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Multivariable Calculus, Partial Derivatives and Vectors

  1. Jul 23, 2012 #1
    I just got to a point in multivariable calculus where I realise I can solve problems in assignments and tests but have no actual idea of what I'm doing. So I started thinking about stuff and came up with a few questions:

    1. Is picturing the derivative as the slope of the tangent line to a point of the graph just a useful analogy? Like imagining the integral is the area under the curve?

    2. What is the meaning of taking the derivative of a vector? I get that you can take the derivative of each component separately with respect to a parameter that they all depend on. But what does that mean?

    3. How is taking the derivative of a vector that describes a curve in space related to taking the derivative (or partial derivatives) of an equation that describes the same curve?
     
  2. jcsd
  3. Jul 23, 2012 #2
    It's not just a useful analogy, it's exactly what the derivative is and what it was invented for. Likewise, the integral was precisely invented to find the area under a curve. So these things are more than just analogies

    They don't take the derivative of a vector, they take the derivative of a vectorial function. There's a difference. But anyway, take the function

    [tex]F(t)=(\cos(t),\sin(t))[/tex]

    This describes a circle. But

    [tex]G(t)=(\cos(2t),\sin(2t))[/tex]

    also describes a circle. What's the difference between those two circles?? Well, the difference is that G goes twice as fast. For example, if we start at t=0, then both F and G are at (1,0). But at [itex]t=\frac{\pi}{2}[/itex], the curve F is at (0,1) and traveled the quarter of the circle. However, G is already at (-1,0) and traveled half of the circle. So G is much faster than F. So the velocity of G is greater.
    This velocity is measured by the derivative. In particular, given [itex]F=(x(t),y(t))[/itex], the derivative is [itex]F^\prime(t)=(x^\prime(t),y^\prime(t))[/itex]. So at any point t, we got a vector [itex](x^\prime(t),y^\prime(t))[/itex]. The direction of this vector tells us which way the curve is headed. The length of the vector is the speed of the curve.

    For example, with our circle. The derivative of F is

    [tex]F^\prime(t)=(-\sin(t),\cos(t))[/tex]

    At t=0, we get (0,1). So we can see that at t=0, the vector is going in the vertical direction with speed 1.

    But the derivative of G is

    [tex]G^\prime(t)=(-2\sin(2t),2\cos(2t))[/tex]

    and at t=0, we get (0,2). So at t=0, the vector is again going in the vertical direction, but with speed 2. So it is indeed going twice as fast, as expected.

    A more geometrical interpretation of [itex]F^\prime(t)=(x^\prime(t),y^\prime(t))[/itex] is that [itex]F^\prime(t)[/itex] is the direction of the tangent line of F at t. For example, the derivative of F (the circle) at t=0 is (0,1). So the tangent line of F at t=0 is parallel to the
    y-axis.
     
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