Saturday, January 21, 2012

Vector fun

In Marsden & Tromba's Vector Calculus, I found the following problem. It involves time-derivatives of the position vector r(t), for which we're using the dot notation of the physicists (and Newton).

In the first panel, we have Newton's second law, then the statement that the force is minus the gradient of the gravitational potential V, followed by a calculation of the total energy in the system. The problem is to compute d/dt of the energy. (hint)


The answer is given in the last panel, below. According to the book, it's a "simple calculation."

The first step is to compute d/dt of the kinetic energy. We use the formula from above, plus a trick to convert the squared term back to the dot product of dr/dt with itself. Then we use the chain rule, and finally, the definition of the force in terms of the gradient of the potential.


Calculation of d/dt of the potential energy puzzled me for quite a while, though it really shouldn't have. My solution was to work backward from the answer, as shown.


We put the two results together, and notice that they cancel. Surprise!

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