Curve through 2 points (6603.03,1502.53) & (6699.01,1504.73)

[discrete]

I need to draw a smooth curve between 2 points. The first point is (6603.03, 1502.53) and the second point is (6699.01, 1504.73).

The line passing through the first point needs to have a slope of 9.39, and the line passing through the second point needs to have a slope of 0.

Help?

Thanks!

 

[stapel]

I need to draw a smooth curve between 2 points. The first point is (6603.03, 1502.53) and the second point is (6699.01, 1504.73). The line passing through the first point needs to have a slope of 9.39, and the line passing through the second point needs to have a slope of 0.

Since you posted this question to the "geometry" cagegory, rather than to the "calculus" category, it would appear that you are needing to use pre-calculus techniques.

What formulas, algorithms, or methods have they given you? How far have you gotten in applying them? Where are you stuck?

Please be complete. Thank you!

Eliz.

 

[Deleted member 4993]

You are given four boundary conditions - 2 end points and 2 slopes. So a cubic curve with four parameters will do:

y = Ax^3 + B x^2 + Cx + D

and

y' = 3Ax^2 + 2Bx + C

Apply your boundary conditions to the above two equations - and solve the 4x4 matrix to define your A, B, C & D.

If I were to do the problem, I would shift the origin to one of the points to make life little easier.

 

[discrete]

Re:

I need to draw a smooth curve between 2 points. The first point is (6603.03, 1502.53) and the second point is (6699.01, 1504.73). The line passing through the first point needs to have a slope of 9.39, and the line passing through the second point needs to have a slope of 0.

Since you posted this question to the "geometry" cagegory, rather than to the "calculus" category, it would appear that you are needing to use pre-calculus techniques.

What formulas, algorithms, or methods have they given you? How far have you gotten in applying them? Where are you stuck?

Please be complete. Thank you!

Eliz.

I haven't been given formulas or methods - if there's an reasonable way to do it with calculus I can do that too... I first tried circle formulas, but quickly found they didn't work (I could make them work changing one of the slopes). I'm trying to work with eliptical and parabola functions but it's not working out for me...)

 

[stapel]

I haven't been given formulas or methods....

Um... This homework question must relate to something in your course, mustn't it...? :shock:

What topics have recently been covered in class? Maybe if we know with what you're currently working, we can figure out what is expected of you on this assignment. :idea:

if there's an reasonable way to do it with calculus I can do that too...

So you have taken calculus...? Then try using the solution method explained in the second reply you received.

I first tried circle formulas, but quickly found they didn't work.... I'm trying to work with eliptical and parabola functions but it's not working out for me.

I'm sorry, but I don't understand what you mean by this...? Please reply showing your work and reasoning.

Thank you!

Eliz.

 

[galactus]

Did you try the derivative of a quadratic, \(\displaystyle y'=2ax+b\)?

Also, the general form of a quadratic is \(\displaystyle ax^{2}+bx+c\). This, along with knowing the slope at a point (as you do) can probably get you close to a parabola that will fit.

Did you mean a slope of 9.39 or 9.39%?. Just clarifying.

Assuming the former, you have two equations you can solve for a and b and thus get the a and b in the quadratic.

From the derivative:

\(\displaystyle 9.39=2a(6603.03)+b\)

\(\displaystyle 0=2a(6699.01)+b\)

Solve for a and b.

Now, once you have those, you can solve for c by using your given points in \(\displaystyle y=ax^{2}+bx+c\)

I came up with an equation that works pretty good. I used the two respective data points and get c varying from each by a very small amount. But it seems OK.

As SK suggested, you may try adjusting your points so you are not working with such large coordinates. Try letting (0,0) be (6600,1500) and go from there.

Let me know how it goes.