# Elements of the Differential and Integral Calculus/Chapter VI part 2

67. Angle between the radius vector drawn to a point on a curve and the tangent to the curve at that point. Let the equation of the curve in polar coördinates be $\rho =f(\theta )$.
Let P be any fixed point $(\rho ,\theta )$ on the curve. If $\theta$, which we assume as the independent variable, takes on an increment $\Delta \theta$, then $\rho$ will take on a corresponding increment $\Delta \rho$. Denote by Q the point $(\rho +\Delta \rho ,\theta +\Delta \theta )$. Draw PR perpendicular to OQ. Then $OQ=\rho +\Delta \rho ,PR=\rho \sin \Delta \theta$, and $OR=\rho \cos \Delta \theta$. Also,
    $\tan PQR={\frac {PR}{RQ}}={\frac {PR}{OQ-OR}}={\frac {\rho \sin \Delta \theta }{\rho \Delta \rho -\rho \cos \Delta \theta }}$.
Denote by $\psi$ the angle between the radius vector OP and the tangent PT. If we now let $\Delta \theta$ approach the limit zero, then
(a) the point Q will approach indefinitely near P;
(b) the secant PQ will approach the tangent PT as a limiting position; and
(c) the angle PQR will approach $\psi$ as a limit.
Hence
| $\tan \ \psi$ | $=\ \lim _{\Delta \theta \to 0}{\frac {\rho \Delta \theta }{\rho \Delta \rho 0\rho \cos \Delta \theta }}$ |
| | $=\ \lim _{\Delta \theta \to 0}{\frac {\rho \Delta \theta }{2\rho \sin ^{2}{\frac {\Delta \theta }{2}}+\Delta \rho }}$ |
| [Since from 39, p. 2, $\rho -\rho \cos \Delta \theta =\rho (1-\cos \Delta \theta )=2\rho \sin ^{2}{\frac {\Delta \theta }{2}}$.] ||
| | $=\ \lim _{\Delta \theta \to 0}{\frac {\frac {\rho \sin \Delta \theta }{\Delta \theta }}{{\frac {2\rho \sin ^{2}{\frac {\Delta \theta }{2}}}{\Delta \theta }}+{\frac {\Delta \rho }{\Delta \theta }}}}$ |
| [Dividing both numerator and denominator by $\Delta \theta$.] ||
| | $=\ \lim _{\Delta \to 0}{\frac {\rho \cdot {\frac {\sin \Delta \theta }{\Delta \theta }}}{\rho \sin {\frac {\Delta \theta }{2}}\cdot {\frac {\sin {\frac {\Delta \theta }{2}}}{\frac {\Delta \theta }{2}}}+{\frac {\Delta \rho }{\Delta \theta }}}}$. |
Since $\lim {\Delta \theta \to 0}\left({\frac {\Delta \rho }{\Delta \theta }}\right)={\frac {d\rho }{d\theta }}$ and $\lim _{\Delta \theta \to 0}\left(\sin {\frac {\Delta \theta }{2}}\right)=0$, also $\lim _{\Delta \theta \to 0}\left({\frac {\sin \Delta \theta }{\Delta \theta }}\right)=1$
and $\lim _{\Delta \theta \to 0}{\frac {\sin {\frac {\Delta \theta }{2}}}{\frac {\Delta \theta }{2}}}=1$ by § 22, p. 21, we have
(A) $\tan \ \psi ={\frac {\rho }{\frac {d\rho }{d\theta }}}$.
From the triangle OPT we get
(B) $\tau =\theta +\psi$.
Having found $\tau$, we may then find $\tan \tau$, the slope of the tangent to the curve at P. Or since, from (B),
    $\tan \tau =\tan(\theta +\psi )={\frac {\tan \theta +\tan \psi }{1-\tan \theta \tan \psi }}$
we may calculate $\tan \psi$ from (A) and substitute in the formula
(C) slope of tangent $=\tan \tau ={\frac {\tan \theta +\tan \psi }{1-\tan \theta \tan \psi }}$
ILLUSTRATIVE EXAMPLE 1. Find $\psi$ and $\tau$ in the cardioid $\psi =a(1-\cos \theta )$. Also find the slope at $\theta ={\frac {\pi }{6}}$.
Solution. ${\frac {d\psi }{d\theta }}=a\sin \theta$. Substituting in (A) gives
    $\tan \psi ={\frac {\rho }{\frac {d\rho }{d\theta }}}={\frac {a(1-\cos \theta )}{a\sin \theta }}={\frac {2a\sin ^{2}{\frac {\theta }{2}}}{2a\sin \theta {\theta }{2}\cos {\frac {\theta }{2}}}}=\tan {\frac {\theta }{2}}$. By 39, p. 2,and 37, p.2
Since $\tan \psi =\tan {\frac {\theta }{2}},\psi ={\frac {\theta }{2}}$ Ans. Substituting in (B), $\tau =\theta +{\frac {\theta }{2}}={\frac {3\theta }{2}}$. Ans.
$\tan \tau =\tan {\frac {\pi }{4}}=1$. Ans.
To find the angle of intersection $\phi$ of two curves C and C' whose equations are given in polar coöordinates, we may proceed as follows:
angle TPT ' = angle OPT ' - angle OPT,
| or, | $\phi$ | $=\psi '-\psi$. Hence |
| (D) | $\ \tan \phi$ | $={\frac {\tan \psi '-\tan \psi }{1+\tan \psi '\tan \psi }}$, |
where $\tan \psi '$ and $\tan \psi$ are calculated by (A) from the two curves and evaluated for the point of intersection.
ILLUSTRATIVE EXAMPLE 2. Find the angle of of intersection of the curves $\rho =a\sin 2\theta ,\rho =a\cos 2\theta$.
Solution. Solving the two equations simultaneously, we get at the point of intersection
    $\tan 2\theta =1$, $2\theta =45$°, $\theta =22{\frac {1}{2}}$°.
From the first curve, using (A),
    $\tan \psi '={\frac {1}{2}}\tan 2\theta ={\frac {1}{2}}$, for $\theta =22{\frac {1}{2}}$°.
From the second curve,
    $\tan \psi ;=-{\frac {1}{2}}\cot 2\theta =-{\frac {1}{2}},$ for $\theta =22{\frac {1}{2}}$°·
Substituting in (D),
    $\tan \psi ={\frac {{\frac {1}{2}}+{\frac {1}{2}}}{1-{\frac {1}{4}}}}={\frac {4}{3}}$. ∴ $\psi =\arctan {\frac {4}{3}}$. Ans.
68. Lengths of polar subtangent and polar subnormal. Draw a line NT through the origin perpendicular to the radius vector of the point P on the curve. If PT is the tangent and PN the normal to the curve at P, then
| | OT = length of polar sub tangent, |
| and | ON = length of polar subnormal |
of the curve at P.
In the triangle OPT, $\tan \psi ={\frac {OT}{\rho }}$. Therefore
(7) $OT=\rho \tan \psi =\rho ^{2}{\frac {d\theta }{d\rho }}$ = length of polar subtangent.
In the triangle $OPN,\tan \psi ={\frac {\rho }{ON}}$. Therefore
(8) $ON={\frac {\rho }{\tan \psi }}={\frac {d\rho }{d\theta }}$ = length of polar subnormal.
The length of the polar tangent (= PT) and the length of the polar normal (= PN) may be found from the figure, each being the hypot" enuse of a right triangle.
ILLUSTRATIVE EXAMPLE 3. Find lengths of polar subtangent and subnormal to the lemniscate $\rho ^{2}=a^{2}\cos 2\theta$.
Solution. Differentiating the equation of the curve as an implicit function with respect to $\theta$,
| or, | $2\rho {\frac {d\rho }{d\theta }}$ | $=\ 2a^{2}\sin 2\theta$, |
| | ${\frac {d\rho }{d\theta }}$ | $=\ -{\frac {a^{2}\sin 2\theta }{\rho }}$. |
Substituting in (7) and (8), we get
| length of polar subtangent | $=-{\frac {\rho ^{3}}{a^{2}\sin 2\theta }}$, |
| length of polar subnormal | $=-{\frac {a^{2}\sin 2\theta }{\rho }}$. |
If we wish to express the results in terms of $\theta$, find $\rho$ in terms of $\theta$ from the given equation and substitute. Thus, in the above, $\rho =\pm a{\sqrt {\cos 2\theta }}$; therefore length of polar subtangent $=\pm a\cot 2\theta {\sqrt {\cos 2\theta }}$.
1. In the circle $\rho =r\sin \theta$, find $\psi$ and $\tau$ in terms of $\theta$. Ans. $\psi =\theta ,\tau =2\theta$.
2. In the parabola $\rho =a\sec ^{\frac {\theta }{2}}$, show that $\tau +\psi =\pi$.
3. In the curve $\psi ^{2}=a^{2}\cos 2\theta$, show that $2\psi =\pi +4\theta$.
4. Show that $\psi$ is constant in the logarithmic spiral $\psi =e^{a\theta }$. Since the tangent makes a constant angle with the radius vector, this curve is also called the equiangular spiral.
5. Given the curve $\rho =asin^{3}{\frac {\theta }{3}}$, prove that $\tau =4\psi$.
6. Show that $\tan \psi =\theta$ in the spiral of Archimedes $\rho =a\theta$. Find values of $\psi$ when $\theta =2\pi$ and $4\pi$. Ans. $\psi$ = 80° 57' and 85° 27'.
7. Find the angle between the straight line $\rho \cos \theta =2a$ and the circle $\rho =5a\sin \theta$. Ans. $\arctan {\frac {3}{4}}$.
8. Show that the parabolas $\rho =asec^{2}{\frac {\theta }{2}}$ and $\rho =bcsc^{2}{\frac {\theta }{2}}$ intersect at right angles.
9. Find the angle of intersection of $\rho =a\sin \theta$ and $\rho =a\sin 2\theta$. Ans. At origin 0°; at two other points $\arctan 3{\sqrt {3}}$.
10. Find the slopes of the following curves at the points designated:
| (a) $\rho =a(l-\cos \theta )$ | $\theta ={\frac {\pi }{2}}$ | -1. |
| (b) $\rho =a\sec ^{2}\theta$. | $\rho =2a$. | 3. |
| (c) $\rho =a\sin 4\theta$. | origin. | $0,1,\inf ,-1$. |
| (d) $\rho ^{2}=a^{2}\sin 4\theta$. | origin. | $0,1,\inf ,-1$. |
| (e) $\rho =a\sin 3\theta$. | origin. | $0,{\sqrt {3}},-{\sqrt {3}}$. |
| (f) $\rho =a\cos 3\theta$. | origin. |
| (g) $\rho =a\cos 2\theta$. | origin. |
| (h) $\rho =a\sin 2\theta$. | $\theta ={\frac {\pi }{4}}$. |
| (i) $\rho =a\sin 3\theta$. | $\theta ={\frac {pi}{6}}$. |
| (j) $\rho =a\theta$. | $\theta ={\frac {\pi }{2}}$. |
| (k) $\rho \theta =a$. | $\theta ={\frac {\pi }{2}}$. |
| (l) $\rho =e^{\theta }$. | $\theta =0$. |
11. Prove that the spiral of Archimedes $\rho =a\theta$, and the reciprocal spiral $\rho ={\frac {a}{\theta }}$, intersect at right angles.
12. Find the angle between the parabola $\rho =asec^{2}{\frac {\theta }{2}}$ and the straight line $\rho \sin \theta =2a$. Ans. 45°.
13. Show that the two cardioids $\rho =a(1+\cos \theta )$ and $\rho =a(1-\cos \theta )$ cut each other perpendicularly.
14. Find lengths of subtangent, subnormal, tangent, and normal of the spiral of Archimedes $\rho =a\theta$.
| Ans. | subt. = ${\frac {\rho ^{2}}{a}}$, | tan. = ${\frac {\rho }{a}}{\sqrt {a^{2}+\rho ^{2}}}$, |
| | subn. = $a$, | nor. = ${\sqrt {a^{2}+\rho ^{2}}}$. |
The student should note the fact that the subnormal is constant.
15. Get lengths of subtangent, subnormal, tangent, and normal in the logarithmic spiral $\rho =a^{\theta }$.
| Ans. | subt. = ${\frac {\rho }{\log a}}$, | tan. = $\rho {\sqrt {1+{\frac {1}{\log ^{2}a}}}}$, |
| | subn. = $\rho \log a$, | nor. = $\rho {\sqrt {1+\log ^{2}a}}$. |
When $a=e$, we notice that subt. = subn., and tan. = nor.
16. Find the angles between the curves $\rho =a(1+\cos \theta ),\rho =b(1-\cos \theta )$.
Ans. 0 and ${\frac {\pi }{2}}$.
17. Show that the reciprocal spiral $\rho ={\frac {a}{\theta }}$ has a constant subtangent.
18. Show that the equilateral hyperbolas $\rho ^{2}\sin 2\theta =a^{2},\rho ^{2}\cos 2\theta =b^{2}$ intersect at right angles.
69. Solution of equations having multiple roots. Any root which occurs more than once in an equation is called a multiple root. Thus 3, 3, 3, -2 are the roots of
(A) $x^{4}-7x^{3}+9x^{2}+27x-54=0$;
hence 3 is a multiple root occurring three times.
Evidently (A) may also be written in the form
    $(x-3)^{3}(x+2)=0$.
Let $f(x)$ denote an integral rational function of $x$ having a multiple root $a$, and suppose it occurs $m$ times. Then we may write
(B) $f(x)=(x-a)^{m}\phi (x)$,
where $\phi (x)$ is the product of the factors corresponding to all the roots of $f(x)$ differing from $a$. Differentiating (B),
    $f'(x)=(x-a)^{\phi '(x)}+\phi (x)m(x-a)^{m-1}$,
or,
(C) $f'(x)=(x-a)^{m-1}[(x-a)\phi '(x)+\phi (x)m]$.
Therefore $f'(x)$ contains the factor $(x-a)$ repeated $m-1$ times and no more; that is, the highest common factor (H.C.F.) of $f(x)$ and $f'(x)$ has $m-1$ roots equal to $a$.
In case $f(x)$ has a second multiple root $\beta$ occurring $r$ times, it is evident that the H.C.F. would also contain the factor $(x-\beta )^{r-1}$ and so on for any number of different multiple roots, each occurring once more in $f(x)$ than in the H.C.F.
We may then state a 'rule for finding the multiple roots of an equation $f(x)=0$ as follows:
FIRST STEP. Find $f'(x)$.
SECOND STEP. Find the H.C.F. of $f(x)$ and $f'(x)$.
THIRD STEP. Find the roots of the H.C.F. Each different root of the H.C.F. will occur once more in $f(x)$ than it does in the H.C.F.
If it turns out that the H.C.F. does not involve $x$, then $f(x)$ has no multiple roots and the above process is of no assistance in the solution of the equation, but it may be of interest to know that the equation has no equal, i.e. multiple, roots.
ILLUSTRATIVE EXAMPLE 1. Solve the equation $x^{3}-8x^{2}+13x-6=0$.
| Solution. Place | $f(x)$ | $=x^{3}-8x^{2}+13x-6$. |
| First step. | $f'(x)$ | $=3x^{2}-16x+13$. |
| Second step. | H.C.F. | $=x-1$. |
| Third step. | $x-1$ | $=0$ ∴ $x=1$. |
Since 1 occurs once as a root in the H.C.F., it will occur twice in the given equation; that is, $(x-1)^{2}$ will occur there as a factor. Dividing $x^{3}-8x^{2}+13x-6$ by $(x-1)^{2}$ gives the only remaining factor $(x-6)$, yielding the root 6. The roots of our equation are then 1, 1, 6. Drawing the graph of the function, we see that at the double root $x=1$ the graph touches OX but does not cross it.
Solve the first ten equations by the method of this section:
| 1. | $x^{3}-7x^{2}+16x-12=0$. | 2, 2, 3. |
| 2. | $x^{4}-6x^{2}-8x-3=0$. |
| 3. | $x^{4}-7x^{3}+9x^{2}+27x-64=0$. | 3, 3, 3, - 2. |
| 4. | $x^{4}-5x^{3}-9x^{2}+81x-108=0$. | 3, 3, 3, -4. |
| 5. | $x^{4}+6x^{3}+x^{2}-24x+16=0$. | 1, 1, -4, -4. |
| 6. | $x^{4}-9x^{3}+23x^{2}-3x-36=0$. | 3, 3, -1, 4. |
| 7. | $x^{4}-6x^{3}+10x^{2}-8=0$. | 2, 2, $1\pm {\sqrt {3}}$. |
| 8. | $x^{5}-x^{4}-5x^{3}+x^{2}+8x+4=0$. |
| 9. | $x^{5}-15x^{3}+10x^{2}+60x-72=0$. | 2, 2, 2, -3, -3. |
| 10. | $x^{5}-3x^{4}-5x^{3}+13x^{2}+24x+l0=0$. |
Show that the following four equations have no multiple (equal) roots:
11. $x^{3}+9x^{2}+2x-48=0$.
12. $x^{4}-15x^{2}-10x+24=0$.
13. $x^{4}-3x^{3}-6x^{2}+14x+12=0$.
14. $x^{n}-a^{n}=0$.
15. Show that the condition that the equation
    $x^{3}+3qx+r=0$
shall have a double root is $4q^{3}+r^{2}=0$.
16. Show that the condition that the equation
    $x^{3}+3px^{2}+r=0$
shall have a double root is $r(4p^{3}+r)=0$.
70. Applications of the derivative in mechanics. Velocity. Rectilinear motion. Consider the motion of a point P on the straight line AB.
Let s be the distance measured from some fixed point as A to any position of P, and let t be the corresponding elapsed time. To each value of t corresponds a position of P and therefore a distance (or space) s. Hence s will be a function of t, and we may write
$s=f(t)$
Now let t take on an increment $\Delta t$; then s takes on an increment $\Delta s$, and
(A) ${\frac {\Delta s}{\Delta t}}$ = the average velocity
of P during the time interval $\Delta t$. If P moves with uniform motion, the above ratio will have the same value for every interval of time and is the velocity at any instant.
For the general case of any kind of motion, uniform or not, we define the velocity (time rate of change of s) at any instant as the limit of the ratio ${\frac {\Delta s}{\Delta t}}$ as $\Delta t$ approaches the limit zero; that is, At
| | $v=\lim _{\Delta t\to 0}{\frac {\Delta s}{\Delta t}}$, |
| or |
| (9) | $v={\frac {ds}{dt}}$ |
The velocity is the derivative of the distance (= space) with respect to the time.
To show that this agrees with the conception we already have of velocity, let us find the velocity of a falling body at the end of two seconds.
By experiment it has been found that a body falling freely from rest in a vacuum near the earth's surface follows approximately the law
(B) $s=16.1t^{2}$
where s = space fallen in feet, t = time in seconds. Apply the General Rule, p. 29 [§31], to (B).
| FIRST STEP. | $s+\Delta s$ | $=16.1(t+\Delta t)^{2}=16.1t^{2}+32.2t\cdot \Delta t+16.1(\Delta t)^{2}$. | $=16.1(t+\Delta t)^{2}=16.1t^{2}+32.2t\cdot \Delta t+16.1(\Delta t)^{2}$. |
| SECOND STEP. | $\Delta s$ | $=32.2t\cdot \Delta t+16.1(\Delta t)^{2}$. | $=32.2t\cdot \Delta t+16.1(\Delta t)^{2}$. |
| THIRD STEP. | ${\frac {\Delta s}{\Delta t}}$ | $=32.2t+16.1\Delta t$ | = average velocity throughout the time interval $\Delta t$. |
| Placing t = 2, |||
| (C) | ${\frac {\Delta s}{\Delta t}}$ | $=64.4+16.1\Delta t$ | = average velocity throughout the time interval $\Delta t$ after two seconds of falling. |
Our notion of velocity tells us at once that (C) does not give us the actual velocity at the end of two seconds; for even if we take $\Delta t$ very small, say ${\frac {1}{100}}$ or ${\frac {1}{1000}}$ of a second, (C) still gives only the average velocity during the corresponding small interval of time. But what we do mean by the velocity at the end of two seconds is the limit of the average velocity when $\Delta t$ diminishes towards zero; that is, the velocity at the end of two seconds is from (C), 64.4 ft. per second.
Thus even the everyday notion of velocity which we get from experience involves the idea of a limit, or in our notation
$v=\lim _{\Delta t\to 0}\left({\frac {\Delta s}{\Delta t}}\right)$
The above example illustrates well the notion of a limiting value. The student should be impressed with the idea that a limiting value is a definite, fixed value, not something that is only approximated. Observe that it does not make any difference how small 16.1 $\Delta t$ may be taken; it is only the limiting value of
$64.4+16.1\Delta t$
when $\Delta t$ diminishes towards zero, that is of importance, and that value is exactly 64.4.
71. Component velocities. Curvilinear motion. The coördinates x and y of a point P moving in the XY-plane are also functions of the time, and the motion may be defined by means of two equations,
$x=f(t),y=\psi (t)$
These are the parametric equations of the path (see § 66, p. 79).
The horizontal component $v_{x}$ of v is the velocity along OX of the projection M of P, and is therefore the time rate of change of x. Hence, from (9), p. 90 [§70], when s is replaced by x, we get
(10) $v_{x}={\frac {dx}{dt}}$.
In the same way we get the vertical component, or time rate of change of y,
(11) $v_{y}={\frac {dy}{dt}}$.
Representing the velocity and its components by vectors, we have at once from the figure
    $v^{2}={v_{x}}^{2}+{v_{y}}^{2}$,
or,
(12) $v={\frac {ds}{dt}}={\sqrt {\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}}}$,
giving the magnitude of the velocity at any instant.
If $\tau$ be the angle which the direction of the velocity makes with the axis of X; we have from the figure, using (9), (10), (11),
(13) $\sin \tau ={\frac {v_{y}}{v}}={\frac {\frac {dy}{dt}}{\frac {ds}{dt}}};\cos \tau ={\frac {v_{x}}{v}}={\frac {\frac {dx}{dt}}{\frac {ds}{ds}}};\tan \tau ={\frac {v_{y}}{v_{x}}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}$.
72. Acceleration. Rectilinear motion. In general, v will be a function of t, and we may write
    $v=\psi (t)$.
Now let t take on an increment $\Delta t$, then v takes on an increment $\Delta v$, and
${\frac {\Delta v}{\Delta t}}$ = the average acceleration of P during the time interval $\Delta t$.
We define the acceleration $\alpha$ at any instant as the limit of the ratio ${\frac {\Delta v}{\Delta t}}$ as $\Delta t$ approaches the limit zero; that is,
    $\alpha =\lim _{\Delta t\to 0}\left({\frac {\Delta v}{\Delta t}}\right)$,
or,
(14) $\alpha ={\frac {dv}{dt}}$
The acceleration is the derivative of the velocity with respect to the time.
73. Component accelerations. Curvilinear motion. In treatises on Mechanics it is shown that in curvilinear motion the acceleration is not, like the velocity, directed along the tangent, but toward the concave side, of the path of motion. It may be resolved into a tangential component, $\alpha _{t}$, and a normal component, $\alpha _{n}$ where
(14a) $\alpha _{t}={\frac {dv}{dt}};\alpha _{n}={\frac {v^{2}}{R}}$.
    (R is the radius of curvature. See §103.)
The acceleration may also be resolved into components parallel to the axes of the path of motion. Following the same plan used in § 71 for finding component velocities, we define the component accelerations parallel to OX and OY,
(15) $\alpha _{x}={\frac {dv_{x}}{dt}};\alpha _{y}={\frac {dv_{y}}{dt}}$. Also,
(16) $\alpha ={\sqrt {\left({\frac {dv_{x}}{dt}}\right)^{2}+\left({\frac {dv_{y}}{dt}}\right)^{2}}}$,
which gives the magnitude of the acceleration at any instant.
1. By experiment it has been found that a body falling freely from rest in a vacuum near the earth's surface follows approximately the law $s=16.1t^{2}$, where s = space (height) in feet, t = time in seconds. Find the velocity and acceleration (a) at any instant; (b) at end of the first second; (c) at end of the fifth second.
| Solution. | (A) $s=16.1t^{2}$. |
| (a) Differentiating, | (B) ${\frac {ds}{dt}}=32.2t$, or, from (9), $v=32.2t$ ft. per sec. |
| Differentiating again, | (C) ${\frac {dv}{dt}}=32.2$, or, from (14), $\alpha =32.2$ ft. per (sec.)2, |
which tells us that the acceleration of a falling body is constant; in other words, the velocity increases 32.2 ft. per sec. every second it keeps on falling.
(b) To find v and $\alpha$ at the end of the first second, substitute t = 1 in (B) and (C);
    $v=32.2$ ft. per sec., $a=32.2$ ft. per (sec.)$^2$.
(c) To find v and $\alpha$ at the end of the fifth second, substitute $t=5$ in (B) and (C);
    $v=161$ ft. per sec., $\alpha =32.2$ ft. per (sec.)$^2$.
2. Neglecting the resistance of the air, the equations of motion for a projectile are
    $x=v_{1}\cos \phi \cdot t,y=v_{1}\sin \phi \cdot t-16.1t^{2}$;
where $v_{1}$ = initial velocity, $\phi$ = angle of projection with horizon, t = time of flight in seconds, x and y being measured in feet. Find the velocity, acceleration, component velocities, and component accelerations (a) at any instant; (b) at the end of the first second, having given $v_{1}=100$ ft. per sec., $\phi$ = 30°; (c) find direction of motion at the end of the first second.
Solution. From (10) and (11),
| (a) | $v_{x}=v_{1}\cos \phi ;\ v_{y}=v_{1}\sin \phi -32.2t$. | $v_{x}=v_{1}\cos \phi ;\ v_{y}=v_{1}\sin \phi -32.2t$. |
| Also, from (12), | $v={\sqrt {{v_{1}}^{2}-64.4tv_{1}\sin \phi +1036.8t^{2}}}$. | $v={\sqrt {{v_{1}}^{2}-64.4tv_{1}\sin \phi +1036.8t^{2}}}$. |
| From (15) and (16), | $\alpha _{x}=0;\alpha _{y}=-32.2;\alpha =-32.2$. | $\alpha _{x}=0;\alpha _{y}=-32.2;\alpha =-32.2$. |
| (b) Substituting | $t=1,v_{1}=100,\phi =30$° in these results, we get | $t=1,v_{1}=100,\phi =30$° in these results, we get |
| | $v_{x}=86.6$ ft. per sec. | $\alpha _{x}=0$. |
| | $v_{y}=17.8$ ft. per sec. | $\alpha _{y}=-32.2$ ft. per (sec.)$^2$. |
| | $v=88.4$ ft. per sec. | $\alpha =-32.2$ ft. per (sec.)$^2$. |
| (c) | $\tau =\arctan {\frac {v_{y}}{v_{x}}}=\arctan {\frac {17.8}{86.6}}$ = 11° 36'.6 = angle of direction of motion with the horizontal. | $\tau =\arctan {\frac {v_{y}}{v_{x}}}=\arctan {\frac {17.8}{86.6}}$ = 11° 36'.6 = angle of direction of motion with the horizontal. |
3. Given the following equations of rectilinear motion. Find the distance, velocity, and acceleration at the instant indicated:
| (a) $s=t^{3}+2t^{2};t=2$. | Ans. | $s=16,v=20,\alpha =16$. |
| (b) $s=t^{2}+2t;t=3$. | (b) $s=t^{2}+2t;t=3$. | $s=15,v=8,\alpha =2$. |
| (c) $s=3-4t;t=4$. | (c) $s=3-4t;t=4$. | $s=-13,v=-4,\alpha =0$. |
| (d) $x=2t-t^{2};t=1$. | (d) $x=2t-t^{2};t=1$. | $x=1,v=0,\alpha =-2$. |
| (e) $y=2t-t^{3};t=0$. | (e) $y=2t-t^{3};t=0$. | $y=0,v=2,\alpha =0$. |
| (f) $h=20t+16t^{2};t=10$. | (f) $h=20t+16t^{2};t=10$. | $h=1800,v=34-,\alpha =32$. |
| (g) $s=2\sin t;t={\frac {\pi }{4}}$. | (g) $s=2\sin t;t={\frac {\pi }{4}}$. | $s={\sqrt {2}},v={\sqrt {2}},\alpha -{\sqrt {2}}$. |
| (h) $y=a\cos {\frac {\pi t}{3}};t=1$. | (h) $y=a\cos {\frac {\pi t}{3}};t=1$. | $y={\frac {a}{2}},v=-{\frac {\pi a{\sqrt {3}}}{6}},\alpha =-{\frac {\pi ^{2}a}{18}}$. |
| (i) $s=2e^{3t};t=0$. | (i) $s=2e^{3t};t=0$. | $s=2,v=6,\alpha =18$. |
| (j) $s=2t^{2}-3t;t=2$. |||
| (k) $x=4+t^{3};t=3$. |||
| (l) $y=5\cos 2t;t={\frac {\pi }{6}}$. |||
| (m) $s=b\sin {\frac {\pi t}{4}};t=2$. |||
| (n) $x=ae^{-2t};t=1$. |||
| (o) $s={\frac {a}{t}}+bt^{2};t=t_{0}$. |||
| (p) $s=10\log {\frac {4}{4+t}};t=1$. |||
4. If a projectile be given an initial velocity of 200 ft, per sec. in a direction inclined 45° with the horizontal, find
(a) the velocity and direction of motion at the end of the third and sixth seconds;
(b) the component velocities at the same instants.
Conditions are the same as for Ex. 2.
| (a) | When $t=3$, | $v=148.3$ ft. per sec., | $\tau$ = 17° 35', |
| | when $t=6$, | $v=150.5$ ft. per sec., | $\tau$ = 159° 53'; |
| (b) | when $t=3$, | $v_{x}=141.4$ ft. per sec., | $v_{y}=44.8$ ft. per sec. |
| | when $t=6$, | $v_{x}=141.4$ ft. per sec., | $v_{y}=-51.8$ ft. per sec. |
5. The height (= s) in feet reached in t seconds by a body projected vertically upwards with a velocity of $v_{1}$ ft. per sec. is given by the formula
    $s=v_{1}t-16.1t^{2}$.
Find (a) velocity and acceleration at any instant; and, if $v_{1}$ = 300 ft. per sec., find velocity and acceleration (b) at end of 2 seconds; (c) at end of 15 seconds. Resistance of air is neglected.
| (a) | $v=v_{1}-32.2t,\alpha =-32.2$; |
| (b) | $v=235.6$ ft. per sec. Upwards, |
| | $\alpha =32.2$ ft. per (sec.)² downwards; |
| (c) $v=183$ ft. per sec. Downwards, |
| | $\alpha =32.2$ ft. per (sec.)² downwards. |
6.A cannon ball is fired vertically upwards with a muzzle velocity of 644 ft. per sec. Find (a) its velocity at the end of 10 seconds; (b) for how long it will continue to rise. Conditions same as for Ex. 5.
| Ans. | (a) 322 ft. per sec. Upwards; |
| | (b) 20 seconds. |
7. A train left a station and in t hours was at a distance (space) of
    $s=t^{3}+2t^{2}+3t$
miles from the starting point. Find its acceleration (a) at the end of t hours; (b) at the end of 2 hours.
| Ans. | (a) $\alpha =6t+4$; |
| | (b) $\alpha =16$ miles per (hour)². |
8. In t hours a train had reached a point at the distance of ${\frac {1}{4}}t^{4}-4t^{3}+16t^{2}$ miles from the starting point. (a) Find its velocity and acceleration. (b) When will the train stop to change the direction of its motion? (c) Describe the motion during the first 10 hours.
| Ans. | (a) $v=t^{3}-12t^{2}+32t,a=3t^{2}-24t+32$; |
| | (b) at end of fourth and eighth hours; |
| | (c) forward first 4 hours, backward the next 4 hours, forward again after 8 hours. |
9. The space in feet described in t seconds by a point is expressed by the formula
    $s=48t-16t^{2}$.
Find the velocity and acceleration at the end of $1{\frac {1}{2}}$ seconds.
Ans. $v=0,\alpha =-32$ ft. per (sec.)².
10. Find the acceleration, having given
| (a) $v=t^{2}+2t;t=3$. | $\alpha =8$. |
| (b) $v=3t-t^{3};t=2$. | $\alpha =-9$ |
| (c) $v=4\sin {\frac {t}{2}};t={\frac {\pi }{3}}$. | $\alpha ={\sqrt {3}}$. |
| (d) $v=a\cos 3t;t={\frac {\pi }{6}}$. | $\alpha =-3a$. |
| (e) $v=5e^{2t};t=1$. | $\alpha =10e^{2}$. |
11. At the end of t seconds a body has a velocity of $3t^{2}+2t$ ft. per sec.; find its acceleration (a) in general; (b) at the end of 4 seconds.
Ans. (a) $\alpha =6t+2$ ft. per (sec.)²; (b) $a=26$ ft. per (sec.)²
12. The vertical component of velocity of a point at the end of t seconds is
    $v_{y}=3t^{2}-2t+6$ ft. per sec.
Find the vertical component of acceleration (a) at any instant; (b) at the end of 2 seconds.
Ans. (a) $\alpha _{y}=6t-2$; (b) 10 ft. per (sec.)²
13. If a point moves in a fixed path so that
    $s={\sqrt {t}}$,
show that the acceleration is negative and proportional to the cube of the velocity.
14. If the space described is given by
    $s=ae^{t}+be^{-t}$,
show that the acceleration is always equal in magnitude to the space passed over.
15. If a point referred to rectangular coördinates moves so that
    $x=a\cos t+b$, and $y=a\sin t+c$,
show that its velocity has a constant magnitude.
16. If the path of a moving point is the sine curve
    ${\begin{cases}x=at,\\y=b\sin at\end{cases}}$
show (a) that the x-component of the velocity is constant; (b) that the acceleration of the point at any instant is proportional to its distance from the axis of X.
17. Given the following equations of curvilinear motion, find at the given instant $v_{x},v_{y},v;\alpha _{x},\alpha _{y},\alpha$; position of point (coördinates); direction of motion. Also find the equation of the path in rectangular coördinates.
| (a) $x=t^{2},y=t;t=2$. | (g) $x=2\sin t,y=3\cos t;t=\pi$. |
| (b) $x=t,y=t^{3};t=1$. | (h) $x=\sin t,y=\cos 2t;t={\frac {\pi }{4}}$. |
| (c) $x=t^{2},y=t^{3};t=3$. | (i) $x=2t,y=3e^{t};t=0$. |
| (d) $x=2t,y=t^{2}+3;t=0$. |
| (e) $x=1-t^{2},y=2t;t=2$. | (j) $x=3t,y=\log t;t=1$. |
| (f) $x=a\sin t,y=a\cos t;t={\frac {3\pi }{4}}$. | (k) $x=t,y=12t^{-1};t=3$. |
* ↑ 'When $\theta$ increases with $\rho ,{\frac {d\theta }{d\rho }}$ is positive and $\rho$ is an acute angle, as in the above figure. Then the subtangent OT is positive and is measured to the right of an observer placed at O and looking along OP. When ${\frac {d\theta }{d\rho }}$ is negative, the subtangent is negative and is measured to the left of the observer.
* ↑ Even number of multiple roots.Since the first derivative vanishes for every multiple root, it follows that the axis of X is tangent to the graph at all points corresponding to multiple roots. If a multiple root occurs an even number of times, the graph will not cross the axis of X at such a point (see figure); if it occurs an odd number of times, the graph will cross.
* ↑ $\Delta s$ being the space or distance passed over in the time $\Delta t$.
* ↑ The equation of the path in rectangular coordinates may be found by eliminating t between their equations.
* ↑ The direction of v is along the tangent to the path.