[Solved] how to find the equation of a curve, given: 2 gradients & 1 x,y coordinate

Given:

when x=c, gradient1 = -2

when x=b, gradient2 = 0 and y=0

x=[0..2]

How would I find the equation for a concave curve?

(note: I have pre-calc experience, but only cursory calculus knowledge. Any help would be greatly appreciated.)

Welcome to the forums!

There are probably infinitely many functions that could satisify the 3 conditions. However, if you want to limit yourself to the lowest degree polynomial, a quadratic should do:

y = ax² + bx + c
y' = 2ax + b

where a, b, c are unknown. The 2nd equation is the gradient (1st derivative). You need 3 equations, which can be written to satisfy the constraints:

y₁ = ax₁² + bx₁ + c
y₂' = 2ax₂ + b
y₃' = 2ax₃ + b

This is a set of 3 simultaneous linear equations that can be solved for a, b, c by Gauss elimination. Actually the 2nd two equations only involve a & b and can be easily solved by substitution. Once a & b are known, c can be solved in the 1st equation.

Thanks for the help VBAhack. I was able to figure it out from there.

a = 1
b = -4
c = 4

y = x^2 - 4x + 4

How did you determine the equation/curve would be a quadratic? I was trying all kinds of formulas (randomly) such as radioactive decay, but I wasn't sure how to tie-in the gradients (derivatives).

Thanks again for replying. Now I can move on to figuring the area below the curve and then the volume as the curve is rotated 360 degrees. :bigyello:

Quote:

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Originally Posted by viknord

How did you determine the equation/curve would be a quadratic?

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Educated guess. Polynomials are often a good place to start - they are simple. Also, since you had 3 pieces of information, a function with 3 parameters is the obvious thing to try first - thus, a polynomial with 3 parameters. :)

Btw, would you recommend any good Calculus books? Self/home-study type books. I've been reading "Calculus for Cats" by Amdahl and Loats (which helps conceptually) and "Calculus: An Intuitive and Physical Approach" by Kline (just started this one).

Thanks again. :)

Quote:

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Originally Posted by viknord

would you recommend any good Calculus books?

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Nope, but you might try here:

http://www.physicsforums.com/forumdisplay.php?f=21