MATLAB Assignment due 11/9

Do we need to generate Phi and Phi_q for each of these joints?  I guess we will eventually.

The reason I ask, is because some of the C(t) functions are going to be constant when we take their derivative.  We were told to set up our matlab functions using handles (i.e. C(t)), but as far as I know we still haven't solved the error when you try and take C(t) when C(t) is constant.

Example:

syms t
Csym = sym(t^2*2);
C = matlabFunction(Csym);
CPrimeSym = diff(Csym);
CPrime = matlabFunction(CPrimeSym);
CDblPrimeSym = diff(diff(Csym));
CDblPrime = matlabFunction(CDblPrimeSym);

returns
Csym = 2*t^2 
C = @(t)t.^2.*2.0
CPrimeSym = 4*t
CPrime = @(t)t.*4.0
CDblPrimeSym = 4
CDblPrime = @()4.0

Now,

if you say t=2

then

eval(Csym) = 8
C(t) = 8

eval(CPrimeSym) = 8
CPrime(t) = 8

eval(CDblPrimeSym) = 4
CDblPrime(t) = error

Should we abandon handles and switch to the eval function and symbolic equations?

Tyler,

does it work if you use

  Code:

valueC = subs(C, 2);
valueCPrime = subs(CPrime, 2);
valueCDblPrime = subs(CDblPrime, 2);

?

- Markus

[Dan Negrut]

This was sent out in an email that never made it over to you:

yourStringThatDefinesC='2*t+1'
cFunction = eval(yourStringThatDefinesC)
cFunctionPrime = diff(yourStringThatDefinesC)
cFunctionPrimePrime = diff(diff(yourStringThatDefinesC))
functionHandleValue = matlabFunction(cFunction)
functionHandleFirstDeriv = matlabFunction(cFunctionPrime)
functionHandleSecondDeriv = matlabFunction(cFunctionPrimePrime)

% say that now you want to evaluate the functions at t=3:
value_tEQ3 = subs(functionHandleValue, 3)
firstDeriv_tEQ3 = subs(functionHandleFirstDeriv, 3)
secondDeriv_tEQ3 = subs(functionHandleSecondDeriv, 3)