When Time Varies

Excerpts

When Time Varies
What do we mean by *rate*? In both these cases we are making a mental comparison of something that is happening, and the length of time that it takes to happen.
When Time Varies
Ten yards is not the same as $600$ yards, nor is one second the same thing as one minute. What we mean by saying that the *rate* is the same, is this: that the proportion borne between distance passed over and time taken to pass over it, is the same in both cases.
When Time Varies
He did not use the notation of the $dy$ and $dx$, and $dt$ (this was due to Leibnitz), but had instead a notation of his own.
When Time Varies
The force necessary to accelerate a mass is proportional to the mass, and it is also proportional to the acceleration which is being imparted.
When Time Varies
Again, if a force is employed to move something (against an equal and opposite counter-force), it does *work*; and the amount of work done is measured by the product of the force into the distance (in its own direction) through which its point of application moves forward.
When Time Varies
When a railway train has just begun to move, its velocity $v$ is small; but it is rapidly gaining speed---it is being hurried up, or accelerated, by the effort of the engine. So its $\dfrac{d^2y}{dt^2}$ is large. When it has got up its top speed it is no longer being accelerated, so that then $\dfrac{d^2y}{dt^2}$ has fallen to zero.
When Time Varies
But this notation does not tell us what is the independent variable with respect to which the differentiation has been effected. When we see $\dfrac{dy}{dt}$ we know that $y$ is to be differentiated with respect to $t$.
When Time Varies
What we mean by saying that the *rate* is the same, is this: that the proportion borne between distance passed over and time taken to pass over it, is the same in both cases.
When Time Varies
It is said that Sandy had not been in London above five minutes when “bang went saxpence.”
When Time Varies
Now the speed was not actually constant all the way: at starting, and during the slowing up at the end of the journey, the speed was less. Probably at some part, when running downhill, the speed was over $60$ miles an hour. If, during any particular element of time $dt$, the corresponding element of distance passed over was $dy$, then at that part of the journey the speed was $\dfrac{dy}{dt}$.
When Time Varies
That is to say, force may be expressed either as mass times acceleration, or as rate of change of momentum.
When Time Varies
In this last sentence the word *rate* is clearly not used in its time-sense, but in its meaning as ratio or proportion.
When Time Varies
(It is the same velocity as the velocity at the middle of the interval, $t = 5$; for, the acceleration being constant, the velocity has varied uniformly from zero when $t = 0$ to $4~\text{ft./sec.}$ when $t = 10$.)

Equations

When Time Varies
v = \dfrac{dy}{dt}
Velocity is the rate at which distance changes with time: the differential coefficient of distance with respect to time.
When Time Varies
a = \dfrac{dv}{dt}
Acceleration is the rate at which velocity changes with time.
When Time Varies
a = \frac{d\left( \dfrac{dy}{dt} \right)}{dt}
Acceleration is the second differential coefficient of distance with respect to time.
When Time Varies
a = \dfrac{d^2y}{dt^2}
The acceleration equals the second derivative of distance with respect to time.
When Time Varies
f = m \frac{dv}{dt}
The force needed to accelerate a mass is the mass times the rate of change of velocity, that is, mass times acceleration.
When Time Varies
w = f × y
For a constant force, the work done equals the force times the distance moved in its own direction.
When Time Varies
v = \dot{x}
In fluxional notation, velocity is the dot over the distance symbol, meaning the derivative of distance with respect to time.
When Time Varies
a = \dot{v} = \ddot{x}
In fluxional notation, acceleration is the dot over velocity, equal to the double dot over distance, with time as the independent variable.
When Time Varies
f = m\dot{v} = m\ddot{x}
Force equals mass times the rate of change of velocity, which is mass times the second time-derivative of distance.
When Time Varies
w = x × m \ddot{x}
In fluxional notation the work is written as distance times mass times the second time-derivative of distance.

Problems

Exercise V

Exercise V, problem 10a, p. 66
A body moves in such a way that the spaces described in the time $t$ from starting is given by $s = t^n$, where $n$ is a constant. Find the value of $n$ when the velocity is doubled from the $5$th to the $10$th second; find it also when the velocity is numerically equal to the acceleration at the end of the $10$th second.
Printed answer:
- $n = 2$, $n = 11$.
unverified: no computed check settled this one (yet)
How it was checked
- other: the check does not fit this problem 2
Exercise V, problem 10b, p. 66
A body moves in such a way that the spaces described in the time $t$ from starting is given by $s = t^n$, where $n$ is a constant. Find the value of $n$ when the velocity is doubled from the $5$th to the $10$th second; find it also when the velocity is numerically equal to the acceleration at the end of the $10$th second.
Printed answer:
- $n = 2$, $n = 11$.
unverified: no computed check settled this one (yet)
How it was checked
- other: the check does not fit this problem 11
Exercise V, problem 1a, p. 64
If $y = a + bt^2 + ct^4$; find $\dfrac{dy}{dt}$ and $\dfrac{d^2y}{dt^2}$. *Ans.* $\dfrac{dy}{dt} = 2bt + 4ct^3$; $\dfrac{d^2y}{dt^2} = 2b + 12ct^2$.
Printed answer:
- *Ans.* $\dfrac{dy}{dt} = 2bt + 4ct^3$; $\dfrac{d^2y}{dt^2} = 2b + 12ct^2$.
unverified: no computed check settled this one (yet)
How it was checked
- differentiate: the record may be misread 2*b*t + 4*c*t**3
Exercise V, problem 1b, p. 64
If $y = a + bt^2 + ct^4$; find $\dfrac{dy}{dt}$ and $\dfrac{d^2y}{dt^2}$. *Ans.* $\dfrac{dy}{dt} = 2bt + 4ct^3$; $\dfrac{d^2y}{dt^2} = 2b + 12ct^2$.
Printed answer:
- *Ans.* $\dfrac{dy}{dt} = 2bt + 4ct^3$; $\dfrac{d^2y}{dt^2} = 2b + 12ct^2$.
unverified: no computed check settled this one (yet)
How it was checked
- differentiate2: the record may be misread 2*b + 12*c*t**2
Exercise V, problem 2, p. 64
A body falling freely in space describes in $t$ seconds a space $s$, in feet, expressed by the equation $s = 16t^2$. Draw a curve showing the relation between $s$ and $t$. Also determine the velocity of the body at the following times from its being let drop: $t = 2$ seconds; $t = 4.6$ seconds; $t = 0.01$ second.
Printed answer:
- 64; 147.2; and 0.32 feet per second.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 32*t
- evaluate: passes 32*t
- evaluate: passes 32*t
- evaluate: passes 32*t
Exercise V, problem 3a, p. 64
If $x = at - \frac{1}{2}gt^2$; find $\dot{x}$ and $\ddot{x}$.
Printed answer:
- $x = a - gt$; $\ddot{x} = -g$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes a - g*t
Exercise V, problem 3b, p. 64
If $x = at - \frac{1}{2}gt^2$; find $\dot{x}$ and $\ddot{x}$.
Printed answer:
- $x = a - gt$; $\ddot{x} = -g$.
verified: the printed answer passed a computed check
How it was checked
- differentiate2: passes -g
Exercise V, problem 4, p. 64
If a body move according to the law s = 12 - 4.5t + 6.2t^2, find its velocity when $t = 4$ seconds; $s$ being in feet.
Printed answer:
- $45.1$ feet per second.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes -4.5 + 12.4*t
- evaluate: passes -4.5 + 12.4*t
Exercise V, problem 5, p. 65
Find the acceleration of the body mentioned in the preceding example. Is the acceleration the same for all values of $t$?
Printed answer:
- $12.4$ feet per second per second. Yes.
verified: the printed answer passed a computed check
How it was checked
- differentiate2: passes 12.4
- evaluate: passes 12.4
Exercise V, problem 6a, p. 65
The angle $\theta$ (in radians) turned through by a revolving wheel is connected with the time $t$ (in seconds) that has elapsed since starting; by the law = 2.1 - 3.2t + 4.8t^2. Find the angular velocity (in radians per second) of that wheel when $1\frac{1}{2}$ seconds have elapsed. Find also its angular acceleration.
Printed answer:
- Angular velocity ${} = 11.2$ radians per second; angular acceleration ${}= 9.6$ radians per second per second.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes -3.2 + 9.6*t
- evaluate: passes -3.2 + 9.6*t
Exercise V, problem 6b, p. 65
The angle $\theta$ (in radians) turned through by a revolving wheel is connected with the time $t$ (in seconds) that has elapsed since starting; by the law = 2.1 - 3.2t + 4.8t^2. Find the angular velocity (in radians per second) of that wheel when $1\frac{1}{2}$ seconds have elapsed. Find also its angular acceleration.
Printed answer:
- Angular velocity ${} = 11.2$ radians per second; angular acceleration ${}= 9.6$ radians per second per second.
verified: the printed answer passed a computed check
How it was checked
- differentiate2: passes 9.6
- evaluate: passes 9.6
Exercise V, problem 7a, p. 65
A slider moves so that, during the first part of its motion, its distance $s$ in inches from its starting point is given by the expression s = 6.8t^3 - 10.8t; $t$ being in seconds. Find the expression for the velocity and the acceleration at any time; and hence find the velocity and the acceleration after $3$ seconds.
Printed answer:
- $v = 20.4t^2 - 10.8$. $a = 40.8t$. $172.8$ in./sec., $122.4~\text{in./sec}^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 20.4*t**2 - 10.8
- evaluate: passes 20.4*t**2 - 10.8
Exercise V, problem 7b, p. 65
A slider moves so that, during the first part of its motion, its distance $s$ in inches from its starting point is given by the expression s = 6.8t^3 - 10.8t; $t$ being in seconds. Find the expression for the velocity and the acceleration at any time; and hence find the velocity and the acceleration after $3$ seconds.
Printed answer:
- $v = 20.4t^2 - 10.8$. $a = 40.8t$. $172.8$ in./sec., $122.4~\text{in./sec}^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate2: passes 40.8*t
- evaluate: passes 40.8*t
Exercise V, problem 8a, p. 65
The motion of a rising balloon is such that its height $h$, in miles, is given at any instant by the expression $h = 0.5 + \frac{1}{10}\sqrt[3]{t-125}$; $t$ being in seconds. Find an expression for the velocity and the acceleration at any time. Draw curves to show the variation of height, velocity and acceleration during the first ten minutes of the ascent.
Printed answer:
- $v = \dfrac{1}{30 \sqrt[3]{(t - 125)^2}}$, $a = - \dfrac{1}{45 \sqrt[3]{(t - 125)^5}}$.
unverified: no computed check settled this one (yet)
How it was checked
- differentiate: FLAG-DOMAIN 1/(30*((t - 125)**2)**Rational(1,3))
Exercise V, problem 8b, p. 65
The motion of a rising balloon is such that its height $h$, in miles, is given at any instant by the expression $h = 0.5 + \frac{1}{10}\sqrt[3]{t-125}$; $t$ being in seconds. Find an expression for the velocity and the acceleration at any time. Draw curves to show the variation of height, velocity and acceleration during the first ten minutes of the ascent.
Printed answer:
- $v = \dfrac{1}{30 \sqrt[3]{(t - 125)^2}}$, $a = - \dfrac{1}{45 \sqrt[3]{(t - 125)^5}}$.
unverified: no computed check settled this one (yet)
How it was checked
- differentiate2: FLAG-DOMAIN -1/(45*((t - 125)**5)**Rational(1,3))
Exercise V, problem 9a, p. 65
A stone is thrown downwards into water and its depth $p$ in metres at any instant $t$ seconds after reaching the surface of the water is given by the expression p = 44+t^2 + 0.8t - 1. 078.png66% Find an expression for the velocity and the acceleration at any time. Find the velocity and acceleration after $10$ seconds.
Printed answer:
- $v = 0.8 - \dfrac{8t}{(4 + t^2)^2}$, $a = \dfrac{24t^2 - 32}{(4 + t^2)^3}$, $0.7926$ and $0.00211$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 0.8 - 8*t/(4 + t**2)**2
- evaluate: passes 0.8 - 8*t/(4 + t**2)**2
Exercise V, problem 9b, p. 65
A stone is thrown downwards into water and its depth $p$ in metres at any instant $t$ seconds after reaching the surface of the water is given by the expression p = 44+t^2 + 0.8t - 1. 078.png66% Find an expression for the velocity and the acceleration at any time. Find the velocity and acceleration after $10$ seconds.
Printed answer:
- $v = 0.8 - \dfrac{8t}{(4 + t^2)^2}$, $a = \dfrac{24t^2 - 32}{(4 + t^2)^3}$, $0.7926$ and $0.00211$.
verified: the printed answer passed a computed check
How it was checked
- differentiate2: passes (24*t**2 - 32)/(4 + t**2)**3
- evaluate: passes (24*t**2 - 32)/(4 + t**2)**3