em2682
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create an equatio for the graph using the y = ae^(bx)cos(cx) for the following graph
View attachment 8085
View attachment 8085
Country Boy said:I am inclined to think that MarkFL's "d" could be 0 but "f" clearly can't be. Without a non-zero f, the first loop of the function would go from a down to -a but you want it to go from 140 to 100. The average of those two is 120 so you want it to go from 120+ 20 to 120- 20.
MarkFL said:If $d=0$ and presumably $b<0$, then what would prevent the amplitude from tending to zero as $x\to\infty$, instead of some value greater than zero?
Country Boy said:I am inclined to think that MarkFL's "d" could be 0 but "f" clearly can't be. Without a non-zero f, the first loop of the function would go from a down to -a but you want it to go from 140 to 100. The average of those two is 120 so you want it to go from 120+ 20 to 120- 20.
MarkFL said:It appears to me that you need something of the form:
$$y=a\left(e^{bx}+d\right)\cos(cx)+f$$
MarkFL said:Well, that's a tiny graph, which isn't very easy to read.
I would begin with the amplitude $$A(x)=a\left(e^{bx}+d\right)$$
Let's say:
$$A(0)=a\left(1+d\right)=25\tag{1}$$
$$A(50)=a\left(e^{50b}+d\right)=15\tag{2}$$
$$A(\infty)=ad=\frac{15}{2}\tag{3}$$
(1) and (3) imply:
$$a=\frac{35}{2},\,d=\frac{3}{7}$$
And so, substituting these values into (2), we find:
$$\frac{35}{2}\left(e^{50b}+\frac{3}{7}\right)=15$$
$$b=\frac{1}{50}\ln\left(\frac{3}{7}\right)$$
And so we have:
$$A(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)$$
The period appears to be about $10\text{ s}$ and the vertical shift is $$f=\frac{245}{2}$$ and so we have:
$$y(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\cos\left(\frac{\pi}{5}x\right)+\frac{245}{2}$$
thank you so much! what were some importnt assumptions made to determine the equation?MarkFL said:Well, that's a tiny graph, which isn't very easy to read.
I would begin with the amplitude $$A(x)=a\left(e^{bx}+d\right)$$
Let's say:
$$A(0)=a\left(1+d\right)=25\tag{1}$$
$$A(50)=a\left(e^{50b}+d\right)=15\tag{2}$$
$$A(\infty)=ad=\frac{15}{2}\tag{3}$$
(1) and (3) imply:
$$a=\frac{35}{2},\,d=\frac{3}{7}$$
And so, substituting these values into (2), we find:
$$\frac{35}{2}\left(e^{50b}+\frac{3}{7}\right)=15$$
$$b=\frac{1}{50}\ln\left(\frac{3}{7}\right)$$
And so we have:
$$A(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)$$
The period appears to be about $10\text{ s}$ and the vertical shift is $$f=\frac{245}{2}$$ and so we have:
$$y(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\cos\left(\frac{\pi}{5}x\right)+\frac{245}{2}$$
em2682 said:thank you so much! what were some importnt assumptions made to determine the equation?