Math 2240 The Ubiquitous Ski Jump Problem Project #1 

Thanks to Elizabeth Kehler for this plot -- It's my favorite! 

Also, a big "Thank you!" to Bryson Stalnaker for the excellent solution presented below. 

The Ski jump problem asks us to use a cubic polynomial to model a ski jump starting at a height of 100 ft. and ending at 10 ft., covering a horizontal distance of 120 ft., and with tangent lines of  0°  and 30° at the start and end respectively.  This problem lends itself to the procedure of polynomial curve fitting which can be used to fit a polynomial function to a set of data point in a plane.  It has been proven that having n number of distinct points there is precisely one polynomial of degree n-1 (or less) that fits the points. The polynomial for this problem will take the form: 

(1)

 

Since we are using a polynomial of the third degree to model the ski jump we will need at least four points to use polynomial curve fitting.  Drawing a simple representation or the ski-jump in the xy-plane helps to visualize this problem. 

The first two points (0,100) and (120,10) and the start and end of the jump.  Putting those points into the original polynomial we get the linear equations: 

//when 0 is put in the polynomial for x and 100 in for f(x)or y it is solved to d=100 

(2)

 

//when 120 is put in the polynomial for x and 10 in for f(x) or y it leaves a linear equation. 

(3)

 

The second two points are solved for using some elementary calculus.  Taking the derivative of our original polynomial gives us: 

//derivative of original polynomial 

(4)

 

(5)

 

The derivative is the slope of the tangent line at a point.  We know from the problem that the slope of the tangent at x=0 is 0 and using a little trig(30°60°90° triangle) we can find the slope of the tangent at x=120 to be Putting these values in the derivative yields: 

//when 0 is put in the derivative for x and 0 in for y' it is solved to c=100 

(6)

 

//when 120 is put in the derivative for x and1/sqrt(3) in for y' it leaves a linear equation. 

(7)

 

Using the four equation we can set up a matrix and solve for a, b, c, d. 

(8)

 

//Swap R1 with R2 

(9)

 

//swap R4 with R2 

(10)

 

//(-43200/1728000)*R1+R2 

(11)

 

//(-1/120)*R2 

(12)

 

//(-14400*R2)+R1 

(13)

 

The matrix is now in a solvable form. a, b, c, d can be found at this point by using backsolving, however if we keep reducing the matrix we can get it in a form so a, b, c, d can be read off more easily. 

//(1/1728000)*R1 

(14)

 

//(-1/60)*R3+R2 

(15)

 

//(1/14400)*R3+R1 

(16)

 

//(-1/4800)*R4+R2 

(17)

 

//(1/864000)*R4+R1 

(18)

 

The matrix is now in reduced row echelon form and the answers for the coefficients of the polynomial can be directly read from the matrix. 

//Value of d 

(19)

 

//Value of c 

(20)

 

//Value of b 

(21)

 

//value of c 

(22)

 

//decimal approximation of a 

(23)

 

//decimal approximation of b 

(24)

 

//decimal approximation of c 

(25)

 

//decimal approximation of d 

(26)

 

//Equation for the polynomial That models our ski jump (the answer!!!!). 

(27)

 

//Graph of the polynomial that was obtained.