# Differentiation of Vectors

If a vector R varies continuously as a scalar variable t changes, then R is said to be a function of t and is written as R = F(t).

Just as in scalar calculus, we define derivative of a vector function R = F(t) as

and write it as dR/dt or dF/dt or F'(t).

General rules of differentiation are similar to those of ordinary calculus provided the order of factors in vector products is maintained.  Thus, if Ф, F, G, H are scalar and vector functions of a scalar variable t, we have

1. Example:  If A= 5t2I + tJ – t3K, B=sintI - cos(t)J then
Find i) d/dt(A.B)
ii) d/dt(AxB)
Solution: i) d/dt(A.B) = A.(dB/dt) + (dA/dt).B
= (5t2I+tJ-t3K)[costI –(-sint)J] + (10tI+J-3t2K)(sintI - costJ)
= (5t2cost-tsint) + (10tsint-cost)
= 5t2 cos(t) + 11tsin(t) – cos(t).
ii) d/dt(AXB) = Ax(dB/dt) + (dA/dt)xB
= (5t2I+tJ-t3K)x(costI +sintJ) + (10tI+J-3t2K)x(sintI -costJ)
= (t3sint-3t2cost)I – t2(tcost+3sint)J + [(5t2-1)sint-11tcost]K

## Derivative of Polar Functions

Let r = r(θ) represent a polar curve, then
dy = dy/dθ = r'sinθ + rcosθ

__     ______     ______   _____
dx    dx/dθ    r'cosθ – rsinθ
Since x=rcosθ
dx/dθ = r(-sinθ) + cosθ. r' = r'cosθ – rsinθ
y = rsinθ
dy/dθ = rcosθ + sinθ. r' = r'sinθ + rcosθ
dy = dy/dθ = r'sinθ + rcosθ

___    ______   ______   ______
dx    dx/dθ    r'cosθ – rsinθ
Example: Find the derivative of r = θcosθ
Example: dy/dx = [θ(-sinθ) + cosθ] sinθ + θ cosθ (cosθ)
________________________________

[θ(-sinθ) + cosθ] cosθ – θ cosθ sinθ

=   -θsin2θ + cosθ sinθ + θcos2θ

_________________________

-θsinθ cosθ + cos2θ – θcosθ sinθ

= cosθ sinθ + θ(cos2θ – sin2θ)

_________________________

cos2θ - 2θ cosθ sinθ

=   cosθ sinθ+ θcos2θ

_______________

cos2θ - 2θ cosθ sinθ

Example: Find the slope of the tangent line to the unit circle x= cost, y = sint (0≤t≤2π) at the point t = π/6
Solution: The slope at a general point on the circle is dy/dx

Thus, the slope at t = π/6 is
dy/dx]t=π/6    = -cot π/6 = -√3
Example: Find the slope of the tangent line to the circle r = 4cosθ at the point where θ = π/4
Solution:
Thus, at the point where θ = π/4 the slope of the tangent line is dy/dx]θ=π/4 = -cotπ/2 = 0 which implies that the circle has a horizontal tangent line at the point where θ = π/4

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