1. ## spline cubic interpolation

hello everyone, I need help with problem 4 getting started. I posted earlier and rcvd help from Chiro for problem 3 and in route to completing

Prob 4 just doesn't make sense, Can anyone help me get started.

problem is attached

2. ## Re: spline cubic interpolation

Hey Paragandhi.

Can you provide the definition for a clamped cubic spline?

3. ## Re: spline cubic interpolation

do u want the definition or the rule we are using to solve?

4. ## Re: spline cubic interpolation

its the third level order using derivatives. I'm really drawing a blank on this problem.. because this has to be entered through MAple soft its very confusing. Each problem is taking more than 5 hours to solve. nd I don't know how to approach this one. If you can help me get started in the right direction with the formula input it would highly appreciated

5. ## Re: spline cubic interpolation

A "cubic spline" is a piecewise cubic function. A "clamped" cubic spline has both values and derivatives given at the endpoints. Here were are given the values (distance) and derivative (speed) at every point.

Any cubic can be written in the form $f_0(t)= a_0t^3+ b_0t^2+ c_0t+ d_0$. At the first point, t= 0, we are given $f_0(0)= d_0= 0$ and $f'(0)= c_0= 75$. At the second point, t= 3, we are given $f_0(3)= 27a_0+ 9b_0+ 3c_0+ d_0= 225$ and $f_0'(3)= 27a_0+ 6b_0+ c_0= 77$. We have four equations to solve for the four coefficients.

Between t= 3 and t= 5 we may have a different cubic which we can write as $f_1(t)= a_1t^3+ b_1t^2+ c_1t+ d_1$. At t= 3, we have $f_1(3)=27a_1+ 9b_1+ 3c_1+ d_1= 225$ and $f_1'(3)= 27a_1+ 6b_1+ c_1= f_1'(3)= 77$. At t= 5 we have $f_1(5)= 125a_1+ 25b_1+ 5c_1+ d_1= 383$ and $f_1'(5)= 75a_1+ 10b_1+ c_1= 83$, again four equations to solve for the four coefficients.

Continue that over the intervals from t= 5 to 8 and from t= 8 to 13.

6. ## Re: spline cubic interpolation

Super, Thank YOU!!!