8.4 Calculus with vectors by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.
vCalc Companion Formulas | ||
vCalc Formulary | 8.4 Calculus with vectors | |
`a=F/m` | Acceleration | |
`Deltax=v_o Deltat +1/2 a * Deltat^2` | Distance w/ Constant Acceleration |
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.
? 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.
? 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
`=b/mt^(42)hatx + c/mt^(137)haty`.
The velocity vector `v` is the integral with respect to time of the acceleration,
`v=?adt`
`=?(b/mt^(42)hatx + c/mt^(137)haty)dt`,
and integrating component by component gives
`=(?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.
? 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)/(?2m)hatx + (?b+ct)/(?2m)haty`.
To find the velocity vector as a function of time, we need to integrate the acceleration vector with respect to time,
`v=?adt`
`=?((b?ct)/(?2m)hatx + (?b+ct)/(?2m)haty)dt`
`=1/(?2m)?[(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/(?2m)?(b?ct)dt + haty/(?2m)?(?b+ct)dt`
`=((bt?1/2ct^2)/(?2m)+"constant" #1)hatx + ((?bt+1/2ct^2)/(?2m)+"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)/(?2m))hatx + ((?bt+1/2ct^2)/(?2m)+v_o)haty `
8.4 Calculus with vectors by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.