Derivatives of Sine and Cosine
Velocity is tangent to the line of motion. If a point rotates around a circle, its velocity is perpendicular to its radius-vector. Assume the point rotates around a unit circle. Then the radius-vector of the point is given by its projections on the two axes, which are cos(t) and sin(t):
(1) | r = (cos(t), sin(t)). |
The derivative of (1)
(2) | r' = (cos'(t), sin'(t)). |
is exactly the velocity of the point. If it to be perpendicular to r, we must have
(3) | cos(t)·cos'(t) + sin(t)·sin'(t) = 0. |
This of course could be derived by differentiating the famous trigonometric form of the Pythagorean theorem
(4) | cos²(t) + sin²(t) = 1. |
The applet below gives a geometric meaning of (3). Note how there are always two equal right triangles, with one rotated 90° with respect to the other. Note also how the direction of this rotation changes four times per revolution. Lastly, pay attention to the relative behavior of cos(t), the blue arrow, and the sin(t), the red arrow. sin(t) grows when cos(t) is positive, which happens in the right half plane. sin(t) decreases in the left half plane when cos(t) is negative. This suggests
What if applet does not run? |
The above serves an additional purpose: it illustrates Euler's formula
e^{it} = cos(t) + i·sin(t).
Indeed, if we knew up front that the complex valued function f(t) = e^{it} could be differentiated with respect to t as if it were a real valued function, i.e. as in
(e^{it})' = ie^{it}.
But then from the geometric meaning of multiplication by i, the direcvative of e^{it} would be perpendicular to e^{it} and would have exactly same constant modulus as the function itself. Must not it be tangent to the curve traced by e^{it}? See [Zeitz, p. 142] and [Needham].
(e^{it})' = -sin(t) + i·cos(t).
How fitting!
References
- T. Needham, Visual Complex Analysis, Oxford University Press; Reprint edition (February 18, 1999)
- P. Zeitz, The Art and Craft of Problem Solving, John Wiley & Sons, 1999
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