4.1: Differentiation and Integration of Vector Valued Functions
( \newcommand{\kernel}{\mathrm{null}\,}\)
The formal definition of the derivative of a vector valued function is very similar to the definition of the derivative of a real valued function.
Let r(t) be a vector valued function, then
r′(t)=limh→0r(t+h)−r(t)h.
Because the derivative of a sum is the sum of the derivative, we can find the derivative of each of the components of the vector valued function to find its derivative.
ddt(3ˆi+sintˆj)=costˆj
ddt(3t2ˆi+cos(4t)ˆj+tetˆk)=6tˆi−4sin(t)ˆj+(et+tet)ˆk
Properties of Vector Valued Functions
All of the properties of differentiation still hold for vector values functions. Moreover because there are a variety of ways of defining multiplication, there is an abundance of product rules.
Suppose that v(t) and w(t) are vector valued functions, f(t) is a scalar function, and c is a real number then
- ddt(v(t)+w(t))=ddtv(t)+ddtw(t),
- ddtcv(t)=cddtv(t),
- ddt(f(t)v(t))=f′(t)v(t)+f(t)v(t)′,
- (v(t)⋅w(t))′=v′(t)⋅w(t)+v(t)⋅w′(t),
- (v(t)×w(t))′=v′(t)×w(t)+v(t)×w′(t),
- ddtv(f(t))=v(t)′(f(t))f′(t).
Show that if r is a differentiable vector valued function with constant magnitude, then
r⋅r′=0.
Solution
Since r has constant magnitude, call its magnitude k,
k2=|r|2=r⋅r.
Taking derivatives of the left and right sides gives
0=(r⋅r)′=r′⋅r+r⋅r′
=r⋅r′+r⋅r′=2r⋅r′.
Divide by two and the result follows
Integration of vector valued functions
We define the integral of a vector valued function as the integral of each component. This definition holds for both definite and indefinite integrals.
Evaluate
∫(sint)ˆi+2tˆj−8t3ˆkdt.
Solution
Just take the integral of each component
∫(sint)dtˆi+∫2tdtˆj−∫8t3dtˆk.
=(−cost+c1)ˆi+(t2+c2)ˆj+(2t4+c3)ˆk.
Notice that we have introduce three different constants, one for each component.
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.