- #1
joebohr
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^topic
joebohr said:^topic
dimension10 said:I'll assume your limits are 1 and e.
[tex]xy=a[/tex]
[tex]y=\frac{a}{x}[/tex]
[tex]\mbox{Area under the curve}=\int\limits_{1}^{e}\frac{a}{x}\mbox{d}x[/tex]
[tex]\mbox{Area under the curve}=a\left(\mbox{ln}(e)-\mbox{ln}(1)\right)[/tex]
[tex]\mbox{Area under the curve}=a(1-0)[/tex]
[tex]\mbox{Area}=a[/tex]
joebohr said:Perhaps I should have specified. I knew how to find the are under the curve from a to b, where a is not equal to zero, but wanted to find the area of the curve in either the 1st or 3rd quadrant (that is, from 0 to b or from -b to 0).
dimension10 said:Is the a you are talking about the same as the a in xy=a?
joebohr said:Nope, sorry about that, I should have used different variables. The second a (as in from a to b) just represents that I wasn't asking about the integral from c to b where c≠0 and was concerned with the integral from 0 to b.
dimension10 said:Ok. So,
[tex]\mbox{Area under the curve}=\int_{c}^{b}\frac{a}{x}\mbox{d}x[/tex]
[tex] \mbox{Area under the curve}= a \left(\mbox{ln} |b|-\mbox{ln} |c| \right) [/tex]
joebohr said:Right, as I said I got this result when I tried myself. But I wanted to know what to do when c=0.
dimension10 said:The area is infinite because ln(0)= -∞
joebohr said:That's the problem, and it's what made me think I needed a new approach. Because obviously a hyperbola in the 1st quadrant doesn't have infinite area as it has asymptotes at the x and y axis. Is there perhaps another way to figure this out?
I like Serena said:I'm afraid not. Your limit approaches infinity instead of 0.
joebohr said:Yeah, your right, but Wolfram was giving me a different result a minute ago, or maybe I made a typing mistake. Anyway, can you help with my other questions (sorry for so many)?
joebohr said:Also, are there any clever approximations for this area? Would translating the figure into say (x-3)(y-3)=a make any difference (no more singularity)? How would I find the area under a different type of hyperbola (is there a method that works for all hyperbolas that aren't in the form xy=1)? Thanks.
joebohr said:Wait, but for the second example, isn't the following true:
[itex]\int[/itex][itex]^{\infty}_{1}[/itex][itex]\frac{1}{x}[/itex]dx[itex]\equiv[/itex]-ln(1)+[itex]lim_{x->{\infty}}[/itex]ln(x)=0
?
Also, are there any clever approximations for this area? Would translating the figure into say (x-3)(y-3)=a make any difference (no more singularity)? How would I find the area under a different type of hyperbola (is there a method that works for all hyperbolas that aren't in the form xy=1)? Thanks.
joebohr said:So if I were to cut of the area at the point 10a, the area would be approximately equal to ln(10a)?
To graph a diagonal hyperbola, you first need to rearrange the equation to solve for y. The resulting equation will have a slope and a y-intercept, which you can use to plot points and draw a line. Then, you can use the symmetry of the hyperbola to plot the other half of the curve.
The constant a in the equation xy=a represents the distance between the center of the hyperbola and the asymptotes. It also affects the shape and orientation of the hyperbola.
To find the area under a diagonal hyperbola, you can use the integral calculus technique known as the "area under a curve" method. This involves finding the definite integral of the hyperbola's equation within a given interval.
No, the area under a diagonal hyperbola cannot be negative. This is because the integral of a function represents the net area between the curve and the x-axis, and the area cannot be below the x-axis.
The area under a diagonal hyperbola is infinite, meaning it extends forever in the positive and negative directions. This is because the diagonal hyperbola has an asymptote at y=0, which means the curve never intersects the x-axis. In contrast, other types of hyperbolas have finite areas because they intersect the x-axis at two points.