater than 9 is

0.0364/(pi/6) = 0.0696, that is, about 7% (agreeing with Chris).

 

As an alternative, the form    $$R\sin(12t+\beta)$$    could have been chosen rather than the cosine.

 Mar 25, 2015
 #1
avatar+98091 
+10

This graph is periodic.....see it here....https://www.desmos.com/calculator/gxygumhhqp

Notice that, the period is 30 degrees

And picking a selected period from about 25.88 degrees to about 55.88 degrees.....the graph rises above 9 units on the approximate interval of (32.34, 34.43) degrees

And this interval is [34.43 - 32.34] degrees wide = about 2.09 degrees wide

So the graph exceeds 9 units in one period ..... 2.09 / 30   = about 7% of the time..... 

 

  

 Mar 25, 2015
 #2
avatar+99272 
+5

Thanks Chris, this was an unusual question,

I like your answer but I am trying to think of another way to do it.

I've played with it a little but I have not found another way.   :/

 Mar 25, 2015
 #3
avatar+890 
+10
Best Answer

I really like Chris's answer to this question, using the graph plotter makes life so much easier.

Here's the 'on paper' method that we had to use before these sophisticated plotters were available.

 

Let

$$7\cos12t+6\sin12t\equiv R\cos(12t-\alpha)=R\cos12t\cos\alpha+R\sin12t\sin\alpha,$$

then

$$R\cos\alpha=7 \text{ and } R\sin\alpha=6.$$

Squaring and adding,       $$R^{2}=7^{2}+6^{2}=85, \text{ so }R=\sqrt{85}.$$

Dividing,                         $$\tan\alpha = 6/7, \text{ so }\alpha = 0.7086 \text{ rad}.$$

 

We have then                 $$7\cos12t+6\sin12t=\sqrt{85}\cos(12t-0.7086)>9.$$

 

The maximum for the cosine occurs when 12t - 0.7086 = 0, that is, when t = 0.0591 (4dp, but I kept greater accuracy for all decimal values during the calculation).

For equality with the 9,           $$12t - 0.7086 = \cos^{-1}(9/\sqrt{85}),\quad t=0.0773.$$

 

It follows that the wave is greater than 9 for the time span 2(0.0773 - 0.0591) = 0.0364.

 

Since the period of the wave is 2pi/12, the proportion of the time that the wave is greater than 9 is

0.0364/(pi/6) = 0.0696, that is, about 7% (agreeing with Chris).

 

As an alternative, the form    $$R\sin(12t+\beta)$$    could have been chosen rather than the cosine.

Bertie Mar 25, 2015
 #4
avatar+98091 
0

That's pretty neat, Bertie....!!!!

 

  

 Mar 25, 2015
 #5
avatar+99272 
0

Yes that looks like a terrific answer Bertie.  I shall find time to study it.  :))

 Mar 26, 2015

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