- #1
em2682
- 4
- 0
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:
\(\displaystyle y=a\left(e^{bx}+d\right)\cos(cx)+f\)
MarkFL said:Well, that's a tiny graph, which isn't very easy to read.
I would begin with the amplitude \(\displaystyle A(x)=a\left(e^{bx}+d\right)\)
Let's say:
\(\displaystyle A(0)=a\left(1+d\right)=25\tag{1}\)
\(\displaystyle A(50)=a\left(e^{50b}+d\right)=15\tag{2}\)
\(\displaystyle A(\infty)=ad=\frac{15}{2}\tag{3}\)
(1) and (3) imply:
\(\displaystyle a=\frac{35}{2},\,d=\frac{3}{7}\)
And so, substituting these values into (2), we find:
\(\displaystyle \frac{35}{2}\left(e^{50b}+\frac{3}{7}\right)=15\)
\(\displaystyle b=\frac{1}{50}\ln\left(\frac{3}{7}\right)\)
And so we have:
\(\displaystyle A(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\)
The period appears to be about $10\text{ s}$ and the vertical shift is \(\displaystyle f=\frac{245}{2}\) and so we have:
\(\displaystyle y(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\cos\left(\frac{\pi}{5}x\right)+\frac{245}{2}\)
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 \(\displaystyle A(x)=a\left(e^{bx}+d\right)\)
Let's say:
\(\displaystyle A(0)=a\left(1+d\right)=25\tag{1}\)
\(\displaystyle A(50)=a\left(e^{50b}+d\right)=15\tag{2}\)
\(\displaystyle A(\infty)=ad=\frac{15}{2}\tag{3}\)
(1) and (3) imply:
\(\displaystyle a=\frac{35}{2},\,d=\frac{3}{7}\)
And so, substituting these values into (2), we find:
\(\displaystyle \frac{35}{2}\left(e^{50b}+\frac{3}{7}\right)=15\)
\(\displaystyle b=\frac{1}{50}\ln\left(\frac{3}{7}\right)\)
And so we have:
\(\displaystyle A(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\)
The period appears to be about $10\text{ s}$ and the vertical shift is \(\displaystyle f=\frac{245}{2}\) and so we have:
\(\displaystyle y(x)=\frac{35}{2}\left(\left(\frac{3}{7}\right)^{\Large\frac{x}{50}}+\frac{3}{7}\right)\cos\left(\frac{\pi}{5}x\right)+\frac{245}{2}\)
em2682 said:thank you so much! what were some importnt assumptions made to determine the equation?
The equation y = ae^(bx)cos(cx) is a mathematical expression that can be used to model various real-world phenomena, such as population growth, radioactive decay, and oscillating systems. By understanding the components of this equation, scientists can make predictions and analyze data related to these phenomena.
The values of a, b, and c can be determined by analyzing the given graph. The value of a represents the amplitude of the graph, which is the maximum value of y. The values of b and c determine the frequency and period of the oscillations, respectively. They can be calculated by finding the horizontal and vertical stretches of the graph.
No, this equation is specifically used for graphs that exhibit exponential growth or decay, combined with oscillations. It is not applicable for linear or polynomial graphs.
The value of a determines the amplitude of the graph, so changing it will affect the height of the oscillations. The values of b and c affect the frequency and period of the oscillations, respectively. Changing these values will result in a change in the number of oscillations per unit and the time it takes for one complete oscillation.
Yes, as long as the values of a, b, and c are properly determined and the graph accurately represents the phenomenon being modeled, the equation can be used to make accurate predictions about the behavior of the system. However, it is important to note that any predictions made using this equation are only as accurate as the data and assumptions used to determine its values.