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About Scotto
Expertise
Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.). I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

Experience
Experience in the area: I have tutored students in all areas of mathematics for over 20 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

Publications
Maybe not a publication, but I have respond to well oveer 3000 questions on the PC. That's around 2,000 in basic math and 1,000 in advanced math.

Education/Credentials
I aquired well over 40 hours of upper division courses. This was well over the number that were required. I graduated with honors in both my BS and MS degree from Oregon State University. I was allowed to jump into a few junior level courses my sophomore year.

Awards and Honors
I have been nominated as the expert of the month several times. All of my scores right now are at least a 9.8 average (out of 10).

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My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you.

You are here: Experts > Teens > Homework/Study Tips > Calculus > Calculate points on a curve

Calculus - Calculate points on a curve

Expert: Scotto - 11/5/2009

Question
I am trying to calculate points on a curve. The length is 90.75" by 6" height. I wanted the below to look more like an arch. Here is a illustration:

<--90.75"--->
__________
         .
^       .
6"   .
v .
.
.

Answer
This picture would look a lot better if added on in a JPG file.

From what I can see, we can approximate the curve with four cubics.
On goes from -a to -3. One goes from -3 to 0. One goes from 0 to 3. One goes from 3 to a.

Let d be the height on the y axis where the curve goes from concave down to concave up.
It should be the same on both sides. Let e be the slope of that tangent line at this point.

The four functions will be f1(x), f2(x), f3(x), and f4(x).
If this was a professional paper, I would make 1, 2, 3, and 4 into subscripts,
but ... , oh well. Let h be the hight of the thing with width 6.

The values for f1 would be f1(-a)=0, f1'(-a)=0, f1(-3)=d, f1'(-3)=e.
The values for f2 would be f2(-3)=d, f2'(-3)=e, f2(0)=h, f2'(0)=0.
The values for f3 would be f3(0)=h, f3'(0)=0, f2(3)=d, f3'(3)=-e.
The values for f4 would be f4(3)=d, f4'(3)=-e, f4(a)=0, f4'(-a)=0.

Given 4 conditions on a cubic, the cubic in each interval can be solved for.

The other thing to do is to approximate the curve with e^(a+(x/b)^2) where x is the height.
The value of a would be ln(h) where h was the height in the middle.
The value of b would depend upon how much the curve was spread.
One way of doing this is noting the 1st derivative is (2x/b²)•e^(a+(x/b)^2) and that x is the point of inflection, so in that way b could be found.

Is that what you needed?

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