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1.11 Formulas and Literal Equations

Formulas • Literal Equations • Subscripts • Solve for Symbol before Substituting Numerical Values

An important application of equations is in the use of formulas that are found in geometry and nearly all fields of science and technology. A formula (or literal equation) is an equation that expresses the relationship between two or more related quantities. For example, Einstein’s famous formula $E = m c^{2}$ $E = m c^{2}$ shows the equivalence of energy E to the mass m of an object and the speed of light c.

We can solve a formula for a particular symbol just as we solve any equation.

NOTE

[That is, we isolate the required symbol by using algebraic operations on literal numbers.]

EXAMPLE 1 Solving for symbol in formula—Einstein

In Einstein’s formula $E = m c^{2},$ $E = m c^{2},$ solve for m.

\begin{array}{rcll} \frac{E}{c^{2}} & = & m & divide both sides by c^{2} \\ m & = & \frac{E}{c^{2}} & switch sides to place m at left \end{array}

$\begin{array}{rcll} \frac{E}{c^{2}} & = & m & divide both sides by c^{2} \\ m & = & \frac{E}{c^{2}} & switch sides to place m at left \end{array}$

The required symbol is usually placed on the left, as shown.

EXAMPLE 2 Symbol with subscript in formula—velocity

A formula relating acceleration a, velocity v, initial velocity $v_{0},$ $v_{0},$ and time is $v = v_{0} + a t .$ $v = v_{0} + a t .$ Solve for t.

\begin{array}{rcll} v - v_{0} & = & a t & v_{0} subtracted from both sides \\ t & = & \frac{v - v_{0}}{a} & both sides divided by a and then sides switched \end{array}

$\begin{array}{rcll} v - v_{0} & = & a t & v_{0} subtracted from both sides \\ t & = & \frac{v - v_{0}}{a} & both sides divided by a and then sides switched \end{array}$

EXAMPLE 3 Symbol in capital and in lowercase—forces on a beam

In the study of the forces on a certain beam, the equation $W = \frac{L (w L + 2 P)}{8}$ $W = \frac{L (w L + 2 P)}{8}$ is used. Solve for P.

\begin{array}{rcll} 8 W & = & \frac{8 L (w L + 2 P)}{8} & multiply both sides by 8 \\ 8 W & = & L (w L + 2 P) & simplify right side \\ 8 W & = & w L^{2} + 2 L P & remove parentheses \\ 8 W - w L^{2} & = & 2 L P & subtract w L^{2} from both sides \\ P & = & \frac{8 W - w L^{2}}{2 L} & divide both sides by 2 L and switch sides \end{array}

$\begin{array}{rcll} 8 W & = & \frac{8 L (w L + 2 P)}{8} & multiply both sides by 8 \\ 8 W & = & L (w L + 2 P) & simplify right side \\ 8 W & = & w L^{2} + 2 L P & remove parentheses \\ 8 W - w L^{2} & = & 2 L P & subtract w L^{2} from both sides \\ P & = & \frac{8 W - w L^{2}}{2 L} & divide both sides by 2 L and switch sides \end{array}$

CAUTION

Be careful. Just as subscripts can denote different literal numbers, a capital letter and the same letter in lowercase are different literal numbers. In this example, W and w are different literal numbers. This is shown in several of the exercises in this section.

EXAMPLE 4 Formula with groupings—temperature and volume

The effect of temperature on measurements is important when measurements must be made with great accuracy. The volume V of a special precision container at temperature T in terms of the volume $V_{0}$ $V_{0}$ at temperature $T_{0}$ $T_{0}$ is given by $V = V_{0} [1 + b (T - T_{0})],$ $V = V_{0} [1 + b (T - T_{0})],$ where b depends on the material of which the container is made. Solve for T.

Because we are to solve for T, we must isolate the term containing T. This can be done by first removing the grouping symbols.

\begin{array}{rcl} V & = & V_{0} [1 + b (T - T_{0})] & original equation \\ V & = & V_{0} [1 + b T - b T_{0}] & remove parentheses \\ V & = & V_{0} + b T V_{0} - b T_{0} V_{0} & remove brackets \\ V - V_{0} + b T_{0} V_{0} & = & b T V_{0} & subtract V_{0} and add b T_{0} V_{0} to both sides \\ T & = & \frac{V - V_{0} + b T_{0} V_{0}}{b V_{0}} & divide both sides by b V_{0} and switch sides \end{array}

$\begin{array}{rcl} V & = & V_{0} [1 + b (T - T_{0})] & original equation \\ V & = & V_{0} [1 + b T - b T_{0}] & remove parentheses \\ V & = & V_{0} + b T V_{0} - b T_{0} V_{0} & remove brackets \\ V - V_{0} + b T_{0} V_{0} & = & b T V_{0} & subtract V_{0} and add b T_{0} V_{0} to both sides \\ T & = & \frac{V - V_{0} + b T_{0} V_{0}}{b V_{0}} & divide both sides by b V_{0} and switch sides \end{array}$

NOTE

[To determine the values of any literal number in an expression for which we know values of the other literal numbers, we should first solve for the required symbol and then evaluate.]

EXAMPLE 5 Solve for symbol before substituting—volume of sphere

The volume V (in ${mm}^{3}$ ${mm}^{3}$ ) of a copper sphere changes with the temperature T (in $° C$ $° C$ ) according to $V = V_{0} + V_{0} β T,$ $V = V_{0} + V_{0} β T,$ where $V_{0}$ $V_{0}$ is the volume at $0 ° C$ $0 ° C$ . For a given sphere, $V_{0} = {6715 mm}^{3}$ $V_{0} = {6715 mm}^{3}$ and $β = 5.10 \times 10^{- 5} / ° C .$ $β = 5.10 \times 10^{- 5} / ° C .$ Evaluate T for $V = {6908 mm}^{3} .$ $V = {6908 mm}^{3} .$

We first solve for T and then substitute the given values.

\begin{array}{rcl} V & = & V_{0} + V_{0} β T \\ V - V_{0} & = & V_{0} β T \\ T & = & \frac{V - V_{0}}{V_{0} β} \end{array}

$\begin{array}{rcl} V & = & V_{0} + V_{0} β T \\ V - V_{0} & = & V_{0} β T \\ T & = & \frac{V - V_{0}}{V_{0} β} \end{array}$

Now substituting, we have

\begin{array}{rcl} T & = & \frac{6908 - 6715}{(6715) (5.10 \times 10^{- 5})} \\ = & 564 ° C \end{array}

$\begin{array}{rcl} T & = & \frac{6908 - 6715}{(6715) (5.10 \times 10^{- 5})} \\ = & 564 ° C \end{array}$

EXERCISES 1.11

In Exercises 1–4, solve for the given letter from the indicated example of this section.

For the formula in Example 2, solve for a.
For the formula in Example 3, solve for w.
For the formula in Example 4, solve for $T_{0} .$ $T_{0} .$
For the formula in Example 5, solve for $β .$ $β .$ (Do not evaluate.)

In Exercises 5–42, each of the given formulas arises in the technical or scientific area of study shown. Solve for the indicated letter.

$E = I R,$ $E = I R,$ for R (electricity)
$P V = n R T,$ $P V = n R T,$ for T (chemistry)
$r L = g_{2} - g_{1},$ $r L = g_{2} - g_{1},$ for $g_{1}$ $g_{1}$ (surveying)
$W = S_{d} T - Q,$ $W = S_{d} T - Q,$ for Q (air conditioning)
$B = \frac{n T W L}{12},$ $B = \frac{n T W L}{12},$ for n (construction management)
$P = 2 π T f,$ $P = 2 π T f,$ for T (mechanics)
$p = p_{a} + d g h,$ $p = p_{a} + d g h,$ for g (hydrodynamics)
$2 Q = 2 I + A + S,$ $2 Q = 2 I + A + S,$ for I (nuclear physics)
$F_{c} = \frac{m v^{2}}{r},$ $F_{c} = \frac{m v^{2}}{r},$ for r (centripetal force)
$P = \frac{4 F}{π D^{2}},$ $P = \frac{4 F}{π D^{2}},$ for F (automotive trades)
$S_{T} = \frac{A}{5 T} + 0.05 d,$ $S_{T} = \frac{A}{5 T} + 0.05 d,$ for A (welding)
$u = - \frac{e L}{2 m},$ $u = - \frac{e L}{2 m},$ for L (spectroscopy)
$c t^{2} = 0.3 t - a c,$ $c t^{2} = 0.3 t - a c,$ for a (medical technology)
$2 p + d v^{2} = 2 d (C - W),$ $2 p + d v^{2} = 2 d (C - W),$ for W (fluid flow)
$T = \frac{c + d}{v},$ $T = \frac{c + d}{v},$ for d (traffic flow)
$B = \frac{μ_{0} I}{2 π R},$ $B = \frac{μ_{0} I}{2 π R},$ for R (magnetic field)
$\frac{K_{1}}{K_{2}} = \frac{m_{1} + m_{2}}{m_{1}},$ $\frac{K_{1}}{K_{2}} = \frac{m_{1} + m_{2}}{m_{1}},$ for $m_{2}$ $m_{2}$ (kinetic energy)
$f = \frac{F}{d - F},$ $f = \frac{F}{d - F},$ for d (photography)
$a = \frac{2 m g}{M + 2 m},$ $a = \frac{2 m g}{M + 2 m},$ for M (pulleys)
$v = \frac{V (m + M)}{m},$ $v = \frac{V (m + M)}{m},$ for M (ballistics)
$C_{0}^{2} = C_{1}^{2} (1 + 2 V),$ $C_{0}^{2} = C_{1}^{2} (1 + 2 V),$ for V (electronics)
$A_{1} = A (M + 1),$ $A_{1} = A (M + 1),$ for M (photography)
$N = r (A - s),$ $N = r (A - s),$ for s (engineering)
$T = 3 (T_{2} - T_{1}),$ $T = 3 (T_{2} - T_{1}),$ for $T_{1}$ $T_{1}$ (oil drilling)
$T_{2} = T_{1} - \frac{h}{100},$ $T_{2} = T_{1} - \frac{h}{100},$ for h (air temperature)
$p_{2} = p_{1} + r p_{1} (1 - p_{1}),$ $p_{2} = p_{1} + r p_{1} (1 - p_{1}),$ for r (population growth)
$Q_{1} = P (Q_{2} - Q_{1}),$ $Q_{1} = P (Q_{2} - Q_{1}),$ for $Q_{2}$ $Q_{2}$ (refrigeration)
$p - p_{a} = d g (y_{2} - y_{1}),$ $p - p_{a} = d g (y_{2} - y_{1}),$ for $y_{1}$ $y_{1}$ (fire science)
$N = N_{1} T - N_{2} (1 - T),$ $N = N_{1} T - N_{2} (1 - T),$ for $N_{1}$ $N_{1}$ (machine design)
$t_{a} = t_{c} + (1 - h) t_{m},$ $t_{a} = t_{c} + (1 - h) t_{m},$ for h (computer access time)
$L = π (r_{1} + r_{2}) + 2 x_{1} + x_{2},$ $L = π (r_{1} + r_{2}) + 2 x_{1} + x_{2},$ for $r_{1}$ $r_{1}$ (pulleys)
$I = \frac{V R_{2} + V R_{1} (1 + μ)}{R_{1} R_{2}},$ $I = \frac{V R_{2} + V R_{1} (1 + μ)}{R_{1} R_{2}},$ for $μ$ $μ$ (electronics)
$P = \frac{V_{1} (V_{2} - V_{1})}{g J},$ $P = \frac{V_{1} (V_{2} - V_{1})}{g J},$ for $V_{2}$ $V_{2}$ (jet engine power)
$W = T (S_{1} - S_{2}) - Q,$ $W = T (S_{1} - S_{2}) - Q,$ for $S_{2}$ $S_{2}$ (refrigeration)
$C = \frac{2 e A k_{1} k_{2}}{d (k_{1} + k_{2})},$ $C = \frac{2 e A k_{1} k_{2}}{d (k_{1} + k_{2})},$ for e (electronics)
$d = \frac{3 L P x^{2} - P x^{3}}{6 E I},$ $d = \frac{3 L P x^{2} - P x^{3}}{6 E I},$ for L (beam deflection)
$V = C (1 - \frac{n}{N}),$ $V = C (1 - \frac{n}{N}),$ for n (property deprecation)
$\frac{p}{P} = \frac{A I}{B + A I},$ $\frac{p}{P} = \frac{A I}{B + A I},$ for B (atomic theory)

In Exercises 43–48, find the indicated values.

For a car’s cooling system, the equation $p (C - n) + n = A$ $p (C - n) + n = A$ is used. If $p = 0.25, C = 15.0 L,$ $p = 0.25, C = 15.0 L,$ and $A = 13.0 L,$ $A = 13.0 L,$ solve for n (in L).
A formula used in determining the total transmitted power $P_{t}$ $P_{t}$ in an AM radio signal is $P_{t} = P_{c} (1 + 0.500 m^{2}) .$ $P_{t} = P_{c} (1 + 0.500 m^{2}) .$ Find $P_{c}$ $P_{c}$ if $P_{t} = 680 W$ $P_{t} = 680 W$ and $m = 0.925.$ $m = 0.925.$
A formula relating the Fahrenheit temperature F and the Celsius temperature C is $F = \frac{9}{5} C + 32.$ $F = \frac{9}{5} C + 32.$ Find the Celsius temperature that corresponds to $90.2 ° F .$ $90.2 ° F .$
In forestry, a formula used to determine the volume V of a log is $V = \frac{1}{2} L (B + b),$ $V = \frac{1}{2} L (B + b),$ where L is the length of the log and B and b are the areas of the ends. Find b (in ${ft}^{2}$ ${ft}^{2}$ ) if $V = 38.6 {ft}^{3}, L = 16.1 ft,$ $V = 38.6 {ft}^{3}, L = 16.1 ft,$ and $B = 2.63 {ft}^{2} .$ $B = 2.63 {ft}^{2} .$ See Fig. 1.17.

Fig. 1.17
The voltage $V_{1}$ $V_{1}$ across resistance $R_{1}$ $R_{1}$ is $V_{1} = \frac{V R_{1}}{R_{1} + R_{2}},$ $V_{1} = \frac{V R_{1}}{R_{1} + R_{2}},$ where V is the voltage across resistances $R_{1}$ $R_{1}$ and $R_{2} .$ $R_{2} .$ See Fig. 1.18. Find $R_{2}$ $R_{2}$ (in $Ω$ $Ω$ ) if $R_{1} = 3.56 Ω, V_{1} = 6.30 V,$ $R_{1} = 3.56 Ω, V_{1} = 6.30 V,$ and $V = 12.0 V .$ $V = 12.0 V .$

Fig. 1.18
The efficiency E of a computer multiprocessor compilation is given by $E = \frac{1}{q + p (1 - q)},$ $E = \frac{1}{q + p (1 - q)},$ where p is the number of processors and q is the fraction of the compilation that can be performed by the available parallel processors. Find p for $E = 0.66$ $E = 0.66$ and $q = 0.83.$ $q = 0.83.$

In Exercises 49 and 50, set up the required formula and solve for the indicated letter.

One missile travels at a speed of $v_{2}$ $v_{2}$ mi/h for 4 h, and another missile travels at a speed of $v_{1}$ $v_{1}$ for $t + 2$ $t + 2$ hours. If they travel a total of d mi, solve the resulting formula for t.
A microwave transmitter can handle x telephone communications, and 15 separate cables can handle y connections each. If the combined system can handle C connections, solve for y.

Answers to Practice Exercises

$θ - k A$ $θ - k A$
$\frac{P + n c}{n}$ $\frac{P + n c}{n}$

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Table of Contents for 1.11 Formulas and Literal Equations

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Table of Contents for
1.11 Formulas and Literal Equations