How do you take this derivative?

  • Thread starter Thread starter msell2
  • Start date Start date
  • Tags Tags
    Derivative
Click For Summary
SUMMARY

The discussion focuses on differentiating the function x(t) = (Cv + D(tv + w))e^(λt) with respect to time t, where C, D, and λ are constants, and v and w are vectors. The product rule for differentiation is applied, specifically (fg)' = f'g + fg', to handle the vector components. The user is also encouraged to explore the implications of treating C and D as constant n×n matrices instead of scalar constants, which could further clarify the differentiation process.

PREREQUISITES
  • Understanding of vector calculus and differentiation
  • Familiarity with the product rule of differentiation
  • Knowledge of eigenvectors and generalized eigenvectors
  • Basic concepts of matrix algebra
NEXT STEPS
  • Practice differentiating vector-valued functions using the product rule
  • Explore the implications of matrix differentiation in linear algebra
  • Learn about eigenvalues and eigenvectors in the context of differential equations
  • Investigate the properties of n×n matrices and their role in linear transformations
USEFUL FOR

Mathematics students, physicists, and engineers who are working with vector calculus, linear algebra, and differential equations.

msell2
Messages
13
Reaction score
0
x(t) = (Cv + D(tv + w))e^(λt)
x'(t) = ?
C, D, λ are constants
x(t), v, w are vectors (v is an eigenvector, w is a generalized eigenvector)
t is the variable

This is coming from a larger question, asking to prove that the above equation satisfies x'(t)=Ax(t), where A is a 2x2 matrix and x'(t) and x(t) are vectors. If you know how to answer this too, that'd be great. I've been trying to figure this out for some time now. Any help would be appreciated.
 
Physics news on Phys.org
x is the product of two functions, f(t)= (Cv + D(tv + w)) and g(t)= e^(λt). Do you know how to differentiate each of those? Do you know the product rule: (fg)'= f'g+ fg'?
 
So everything is the same even though there are vectors?
 
msell2 said:
So everything is the same even though there are vectors?

Sure: just write it out component-by-component:
x_1(t) = (Cv_1 + D(v_1 t + w_1))e^{\lambda t}\\<br /> x_2(t) = (Cv_2 + D(v_2 t + w_2))e^{\lambda t}\\<br /> \vdots \\<br /> x_n(t) = (Cv_n + D(v_n t + w_n))e^{\lambda t}

Just for practice (and to help clarify some of the issues), try to see what would happen if C and D are constant n×n matrices instead of scalar constants.

RGV
 
Last edited:

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
Replies
9
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K