Problem :
Find the derivative of the vectorvalued function,
f(x) = (3x^{2} +2x + 23, 2x^{3} +4x, x^{5} +2x^{2} + 12)
We take the derivative of a vectorvalued function
coordinate by
coordinate:
f'(x) = (6x + 2, 6x^{2} +4, 5x^{4} + 4x)
Problem :
The motion of a creature in three dimensions can be described by the
following equations for position in the x, y, and zdirections.
x(t)  =  3t^{2} + 5 

y(t)  =   t^{2} + 3t  2 

z(t)  =  2t + 1 

Find the magnitudes** of the acceleration, velocity, and
position vectors at times
t = 0,
t = 2, and
t =  2.
The first order of business is to write the above equations in vector form.
Because they are all (at most quadratic) polynomials in
t, we can write
them together as:
x(t) = (3, 1, 0)t^{2} + (0, 3, 2)t + (5,  2, 1)
We are now in a position to compute the velocity and acceleration functions.
Using the rules established in this section we find that,
v(t)  =  2(3,  1, 0)t + (0, 3, 2) = (6,  2, 0)t + (0, 3, 2) 

a(t)  =  (6,  2, 0) 

Notice that the acceleration function
a(t) is constant; therefore
the magnitude (and direction!) of the acceleration vector will be the same at
all times:

a = (6, 2, 0) =
= 2
All that's left now is to compute the magnitudes of the position and velocity
vectors at times
t = 0, 2,  2:
 At t = 0, x(0) = (5, 2, 1) = , and
v(0) = (0, 3, 2) =
 At t = 2, x(2) = (17, 0, 5) = , and
v(2) = (12, 1, 2) =
 At t =  2, x( 2) = (17, 12, 3) = , and
v( 2) = ( 12, 7, 2) =
Notice that the magnitude of the creature's velocity (i.e. the speed at which
the creature is traveling) is high at
t =  2, decreases considerably at
t = 0,
and goes back up again at
t = 2, even though the acceleration is constant! This
is because the acceleration is causing the creature to slow down and
change
directionin the same way that a ball thrown upwards (which experiences
constant acceleration due to the earth's gravity) slows down to zerovelocity as
it reaches its maximum height, and then changes direction to fall back down.