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Problem : Find the derivative of the vector-valued function,

f(x) = (3x2 +2x + 23, 2x3 +4x, x-5 +2x2 + 12)

We take the derivative of a vector-valued function coordinate by coordinate:

f'(x) = (6x + 2, 6x2 +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 z-directions.


x(t)=3t2 + 5  
y(t)=- t2 + 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)t2 + (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 direction--in the same way that a ball thrown upwards (which experiences constant acceleration due to the earth's gravity) slows down to zero-velocity as it reaches its maximum height, and then changes direction to fall back down.
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