# 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

Example: If A= 5t^{2}I + tJ – t^{3}K, 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

= (5t^{2}I+tJ-t^{3}K)[costI –(-sint)J] + (10tI+J-3t^{2}K)(sintI - costJ)

= (5t^{2}cost-tsint) + (10tsint-cost)

= 5t^{2}cos(t) + 11tsin(t) – cos(t).

ii) d/dt(AXB) = Ax(dB/dt) + (dA/dt)xB

= (5t^{2}I+tJ-t^{3}K)x(costI +sintJ) + (10tI+J-3t^{2}K)x(sintI -costJ)

= (t^{3}sint-3t^{2}cost)I – t^{2}(tcost+3sint)J + [(5t^{2}-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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