Asymptote Ramp

[KennyD94]

Hi All,

If my formula is:
\(\displaystyle y=100−6.101\)x\(\displaystyle 10^{39}e^{−x}\)
\(\displaystyle y(90)=95.00\)
\(\displaystyle y(91)=98.16\)
\(\displaystyle y(92)=99.34\)
\(\displaystyle y(93)=99.91\)
\(\displaystyle y(94)=99.99\)
.
.
.
\(\displaystyle y(100)=99.99\)

how do I change the formula to get a slower ramp towards 99.99 as x -> 100
example
\(\displaystyle y(90)=95.00\)
\(\displaystyle y(95)=97.50\)
\(\displaystyle y(100)=99.99 \)

 

[Dr.Peterson]

Hi All,

If my formula is:
\(\displaystyle y=100−6.101\)x\(\displaystyle 10^{39}e^{−x}\)
\(\displaystyle y(90)=95.00\)
\(\displaystyle y(91)=98.16\)
\(\displaystyle y(92)=99.34\)
\(\displaystyle y(93)=99.91\)
\(\displaystyle y(94)=99.99\)
.
.
.
\(\displaystyle y(100)=99.99\)

how do I change the formula to get a slower ramp towards 99.99 as x -> 100
example
\(\displaystyle y(90)=95.00\)
\(\displaystyle y(95)=97.50\)
\(\displaystyle y(100)=99.99 \)

Your formula is [MATH]y=100−6.101\times10^{39}e^{−x}[/MATH]. Keeping the horizontal asymptote at 100 and retaining the general form of the equation, there are two parameters you could manipulate, only one of which you have used.

Make it [MATH]y=100−ke^{−rx}[/MATH], and you can adjust both k and r. Given any two desired points (e.g. [MATH]y(90)=95.00[/MATH] and [MATH]y(100)=99.99[/MATH]), you can solve for k and r. The result may or may not fit the particular shape you want (e.g. your third suggested point), but it should be acceptable if all you want is a slower approach.

 

[KennyD94]

Your formula is [MATH]y=100−6.101\times10^{39}e^{−x}[/MATH]. Keeping the horizontal asymptote at 100 and retaining the general form of the equation, there are two parameters you could manipulate, only one of which you have used.

Make it [MATH]y=100−ke^{−rx}[/MATH], and you can adjust both k and r. Given any two desired points (e.g. [MATH]y(90)=95.00[/MATH] and [MATH]y(100)=99.99[/MATH]), you can solve for k and r. The result may or may not fit the particular shape you want (e.g. your third suggested point), but it should be acceptable if all you want is a slower approach.

Ah I see,

I didn't think of the variable r, I have that set to 0.3 and solved for k and it gave the result I was after
\(\displaystyle y = 2.66\)x\(\displaystyle 10^{12}e^{-0.3x}\)

Thank you!

 

[Dr.Peterson]

I think you mean [MATH]y = 100 - 2.66\times10^{12}e^{-0.3x}[/MATH], and this gives y(100) = 99.75, not 99.99, I think. But I think you've got it.
[/QUOTE]

 

[KennyD94]

I think you mean [MATH]y = 100 - 2.66\times10^{12}e^{-0.3x}[/MATH], and this gives y(100) = 99.75, not 99.99, I think. But I think you've got it.

[/QUOTE]
correct! typing error. yeah I created an excel sheet that calculates a new valve of k for a variable r and 0.3 seems to give the best for what i'm looking for, thank you!