Next Stage. What to do with Constants

Excerpts

Next Stage. What to do with Constants
So if we take the letter $a$, or $b$, or $c$ to represent any constant, it will simply disappear when we differentiate.
Next Stage. What to do with Constants
If we had begun with $y = ax^n$, we should have had $\dfrac{dy}{dx} = a×nx^{n-1}$.
Next Stage. What to do with Constants
And, what is true about multiplication is equally true about *division*: for if, in the example above, we had taken as the constant $\frac{1}{7}$ instead of $7$, we should have had the same $\frac{1}{7}$ come out in the result after differentiation.
Next Stage. What to do with Constants
We usually think of $x$ as a quantity that we can vary; and, regarding the variation of $x$ as a sort of *cause*, we consider the resulting variation of $y$ as an *effect*. In other words, we regard the value of $y$ as depending on that of $x$.
Next Stage. What to do with Constants
So the $5$ has quite disappeared. It added nothing to the growth of $x$, and does not enter into the differential coefficient.
Next Stage. What to do with Constants
So that any mere multiplication by a constant reappears as a mere multiplication when the thing is differentiated. And, what is true about multiplication is equally true about *division*: for if, in the example above, we had taken as the constant $\frac{1}{7}$ instead of $7$, we should have had the same $\frac{1}{7}$ come out in the result after differentiation.
Next Stage. What to do with Constants
As a rule an expression of this kind will need a little more knowledge than we have acquired so far; it is, however, always worth while to try whether the expression can be put in a simpler form.
Next Stage. What to do with Constants
If $r = 5.5$ in. and $h=20$ in. this becomes $690.8$. It means that a change of radius of $1$ inch will cause a change of volume of $690.8$ cub. inch. This can be easily verified, for the volumes with $r = 5$ and $r = 6$ are $1570$ cub. in. and $2260.8$ cub. in. respectively, and $2260.8 - 1570 = 690.8$.
Next Stage. What to do with Constants
The sensitiveness is approximately doubled from $800°$ to $1000°$, and becomes three-quarters as great again up to $1200°$.

Equations

Next Stage. What to do with Constants
\frac{dy}{dx} = 3x^2.
The derivative of x^3 + 5 with respect to x is 3x^2, so the added constant 5 drops out.
Next Stage. What to do with Constants
\frac{dy}{dx} = 14x.
The derivative of 7x^2 with respect to x is 14x, so the constant factor 7 is carried through.
Next Stage. What to do with Constants
\frac{dy}{dx} = a × 2x.
For y = ax^2, the derivative is a times 2x, so a constant multiplier reappears unchanged in the derivative.
Next Stage. What to do with Constants
\dfrac{dy}{dx} = a×nx^{n-1}
For y = ax^n, the derivative is a n x^(n-1): a constant multiplier passes through and the power drops by one.
Next Stage. What to do with Constants
\frac{dy}{dx} = \frac{5}{7} x^4.
The derivative of x^5/7 - 3/5 is 5x^4/7; the constant 3/5 vanishes and the 1/7 stays as a factor.
Next Stage. What to do with Constants
\frac{dy}{dx} = \frac{a}{2\sqrt{x}}.
The derivative of a times the square root of x, minus the constant term (1/2)sqrt(a), is a divided by 2 sqrt(x).
Next Stage. What to do with Constants
\frac{dy}{dx} = \sqrt{\frac{a+b}{a-b}}.
After squaring and simplifying the given relation, y is proportional to x and the derivative of y with respect to x is the square root of (a+b)/(a-b).
Next Stage. What to do with Constants
V = \pi r^2 h
The volume of a cylinder equals pi times the square of its radius times its height.
Next Stage. What to do with Constants
\frac{dV}{dr} = 2 \pi r h.
The rate of change of the cylinder's volume with its radius is 2 pi r h.
Next Stage. What to do with Constants
\dfrac{dV}{dr} = 2\pi r^2 = 400
When r = h, the rate of change of volume with radius equals 2 pi r^2, and this is set equal to 400 cubic inches per inch.
Next Stage. What to do with Constants
r = h = \sqrt{\dfrac{400}{2\pi}} = 7.98~\text{in}.
Solving 2 pi r^2 = 400 with r = h gives a radius and height of about 7.98 inches.
Next Stage. What to do with Constants
\dfrac{\theta}{\theta_1} = \left(\dfrac{t}{t_1}\right)^4
The pyrometer reading is proportional to the fourth power of the Centigrade temperature, relative to a reading taken at a known temperature.
Next Stage. What to do with Constants
\dfrac{d\theta}{dt} = \dfrac{100t^3}{1000^4} = \dfrac{t^3}{10,000,000,000}.
With the reading fixed at 25 at 1000 degrees C, the sensitiveness d(theta)/dt of the pyrometer equals t^3 divided by 10^10.

Problems

Exercise II

Exercise II, problem 1, p. 33
$y = ax^3 + 6$.
Printed answer:
- $\dfrac{dy}{dx} = 3ax^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 3*a*x**2
Exercise II, problem 10, p. 34
The greatest external pressure $P$ which a tube can support without collapsing is given by P = (2E1-^2) t^3D^3, where $E$ and $\sigma$ are constants, $t$ is the thickness of the tube and $D$ is its diameter. (This formula assumes that $4t$ is small compared to $D$.) Compare the rate at which $P$ varies for a small change of thickness and for a small change of diameter taking place separately.
Printed answer:
- $\dfrac{\text{Rate of change of~$P$ when $t$~varies}} {\text{Rate of change of~$P$ when $D$~varies}} = - \dfrac{D}{t}$.
verified: the printed answer passed a computed check
How it was checked
- partial_ratio: passes -D/t
Exercise II, problem 11a, p. 34
Find, from first principles, the rate at which the following vary with respect to a change in radius: SubProbs [(*a*)] the circumference of a circle of radius $r$; [(*b*)] the area of a circle of radius $r$; [(*c*)] the lateral area of a cone of slant dimension $l$; [(*d*)] the volume of a cone of radius $r$ and height $h$; [(*e*)] the area of a sphere of radius $r$; [(*f*)] the volume of a sphere of radius $r$. SubProbs
Printed answer:
- $2\pi$, $2\pi r$, $\pi l$, $\frac{2}{3}\pi rh$, $8\pi r$, $4\pi r^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes, with the problem read into an equation 2*pi
Exercise II, problem 11b, p. 34
Find, from first principles, the rate at which the following vary with respect to a change in radius: SubProbs [(*a*)] the circumference of a circle of radius $r$; [(*b*)] the area of a circle of radius $r$; [(*c*)] the lateral area of a cone of slant dimension $l$; [(*d*)] the volume of a cone of radius $r$ and height $h$; [(*e*)] the area of a sphere of radius $r$; [(*f*)] the volume of a sphere of radius $r$. SubProbs
Printed answer:
- $2\pi$, $2\pi r$, $\pi l$, $\frac{2}{3}\pi rh$, $8\pi r$, $4\pi r^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes, with the problem read into an equation 2*pi*r
Exercise II, problem 11c, p. 34
Find, from first principles, the rate at which the following vary with respect to a change in radius: SubProbs [(*a*)] the circumference of a circle of radius $r$; [(*b*)] the area of a circle of radius $r$; [(*c*)] the lateral area of a cone of slant dimension $l$; [(*d*)] the volume of a cone of radius $r$ and height $h$; [(*e*)] the area of a sphere of radius $r$; [(*f*)] the volume of a sphere of radius $r$. SubProbs
Printed answer:
- $2\pi$, $2\pi r$, $\pi l$, $\frac{2}{3}\pi rh$, $8\pi r$, $4\pi r^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes, with the problem read into an equation pi*l
Exercise II, problem 11d, p. 34
Find, from first principles, the rate at which the following vary with respect to a change in radius: SubProbs [(*a*)] the circumference of a circle of radius $r$; [(*b*)] the area of a circle of radius $r$; [(*c*)] the lateral area of a cone of slant dimension $l$; [(*d*)] the volume of a cone of radius $r$ and height $h$; [(*e*)] the area of a sphere of radius $r$; [(*f*)] the volume of a sphere of radius $r$. SubProbs
Printed answer:
- $2\pi$, $2\pi r$, $\pi l$, $\frac{2}{3}\pi rh$, $8\pi r$, $4\pi r^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes, with the problem read into an equation Rational(2,3)*pi*r*h
Exercise II, problem 11e, p. 34
Find, from first principles, the rate at which the following vary with respect to a change in radius: SubProbs [(*a*)] the circumference of a circle of radius $r$; [(*b*)] the area of a circle of radius $r$; [(*c*)] the lateral area of a cone of slant dimension $l$; [(*d*)] the volume of a cone of radius $r$ and height $h$; [(*e*)] the area of a sphere of radius $r$; [(*f*)] the volume of a sphere of radius $r$. SubProbs
Printed answer:
- $2\pi$, $2\pi r$, $\pi l$, $\frac{2}{3}\pi rh$, $8\pi r$, $4\pi r^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes, with the problem read into an equation 8*pi*r
Exercise II, problem 11f, p. 34
Find, from first principles, the rate at which the following vary with respect to a change in radius: SubProbs [(*a*)] the circumference of a circle of radius $r$; [(*b*)] the area of a circle of radius $r$; [(*c*)] the lateral area of a cone of slant dimension $l$; [(*d*)] the volume of a cone of radius $r$ and height $h$; [(*e*)] the area of a sphere of radius $r$; [(*f*)] the volume of a sphere of radius $r$. SubProbs
Printed answer:
- $2\pi$, $2\pi r$, $\pi l$, $\frac{2}{3}\pi rh$, $8\pi r$, $4\pi r^2$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes, with the problem read into an equation 4*pi*r**2
Exercise II, problem 12, p. 34
The length $L$ of an iron rod at the temperature $T$ being given by $L = l_t\bigl[1 + 0.000012(T-t)\bigr]$, where $l_t$ is the length at the temperature $t$, find the rate of variation of the diameter $D$ of an iron tyre suitable for being shrunk on a wheel, when the temperature $T$ varies.
Printed answer:
- (12) $\dfrac{dD}{dT} = \dfrac{0.000012l_t}{\pi}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes, with the problem read into an equation 0.000012*l_t/pi
Exercise II, problem 2, p. 33
$y = 13x^{\efrac{3}{2}} - c$.
Printed answer:
- $\dfrac{dy}{dx} = 13 × \frac{3}{2}x^{\efrac{1}{2}}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 13*Rational(3,2)*x**Rational(1,2)
Exercise II, problem 3, p. 33
$y = 12x^{\efrac{1}{2}} + c^{\efrac{1}{2}}$.
Printed answer:
- $\dfrac{dy}{dx} = 6x^{-\efrac{1}{2}}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 6*x**Rational(-1,2)
Exercise II, problem 4, p. 33
$y = c^{\efrac{1}{2}} x^{\efrac{1}{2}}$.
Printed answer:
- $\dfrac{dy}{dx} = \dfrac{1}{2}c^{\efrac{1}{2}} x^{-\efrac{1}{2}}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes Rational(1,2)*c**Rational(1,2)*x**Rational(-1,2)
Exercise II, problem 5, p. 33
$u = \dfrac{az^n - 1}{c}$.
Printed answer:
- $\dfrac{du}{dz} = \dfrac{an}{c} z^{n-1}$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes a*n/c*z**(n-1)
Exercise II, problem 6, p. 33
$y = 1.18t^2 + 22.4$.
Printed answer:
- $\dfrac{dy}{dt} = 2.36t$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 2.36*t
Exercise II, problem 7, p. 33
If $l_t$ and $l_0$ be the lengths of a rod of iron at the temperatures $t°$C. and $0°$C. respectively, then $l_t = l_0(1 + 0.000012t)$. Find the change of length of the rod per degree Centigrade.
Printed answer:
- $\dfrac{dl_t}{dt} = 0.000012×l_0$.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 0.000012*l_0
Exercise II, problem 8, p. 33
It has been found that if $c$ be the candle power of an incandescent electric lamp, and $V$ be the voltage, $c = aV^b$, where $a$ and $b$ are constants. Find the rate of change of the candle power with the voltage, and calculate the change of candle power per volt at $80$, $100$ and $120$ volts in the case of a lamp for which $a = 0.5×10^{-10}$ and $b=6$.
Printed answer:
- $\dfrac{dC}{dV} = abV^{b-1}$, $0.98$, $3.00$ and $7.47$ candle power per volt respectively.
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes a*b*V**(b-1)
- evaluate: passes a*b*V**(b-1)
- evaluate: passes a*b*V**(b-1)
- evaluate: PASS-LOOSE a*b*V**(b-1)
Exercise II, problem 91, p. 33
The frequency $n$ of vibration of a string of diameter $D$, length $L$ and specific gravity $\sigma$, stretched with a force $T$, is given by n = 1DL gT. Find the rate of change of the frequency when $D$, $L$, $\sigma$ and $T$ are varied singly.
Printed answer:
- $\begin{aligned}[t] \dfrac{dn}{dD} &= -\dfrac{1}{LD^2} \sqrt{\dfrac{gT}{\pi \sigma}}, & \dfrac{dn}{dL} &= -\dfrac{1}{DL^2} \sqrt{\dfrac{gT}{\pi \sigma}}, \\ % \dfrac{dn}{d \sigma} &= -\dfrac{1}{2DL} \sqrt{\dfrac{gT}{\pi \sigma^3}}, & \dfrac{dn}{dT} &= \dfrac{1}{2DL} \sqrt{\dfrac{g}{\pi \sigma T}}. \end{aligned}$
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes -1/(L*D**2)*sqrt(g*T/(pi*sigma))
Exercise II, problem 92, p. 33
The frequency $n$ of vibration of a string of diameter $D$, length $L$ and specific gravity $\sigma$, stretched with a force $T$, is given by n = 1DL gT. Find the rate of change of the frequency when $D$, $L$, $\sigma$ and $T$ are varied singly.
Printed answer:
- $\begin{aligned}[t] \dfrac{dn}{dD} &= -\dfrac{1}{LD^2} \sqrt{\dfrac{gT}{\pi \sigma}}, & \dfrac{dn}{dL} &= -\dfrac{1}{DL^2} \sqrt{\dfrac{gT}{\pi \sigma}}, \\ % \dfrac{dn}{d \sigma} &= -\dfrac{1}{2DL} \sqrt{\dfrac{gT}{\pi \sigma^3}}, & \dfrac{dn}{dT} &= \dfrac{1}{2DL} \sqrt{\dfrac{g}{\pi \sigma T}}. \end{aligned}$
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes -1/(D*L**2)*sqrt(g*T/(pi*sigma))
Exercise II, problem 93, p. 33
The frequency $n$ of vibration of a string of diameter $D$, length $L$ and specific gravity $\sigma$, stretched with a force $T$, is given by n = 1DL gT. Find the rate of change of the frequency when $D$, $L$, $\sigma$ and $T$ are varied singly.
Printed answer:
- $\begin{aligned}[t] \dfrac{dn}{dD} &= -\dfrac{1}{LD^2} \sqrt{\dfrac{gT}{\pi \sigma}}, & \dfrac{dn}{dL} &= -\dfrac{1}{DL^2} \sqrt{\dfrac{gT}{\pi \sigma}}, \\ % \dfrac{dn}{d \sigma} &= -\dfrac{1}{2DL} \sqrt{\dfrac{gT}{\pi \sigma^3}}, & \dfrac{dn}{dT} &= \dfrac{1}{2DL} \sqrt{\dfrac{g}{\pi \sigma T}}. \end{aligned}$
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes -1/(2*D*L)*sqrt(g*T/(pi*sigma**3))
Exercise II, problem 94, p. 33
The frequency $n$ of vibration of a string of diameter $D$, length $L$ and specific gravity $\sigma$, stretched with a force $T$, is given by n = 1DL gT. Find the rate of change of the frequency when $D$, $L$, $\sigma$ and $T$ are varied singly.
Printed answer:
- $\begin{aligned}[t] \dfrac{dn}{dD} &= -\dfrac{1}{LD^2} \sqrt{\dfrac{gT}{\pi \sigma}}, & \dfrac{dn}{dL} &= -\dfrac{1}{DL^2} \sqrt{\dfrac{gT}{\pi \sigma}}, \\ % \dfrac{dn}{d \sigma} &= -\dfrac{1}{2DL} \sqrt{\dfrac{gT}{\pi \sigma^3}}, & \dfrac{dn}{dT} &= \dfrac{1}{2DL} \sqrt{\dfrac{g}{\pi \sigma T}}. \end{aligned}$
verified: the printed answer passed a computed check
How it was checked
- differentiate: passes 1/(2*D*L)*sqrt(g/(pi*sigma*T))