I am trying to figure out how
$$ \frac{\partial}{\partial \dot{q_i}} \frac{1}{2} \sum_{j,k} A_{jk} \dot{q_j}\dot{q_k}$$
is equal to
$$\sum_{j} A_{ij} \dot{q_j}$$
Does anyone have a clear way to take this partial derivative?
Partial Derivative of Double Summation
-2
$\begingroup$
derivatives
pde
summation
partial-derivative
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0What have you tried with so far? Please add what you have tried with so far in you question body, this would be appreciated. – 2017-02-25
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0Only if $A$ is symmetric, i.e. if $A_{jk} = A_{kj}$. – 2017-02-25
1 Answers
1
Hint: $\frac{\partial}{\partial \dot{q_l}}(\dot{q_j}\dot{q_k})=\delta_{l,j}\dot{q_k}+\dot{q_j}\delta_{l,k}$, were $\delta$ is the kronecker's delta.