My Math Forum Scalar Products, Derivatives and Unit Vectors

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 March 4th, 2015, 09:44 PM #1 Newbie     Joined: Mar 2015 From: Auckland, NZ Posts: 7 Thanks: 0 Scalar Products, Derivatives and Unit Vectors In my text there is a section on Conservation of Energy of a Point Mass. Starting with Newton's Second Law as: F = ma = m dv/dt = mg Gravitational field close to Earth is approximately -gz, where z is the unit vector on the z-axis. m dv/dt = -mgz Then take the scalar (dot) product of both sides mv . dv/dt = - mgv . z Now the book jumps (with a simple "This equation can be simplified") to d/dt 1/2 mv . v = d/dt (1/2 mv^2) = -mg dz/dt and I'm trying to shake the rust off a brain which hasn't worked with this type of math since the late 90s by filling in the gaps over which the book jumps. I know what's happening on the left, but the right side has me scratching my head. If I recall correctly, change in position z relative to time is v, so dz/dt = v But then I don't know where that z goes. Is it just that it somehow drops out with a value of 1 since it is the unit vector? Am I just wrong that we can simply say dz/dt = v? Do we actually need to retain that z so it should it be dz/dt = v . z? What is the importance of that unit vector z in operational terms (I know what it means - giving the direction of the action without affecting the value - I just don't quite get how it operates)? Is it something to do with dot products giving a scalar value? I'm sure I'll slap my head when someone points out how simple the answer is but I can take it. Last edited by Ben Hunto; March 4th, 2015 at 09:47 PM.
 March 5th, 2015, 01:57 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions If $\displaystyle v = v(t)$, then using implicit differentiation, $\displaystyle \frac{d}{dt}\left(\frac{1}{2}mv^2\right) = mv \frac{dv}{dt}$. So the reverse logic is applied to simplifying your LHS. Also, z is a unit vector, which means that it always has a magnitude of 1. Therefore, it should only be used to indicate direction. This implies $\displaystyle \frac{dz}{dt} = 0$. A new scalar variable should be used, such as h, where $\displaystyle v = \frac{dh}{dt} \cdot z$ Then integrating both sides with respect to $\displaystyle t$ will yield $\displaystyle \frac{1}{2} mv^2 = mg \Delta h$ where we have removed the unit vectors representing up and all quantities are scalars. Thanks from Ben Hunto
March 5th, 2015, 03:10 AM   #3
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Quote:
 Originally Posted by Ben Hunto In my text there is a section on Conservation of Energy of a Point Mass. Starting with Newton's Second Law as: F = ma = m dv/dt = mg Gravitational field close to Earth is approximately -gz, where z is the unit vector on the z-axis. m dv/dt = -mgz Then take the scalar (dot) product of both sides mv . dv/dt = - mgv . z Now the book jumps (with a simple "This equation can be simplified") to d/dt 1/2 mv . v = d/dt (1/2 mv^2) = -mg dz/dt and I'm trying to shake the rust off a brain which hasn't worked with this type of math since the late 90s by filling in the gaps over which the book jumps. I know what's happening on the left, but the right side has me scratching my head.
Their notation doesn't make sense to me either. Let's review what they're saying.

mv . dv/dt = - mgv . z

Let's call $\displaystyle V_\parallel$ the part of V that's inline with F, and $\displaystyle V_\perp$ the part that isn't inline with F.

We get a scalar, $\displaystyle |V_\parallel||F|$

d/dt 1/2 mv . v = d/dt (1/2 mv^2)

Same scalar.
v . d/dt 1/2 mv + (d/dt 1/2 mv) . v = v . m dv/dt = $\displaystyle |V_\parallel||F|$

I suppose they put the dz/dt (not dz/dt) there as a reminder that this scalar represents something in the z direction. I can see how you'd be confused by it. It doesn't make sense.

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