2.4: Formulae
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The formulae of science usually contain variable letters other than the variable “Formulae” is the plural for “formula.”x . Indeed, formulae in science typically use several letters. For example, Isaac Newton’s Universal Law of Gravitation says that the magnitude of the force of attraction between two celestial bodies is given by the formula:
F=GMmr2
where m usually denotes the mass of the smaller body, M the mass of the larger body, and r is the distance between the two bodies. The letter G represents the universal gravitational constant, having value 6.67428×10−11 N(m/kg)2.
Variable case
Note the use of upper and lower case letter M’s in Newton’s Law of Gravitation. When working with scientific formulae, you must maintain the case of the given letters. You are not allowed to substitute lower for upper case, or upper for lower case in your work.
In Section 2.2, we described the goal that must be met when we are asked to “solve an equation for " x.
Solve for x
When asked to solve an equation for x, the goal is to manipulate the equation into the final form
x=“Stuff”
where “Stuff” is a valid mathematical expression that may contain other variables, mathematical symbols, etc., but it must not contain any occurrence of the variable x.
Thus, to solve an equation for x, we need to isolate the terms containing x on one side of the equation, and all remaining terms on the other side of the equation.
Example 2.4.1
Solve for x:x+a=b.
Solution
To undo the effects of adding a, subtract a from both sides of the equation.
x+a=b Original equation. x+a−a=b−a Subtract a from both sides. x=b−a Simplify.
Exercise 2.4.1
Solve for x:x−c=d.
- Answer
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x=c+d
In Example 2.4.1, note that the answer x=b−a has the required form, x=“Stuff”, where “Stuff” is a valid mathematical expression that contains other variables, mathematical symbols, etc., but it does not contain any occurrence of the variable x. Now, what if we were asked to solve the same equation for a, instead of x?
Example 2.4.2
Solve for a:x+a=b.
Solution
We are instructed to solve the equation x+a=b for a. This means that our final answer must have the form a=“Stuff”, where “Stuff” is a valid mathematical expression that contains other variables, mathematical symbols, etc., but it does not contain any occurrence of the variable a. This means that we must isolate all terms containing the variable a on one side of the equation, and all remaining terms on the other side of the equation. Now, to undo the effect of adding x, subtract x from both sides of the equation.
x+a=b Original equation. x+a−x=b−x Subtract x from both sides. a=b−x Simplify.
Note that we have a=“Stuff”, where “Stuff” contains no occurrence of a, the variable we are solving for.
Exercise 2.4.2
Solve for x:x−c=d.
- Answer
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c=x−d
Example 2.4.3
The formula F=kx, known as “Hooke’s Law”, predicts the force F required to stretch a spring x units. Solve the equation for k.
Solution
We are instructed to solve the equation F=kx for k. This means that our final answer must have the form k=“Stuff”, where “Stuff” is a valid mathematical expression that may contain other variables, mathematical symbols, etc., but it may not contain any occurrence of the variable k. This means that we must isolate all terms containing the variable k on one side of the equation, and all remaining terms on the other side of the equation. However, note that all terms containing the variable k are already isolated on one side of the equation. Terms not containing the variable k are isolated on the other side of the equation. Now, to “undo” the effect of multiplying by x, divide both sides of the equation by x.
F=kx Original equation. Fx=kxx Divide both sides by xFx=k Simplify.
Saying that F/x=k is equivalent to saying that k=F/x. We can leave our answer in the form shown in the last step, but some instructors insist that we write the answer as follows:
k=FxFx=k is equivalent to k=Fx
Note that we have k=“Stuff”, where “Stuff” contains no occurrence of k, the variable we are solving for.
Exercise 2.4.3
Solve for m:E=mc2.
- Answer
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m=Ec2
Example 2.4.4
The formula V=RI is called “Ohm’s Law.” It helps calculate the voltage drop V across a resistor R in an electric circuit with current I. Solve the equation for R.
Solution
We are instructed to solve the equation V=RI for R. This means that our final answer must have the form R=“Stuff”, where “Stuff” is a valid mathematical expression that may contain other variables, mathematical symbols, etc., but it may not contain any occurrence of the variable R. This means that we must isolate all terms containing the variable R on one side of the equation, and all remaining terms on the other side of the equation. However, note that all terms containing the variable R are already isolated on one side of the equation. Terms not containing the variable R are isolated on the other side of the equation. Now, to “undo” the effect of multiplying by I, divide both sides of the equation by I.
V=RI Original equation. VI=RII Divide both sides by IVI=R Simplify.
This can also be written in the following form:
R=VIV/I=R is equivalent to R=V/I
Note that we have R=“Stuff”, where “Stuff” contains no occurrence of R, the variable we are solving for.
Exercise 2.4.4
Solve for t:d=st.
- Answer
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t=ds
Clearing Fractions
If fractions occur in a formula, clear the fractions from the formula by multiplying both sides of the formula by the common denominator.
Example 2.4.5
The formula K=12mv2 is used to calculate the kinetic energy K of a particle of mass m moving with velocity v. Solve the equation for m.
Solution
We’re asked to solve the equation K=(1/2)mv2 for m. First, clear the fractions by multiplying both sides by the common denominator.
K=12mv2 Original equation. 2(K)=2(12mv2) Multiply both sides by 22K=mv2 Simplify. Cancel 2′ s.
Note that all terms containing m, the variable we are solving for, are already isolated on one side of the equation. We need only divide both sides by v2 to complete the solution.
2Kv2=mv2v2 Divide both sides by v22Kv2=m Simplify. Cancel v2 for v2
Note that the final answer has the form m=“Stuff”, where “Stuff” contains no occurrence of the variable m.
Exercise 2.4.5
Solve for g:s=12gt2.
- Answer
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g=2st2
Example 2.4.6
As mentioned earlier, Newton’s Universal Law of Gravitation is described by the formula F=GMmr2Solve this equation for m.
Solution
We’re asked to solve the equation F=GMm/r2 for m. First, clear the fractions by multiplying both sides by the common denominator.
F=GMmr2 Original equation. r2(F)=r2(GMmr2) Multiply both sides by r2r2F=GMm Simplify. Cancel r2 for r2
Note that all terms containing m, the variable we are solving for, are already isolated on one side of the equation. We need only divide both sides by GM to complete the solution.
r2FGM=GMmGM Divide both sides by GMr2FGM=m Simplify. Cancel GM for GM
Note that the final answer has the form m=“Stuff”, where “Stuff” contains no occurrence of the variable m.
Exercise 2.4.6
Solve for q2:F=kq1q2r2.
- Answer
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q2=Fr2kq1
Geometric Formulae
Let’s look at a few commonly used formulae from geometry.
Example 2.4.7
Let W and L represent the width and length of a rectangle, respectively, and let P represent its perimeter.
The perimeter (distance around) of the rectangle is found by summing its four sides, then combining like terms.
P=L+W+L+W Summing the four sides. P=2W+2L Combine like terms.
Solve P=2W+2L for L. Then, given that the perimeter is 300 feet and the width is 50 feet, use your result to calculate the length.
Solution
We’re first asked to solve P=2W+2L for L. First, isolate all terms containing the variable L on one side of the equation.
P=2W+2L Original equation. P−2W=2W+2L−2W Subtract 2W from both sides. P−2W=2L Simplify. P−2W2=2L2 Divide both sides by 2P−2W2=L Simplify.
Note that the final result has L=“Stuff", where “Stuff” contains no occurrence of the variable L.
The second part of this example requests that we find the length of the rectangle, given that the perimeter is P=300 feet and the width is W=50 feet. To calculate the length, substitute P=300 and W=50 in L=(P−2W)/2.
L=P−2W2 Perimeter formula solved for LL=300−2(50)2300 for P,50 for WL=300−1002 Multiply: 2(50)=100L=2002 Subtract: 300−100=200L=100 Divide: 200/2=100
Hence, the length of the rectangle is 100 feet.
Exercise 2.4.7
The perimeter of a rectangle is 160 meters and its width is 30 meters. Finds its length.
- Answer
-
L=50 meters
Example 2.4.8
Let b and h represent the length of the base and the height of a triangle, respectively, and let A represent the area of the triangle.
The area of the triangle is computed using the formula: A=12bhThat is, the area of a triangle is “one-half the base times the height.”
Solve the formula A=12bh for h. Secondly, given that the area is A=90 in2 (90 square inches) and the length of the base is 15 in (15 inches), find the height of the triangle.
Solution
We’re first asked to solve A=(1/2)bh for h. Because the equation has fractions, the first step is to clear the fractions by multiplying both sides by the least common denominator.
A=12bh Area of a triangle formula. 2(A)=2(12bh) Multiply both sides by 22A=bh Simplify.
Now, we already have all terms containing the variable h on one side of the equation, so we can solve for h by dividing both sides of the equation by b.
2Ab=bhb Divide both sides by b2Ab=h Simplify.
Note that the final result has h=“Stuff", where “Stuff” contains no occurrence of the variable h.
The second part of this example requests that we find the height of the triangle, given that the area is A=90 in 2 and the length of the base is b=15 in. To calculate the height of the triangle, substitute A=90 and b=15 in h=2A/b.
h=2Ab Area formula solved for hh=2(90)15Substitue 90 for A,15 for bh=18015 Multiply: 2(90)=180h=12Divide: 180/15=12
Hence, the height of the triangle is 12 inches.
Exercise 2.4.8
The area of a triangle is 140 cm2 and the length of its base is 70 cm. Find the height of the triangle. .
- Answer
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4 cm