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# Solve for a and b

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y=axsin(bx)+1.2 given the maximum point (9, 5.5) and the point (12, 3.25) along domain [0, 12]

Guest Nov 29, 2014

#2
+80990
+5

y=axsin(bx)+1.2

We need to solve the system  (if there is a solution)

5.5  = (9a)sin(9b) + 1.2   →       4.3 = (9a)sin(9b)

3.25 = (12a)sin(12b) + 1.2  →    2.05 = (12a)sin(12b)

WolframAlpha couldn't generate a solution for the above.......nor could it generate a solution  when the first "x" was a "times".......as Melody might have rightly assumed.

CPhill  Nov 30, 2014
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#1
+91451
+5

I have been mucking around with this one this afternoon and no solution has jumped out at me.

A part of the problem is I am not sure how the question should be interpreted but even so the things I try just lead to added confusion.

I got a new idea - Maybe that first x is not an x at all but a times sign - then it is easy!

maximum point (9, 5.5) and the point (12, 3.25) along domain [0, 12]

$$y=asin(bx)+1.2$$

360/b will give the wavelength (in degrees)

360/b=4*9

360/36=b

b=10

$$y=asin(10x)+1.2$$

Sub in (12,3.25)

$$\\3.25=asin120+1.2\\ 2.05=asin60\\ 2.05=a*\frac{\sqrt3}{2}\\ \mathbf{a=\frac{4.1}{\sqrt3}}\\$$

$$y=\frac{4.1}{\sqrt3}\;sin(10x)+1.2$$

Check by subbing in (9,5.5)

$$\\RHS=\frac{4.1}{\sqrt3}\;sin(90)+1.2\\ RHS=\frac{4.1}{\sqrt3}+1.2\\$$

$${\frac{{\mathtt{4.1}}}{{\sqrt{{\mathtt{3}}}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.2}} = {\mathtt{3.567\: \!136\: \!103\: \!677\: \!465\: \!6}}$$

Well now that did not work - bummer!

It does not work because it cannot work - those conditions cannot be met on the same graph.

Maybe my interpretation is wrong again.!

Her is the graph I have.

(It passes through (12,3.25) and it has a maximum at x=9 BUT the maximum value is not 5.5 )

https://www.desmos.com/calculator/8xnxca7qcp

Melody  Nov 30, 2014
#2
+80990
+5

y=axsin(bx)+1.2

We need to solve the system  (if there is a solution)

5.5  = (9a)sin(9b) + 1.2   →       4.3 = (9a)sin(9b)

3.25 = (12a)sin(12b) + 1.2  →    2.05 = (12a)sin(12b)

WolframAlpha couldn't generate a solution for the above.......nor could it generate a solution  when the first "x" was a "times".......as Melody might have rightly assumed.

CPhill  Nov 30, 2014

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