What is calculus used for excluding physics?
What is calculus used for excluding physics?
.png "What is calculus used for excluding physics?")
What is calculus used for excluding physics? Off the top of my head, here are a few:
- It's used in biology to study the growth and decay of populations.
- Chemists use it, since chemistry often involves a significant amount of math.
- The technique of radiocarbon dating in history and geology is based upon calculus.
- It is widely used in electronics and communications for signal processing stuff (think Fourier/Laplace transforms).
- Economics uses calculus for various purposes (such as the analysis of curves/trends).
- Computer scientists use ideas borrowed from calculus to predict the efficiency of algorithms.
- I'm considering that subjects such as mechanical engineering, astronomy, etc are applied physics - if not, I could easily come up with several more examples.
How can I apply calculus to solving problems in physics?
You can apply calculus to solving problems in physics:
- by understanding and knowing what calculus is
- by understanding the essence of the physics of the problem at hand and
- by understanding where and how the two can be joined together in a productive manner
- Example 1/3: a resting point mass m located on a horizontal plane begins to move without friction under the influence of a force F whose magnitude varies over time as kt, for some constant k, and whose direction with respect to the horizon is kept at the constant angle α.
Recover: the velocity of separation of the said mass from the said plane.
Solution: when will the point mass separate from the plane to begin with, in this particular case?
The said point mass will separate from the said plane when the projection of the force F acting on m on the normal to the given surface, a plane, will become larger than the force due to gravity, pictorially (Fig. 1):
or symbolically:
ktsinα⩾mg(1)
At any instance of time t that satisfies the strict inequality shown in (1) the point mass m will not be in contact with the given plane.
The exact instance in time ts when the separation of m from the plane does take place is determined from the following equation:
ktssinα=mg(2)
from where:
ts=mgksinα(3)
However, for as long as the point mass m stays on the plane, the horizontal projection of the above force F onto that plane, ktcosα that is, will, as per the Newton’s Second Law (Axiom) Of Motion, be responsible for the acceleration of that point there because the mass m remains constant at all times (and we translate the change in linear momentum over time into the proverbial am ay):
ktcosα=mdvdt(4)
It so happens that the variables in the differential equation shown in (4) are nicely separable and because you know what integral calculus is, you separate these variables like so:
kcosαtdt=mdv(5)
and integrate the LHS of (5) from 0 to ts and integrate the RHS of (5) from 0 to vs:
kcosα∫0tstdt=m∫0vsdv(6)
Because you know your integrals, you quickly compute:
kcosαt2s2=mvs(7)
from where you recover vs, via (3):
vs=mg2cosα2ksin2α(8)
and we also see that in general, via (7), for as long as the point m is located on the plane, the point’s speed v varies over time t as follows:
v(t)=kcosα2mt2(9)
Not only that, since you know your differential calculus, you know that the velocity of a material point is the first derivative of its radius vector - which in this case is simply:
v=dxdt(10)
and you can also recover the distance xs traveled by the point m prior to lift off:
dx=v(t)dt=kcosα2mt2dt
by integrating the above LHS from 0 to xs and by integrating the above RHS from 0 to ts:
xs=kcosαt3s6m
which, via (3), is to say that:
xs=m2g3cosα6k2sin3α(11)
Example 2/3: at first, a point mass m is lifted from a mine whose depth is αR to the Earth’s surface, where α is some real constant number such that 0<α<1 and R is the Earth’s idealized radius.
Next, the same point mass m is lifted above the Earth’s surface to the height βR, where β is also some real constant number such that 0<β<1.
Recover: the ratios of the works done during each lifting of the point mass.
Solution: of this problem is straightforward, as long as we remember a few basic facts about work in physics and the nature of the force due to gravity.
Namely, wiki actually has a reasonable definition of the work done by a variable force as that of:
the line integral of its scalar tangential component along the path of its application point
In our case, the force due to gravity is conservative. Thus, the only points that matter to us are the initial, I, and the final, F, points of m on its lifting journeys and we will assume that both of these journeys took place along the same radial straight line (Fig. 2):
In order to keep things under control, we will roll the equations numbering in each problem separately.
In general, with a reasonable degree of approximation, the mass of the spherical Earth Mx captured inside of the interior sphere of radius x measured from the Earth’s center O is given by:
Mx=ρVx=ρ4πx33(1)
where the Earth’s average density ρ can be recovered from the following obvious relationship:
M=ρV=ρ4πR33(2)
implying that:
ρ=3M4πR3(3)
The said mass Mx, then, will be given by:
Mx=MR3x3(4)
If we model the interior portion of Earth with thin spherical shells, then due to the theorem proved by Newton, the gravitational force that acts on a point mass located strictly inside of such a spherical shell is zero, while the gravitational force that acts on a point mass located strictly outside of such a shell is equal to the good old gravitational interaction between the massive points m and Mx, where the mass Mx is concentrated at the center of the said shell (sphere).
Thus, for as long as our point mass is located inside of the deep mine, the gravitational force Fx that acts on it will be given by:
Fx=GmMxx3(5)
implying, via (4), that the said force varies in x linearly:
Fx(x)=GmMR3x(6)
and the game is essentially over.
Why.
Because in order to obtain the work Aα done by the force Fx while lifting m to the Earth’s surface, we integrate the RHS of (6) from the point x=(1−α)R to the point x=R:
Aα=∫(1−α)RRGmMR3xdx=GmM2R3x2∣∣∣R(1−α)R
and find that:
Aα=GmM2Rα(2−α)(7)
Next, once the point mass m is on the Earth’s surface, the entire mass of Earth, M, will generate the respective gravitational force that will act on m from the point x=R to the point x=(1+β)R:
Aβ=∫R(1+β)RGmMx2dx=GmM[−1x](1+β)RR
which, because you know your integrals, is to say that:
Aβ=GmMRβ1+β(8)
Dividing (7) by (8), we find:
AαAβ=α2β(2−α)(1+β)(9)
For example, say, in some naive way we drilled the above hypothetical radial tunnel of length α=0.5R and then we lifted the point mass m by as much above the Earth’s surface:
α=β=R2
Then, according to (9), we shall have the 9 to 8 ratio:
AαAβ=98
Lastly, a slightly more sophisticated or a grown-up, theoretical physics, example.
Example 3/3: A point mass m is sliding without friction and initial velocity off of a positive branch of a horizontal parabola y2=ax from a known initial height h0 (Fig 3):
Recover: the separation height hs of the point mass at which m will jump off of the said parabola.
Solution. In terms of Lagrangian mechanics as it is applicable to mechanical systems with constraints, the symbolic form f(x,y) of the geometry of this particular constraint is given by:
f(x,y)=y2−ax=0(1)
differentiating which with respect to time once and twice, we find:
2yy˙−ax˙=0(2)
and:
2y˙2+2yy¨−ax¨=0(3)
From the law of conservation of the total mechanical energy we recover the square of the point’s velocity:
x˙2+y˙2=2g(h0−y)(4)
Using the relation in (1) as the fodder for the Lagrange equations of the first kind where λ is the Lagrange multiplier, we obtain:
my¨=−mg+2λy(5)
because the vector of the force of gravity points in the direction opposite to that of the y−axis, and:
mx¨=−aλ(6)
The overall goal being the recovery of λ as a function of m,g,a,y and h0, take the image of x˙ from (2):
x˙=2ya⋅y˙
and insert it into (4):
4y2a2⋅y˙2+y˙2=2g(h0−y)(7)
Solve (7) for y˙2:
y˙2=2ga2(h0−y)4y2+a2(8)
Now exploit the image of y˙2 in (3):
y˙2=a2⋅x¨−yy¨(9)
Into the RHS of (9) insert the corresponding images of x¨ from (6):
x¨=−aλm
and of y¨ from (5):
y¨=−g+2λmy
to obtain:
y˙2=−a2λ2m+gy−2λmy2(10)
But the magnitude of y˙2 in (8) and (10) is one and the same. Hence, equating the RHSs of (8) and (10), we recover the shape of λ:
λ=2m4y2+a2⋅(2ga2(y−h0)4y2+a2+gy)(11)
At separation time the said λ vanishes. The factor in front of the parenthesis on the RHS of (11) is a strictly positive real number. Therefore, the only way that λ can vanish is when the parenthesised expression on the RHS of (11) vanishes:
2ga2y−2ga2h0+4gy3+ga2y=0
But we agreed earlier that our experiment unfolds in near-Earth gravity. Therefore, g≠0.
Therefore, the separation height hs sought-after must be the only (!) real (!) positive (!) root of the following equation which is a depressed cubic in y:
4y3+3a2y−2a2h0=0(12)
Game over. Problem solved - the rest is known arithmetic due to Cardano:
y3+py+q=0(13)
where
p=3a24,q=−a2h02
and, thus, the magnitude 4p3+27q2 is always strictly positive:
4p3+27q2=27a44(a24+h20)>0
and the real root of (12, 13) is given by:
hs=a2h04+a4h2016+a664−−−−−−−−−√−−−−−−−−−−−−−−−−−⎷3+a2h04−a4h2016+a664−−−−−−−−−√−−−−−−−−−−−−−−−−−⎷3(14)
which is the answer.
The point of the above discussion is: in order to apply calculus to the solutions of physics problems successfully, you need to analyze each problem from scratch and see where and how the machineries of the differential and/or the integral calculus can be used in the given setting.
Related Posts:
Leave a Comment