If *P* is a point,
then the difference Δ*P* is a vector
(because it's the result of subtracting two points),
and then the differential d*P* is an infinitesimal vector.
If *P* is a function of some scalar quantity *t*,
then d*P*/d*t* makes sense,
because it's a vector divided by a scalar,
but now it's no longer infinitesimal (unless it happens to be zero).
In other words,
*the derivative of a point with respect to a scalar is a vector*.
Another way to say this is that if *f* is a point-valued function,
then its derivative *f*′ is a vector-valued function:

first subtract two points to get a vector, then divide by the scalarf′(t) = lim_{h→0}((f(t+h) −f(t))/h);

For example, if *P* gives the position of some object at time *t*,
then *P* is a point,
but d*P*/d*t*, the *velocity* of the object, is a vector.
(Note that the magnitude of this vector is the object's *speed*.)
If we write **v** for d*P*/d*t*,
then d**v**/d*t* is the acceleration of the object,
which is also a vector.
Physicists and mechanical engineers
use the word ‘acceleration’ like this,
to indicate any change in velocity ―speed or direction— over time.
In everyday language,
this word means something more like d(|**v**|)/d*t*,
the derivative of speed with respect to time,
which is the same as
the scalar component of the acceleration in the direction of the velocity.
(This is positive if the object is speeding up
and negative if the object is slowing down, or decelerating.)

Reversing this,
if you take the indefinite integral of a vector,
then the result may be either a point or a vector,
because differentiating either of these yields a vector.
This ambiguity is all packaged into the constant of integration.
For example,
∫ ⟨2*t*, 3⟩ d*t* =
⟨*t*^{2}, 3*t*⟩ + *C*,
which is a point if *C* is a point and a vector if *C* is a vector.
The definite integral of a vector is always a vector:
fundamentally, you get it by adding up infinitely many infinitesimal vectors
(or approximate it by adding up a large number of small vectors),
and adding up vectors yields a vector;
in practice, you usually calculate it by subtracting indefinite integrals,
and regardless of whether you view the indefinite integrals
as points or as vectors,
subtracting them yields a vector.
For example,
∫^{1}_{t=0} ⟨2*t*, 3⟩ d*t* =
⟨*t*^{2}, 3*t*⟩|^{1}_{t=0} =
⟨1, 3⟩ − ⟨0, 0⟩ =
⟨1, 3⟩,
or
∫^{1}_{t=0} ⟨2*t*, 3⟩ d*t* =
(*t*^{2}, 3*t*)|^{1}_{t=0} =
(1, 3) − (0, 0) =
⟨1, 3⟩.

Putting this all together,
consider the initial-value problem
in which the acceleration of an object
is −32**k** = ⟨0, 0, −32⟩
(which is the acceleration of a freely falling object near Earth's surface,
if we use units of feet and seconds),
the object's initial velocity is ⟨3, 0, 4⟩
(so a speed of 5 ft/s eastward and upward with a slope of 4/3),
and the object's initial position is (0, 0, 100)
(so 100 feet above the origin on the ground).
Then you can calculate a general formula
for the object's position *P* as a function of the elapsed time *t*
by integrating:

- d
**v**/d*t*= ⟨0, 0, −32⟩; - d
**v**= ⟨0, 0, −32⟩ d*t*; - ∫
_{v=⟨3,0,4⟩}d**v**= ∫_{t=0}⟨0, 0, −32⟩ d*t*; **v**− ⟨3, 0, 4⟩ = ⟨0, 0, −32*t*⟩ − ⟨0, 0, −32(0)⟩;**v**= ⟨3, 0, 4⟩ + ⟨0, 0, −32*t*⟩;- d
*P*/d*t*= ⟨3, 0, 4 − 32*t*⟩; - d
*P*= ⟨3, 0, 4 − 32*t*⟩ d*t*; - ∫
_{P=(0,0,100)}d*P*= ∫_{t=0}⟨3, 0, 4 − 32*t*⟩ d*t*; *P*− (0, 0, 100) = ⟨3*t*, 0, 4*t*− 16*t*^{2}⟩ − ⟨3(0), 0, 4(0) − 16(0)^{2}⟩;*P*= (0, 0, 100) + ⟨3*t*, 0, 4*t*− 16*t*^{2}⟩;*P*= (3*t*, 0, 100 + 4*t*− 16*t*^{2}).

In the course of solving this, I've used the *semidefinite integral*:

∫The Fundamental Theorem of Calculus allows us to calculate these integrals easily:_{t=a}f(t) dt= ∫^{t}_{τ=a}f(τ) dτ.

∫This is very handy when solving initial-value problems. Since_{t=a}F′(t) dt=F(t) −F(a).

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This web page was written in 2016 by Toby Bartels, last edited on 2016 January 19. Toby reserves no legal rights to it.

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