LM 8_4 Calculus with vectors Collection

8.4 Calculus with vectors  by Benjamin Crowell, Light and Matter  licensed under the Creative Commons Attribution-ShareAlike license.

8.4 Calculus with vectors (optional calculus-based section)

Using the unit vector notation introduced in section 7.4, the definitions of the velocity and acceleration components given in chapter 6 can be translated into calculus notation as

`v=(dx)/(dt)hatx + (dy)/(dt)haty + (dz)/(dt)hatz`

and

`a=(dv_x)/(dt)hatx + (dv_y)/(dt)haty + (dv_z)/(dt)hatz`.  

To make the notation less cumbersome, we generalize the concept of the derivative to include derivatives of vectors, so that we can abbreviate the above equations as

   `v=(dr)/(dt)`
and
  `a =(dv)/(dt)`.

In words, to take the derivative of a vector, you take the derivatives of its components and make a new vector out of those. This definition means that the derivative of a vector function has the familiar properties

`(d(cf))/(dt) = c (df)/(dt)`  [c is a constant]
and
`(d(f+g))/(dt) = {df}/{dt} + {dg}/{dt}` 

The integral of a vector is likewise defined as integrating component by component.

Example 8: The second derivative of a vector

`=>` Two objects have positions as functions of time given by the equations

`r_1=3t^2hatx + thaty`
and
`r^2=3t^4hatx + thaty`.

Find both objects' accelerations using calculus. Could either answer have been found without calculus?

`=>` Taking the first derivative of each component, we find

`v_1=6thatx + haty`

`v_2=12t^3hatx + haty`,

and taking the derivatives again gives acceleration,

`a_1=6hatx`

`a_2=36t^2hatx`.

The first object's acceleration could have been found without calculus, simply by comparing the `x` and `y` coordinates with the constant-acceleration equation `Deltax=v_o Deltat +1/2 a * Deltat^2`. The second equation, however, isn't just a second-order polynomial in `t`, so the acceleration isn't constant, and we really did need calculus to find the corresponding acceleration.

Example 9: The integral of a vector

`=>` Starting from rest, a flying saucer of mass `m` is observed to vary its propulsion with mathematical precision according to the equation

`F=bt^(42)hatx + ct^(137)haty`.

(The aliens inform us that the numbers 42 and 137 have a special religious significance for them.) Find the saucer's velocity as a function of time.

`=>` From the given force, we can easily find the acceleration

`a=F/m`

`=b/mt^(42)hatx + c/mt^(137)haty`.

The velocity vector `v` is the integral with respect to time of the acceleration,

`v=intadt`
`=int(b/mt^(42)hatx + c/mt^(137)haty)dt`,
and integrating component by component gives
`=(int b/mt^(42)dt)hatx + (c/mt^(137)dt)haty`

`=b/(43m)t^(43)hatx + c/(138m)t^(138)haty`,

where we have omitted the constants of integration, since the saucer was starting from rest.

Example 10: A fire-extinguisher stunt on ice

`=>` Prof. Puerile smuggles a fire extinguisher into a skating rink. Climbing out onto the ice without any skates on, he sits down and pushes off from the wall with his feet, acquiring an initial velocity `v_ohaty`. At `t=0`, he then discharges the fire extinguisher at a 45-degree angle so that it applies a force to him that is backward and to the left, i.e., along the negative `y` axis and the positive `x` axis. The fire extinguisher's force is strong at first, but then dies down according to the equation `|F|=b-ct`, where `b` and `c` are constants. Find the professor's velocity as a function of time.

`=>` Measured counterclockwise from the `x` axis, the angle of the force vector becomes 315°. Breaking the force down into `x` and `y` components, we have

`F_x=|F|cos315°`
`=(b-ct)`
`F_y=|F|sin315°`
`=(-b+ct).`

In unit vector notation, this is

`F=(b-ct)hatx + (-b+ct)haty`.

Newton's second law gives

`a =F/m`
`=(b-ct)/(sqrt2m)hatx + (-b+ct)/(sqrt2m)haty`.

To find the velocity vector as a function of time, we need to integrate the acceleration vector with respect to time,

`v=intadt`
`=int((b-ct)/(sqrt2m)hatx + (-b+ct)/(sqrt2m)haty)dt`
`=1/(sqrt2m)int[(b-ct)hatx + (-b+ct)haty]dt`

A vector function can be integrated component by component, so this can be broken down into two integrals,

`v=hatx/(sqrt2m)int(b-ct)dt + haty/(sqrt2m)int(-b+ct)dt`

`=((bt-1/2ct^2)/(sqrt2m)+"constant" #1)hatx + ((-bt+1/2ct^2)/(sqrt2m)+"constant" #2)haty`

Here the physical significance of the two constants of integration is that they give the initial velocity. Constant #1 is therefore zero, and constant #2 must equal `v_o`. The final result is

`v=((bt-1/2ct^2)/(sqrt2m))hatx +  ((-bt+1/2ct^2)/(sqrt2m)+v_o)haty`

8.4 Calculus with vectors  by Benjamin Crowell, Light and Matter  licensed under the Creative Commons Attribution-ShareAlike license.