Formulae
A formula is a mathematical expression that shows how two or more quantities are related to one another.
An example is the relationship between the amount saved in a bank, the rate at which interest is being paid, the length of time for which the money is saved and the simple interest. This relationship is given by the formula below:
I = P x R X T where P = Principal, R = Rate (in %) and T = Time
100
The formula will help a person who saves money in a bank know how much interest he/she would have after a period of time based on the bank’s interest rate.
Substitution of Values in Formulae
We can replace quantities in a formula by putting values in their place so that we would be able to calculate their values.
Example: The perimeter p cm of a rhombus of side d cm is given by the formula p = 4d.
(a) Find the perimeter of a rhombus of side 3.2cm.
(b) Find the length of a rhombus of perimeter 14cm
Solution:
(a) Substitute for d = 3.2cm in p = 4d; p = 4 (3.2) = 12.8cm
(b) If p = 14cm and p = 4d then 14cm = 4d; d = 14/4 = 3.5cm
Example 2: A circuit of voltage V volts and resistance R ohms has a current of I amps, where I = V/R. (a) Find the current when the voltage is 240 volts and the resistance is 80 ohms
(b) Find the voltage when the current is 0.6 amps and the resistance is 5 ohms
(c) Find the resistance when the current is 0.1 amps and the voltage is 9 volts.
Solution:
(a) V = 240 volts, R = 80 ohms, I = V/R; I = 240/80 = 3 amps
(b) If I = V/R then V = IR; for I = 0.6 amps, R = 5 ohms; V = 0.6 x 5 = 3 Volts
(c) I = V/R; 0.1 = 9/R ===> R = V/I = 9/0.1 = 90 ohms
TRY THIS:
The circumference, C units, of a circle of radius r units is given by the formula C = 2πr, π = 22/7
(a) What is the circumference of a circle of radius 7cm? [fusion_builder_container hundred_percent=”yes” overflow=”visible”][fusion_builder_row][fusion_builder_column type=”1_1″ background_position=”left top” background_color=”” border_size=”” border_color=”” border_style=”solid” spacing=”yes” background_image=”” background_repeat=”no-repeat” padding=”” margin_top=”0px” margin_bottom=”0px” class=”” id=”” animation_type=”” animation_speed=”0.3″ animation_direction=”left” hide_on_mobile=”no” center_content=”no” min_height=”none”][ANS: 44cm]
(b) What is the radius of a circle whose circumference is 22cm? [ANS: 3.5cm]
(c) What is the circumference of a circle of radius one metre? [ANS: 62/7 m or 628.57cm]
(d) What is the radius of a circle of circumference 2.75m? [ANS: 43¾cm]
Change of Subject of Formulae
In a formula, the quantity represented by the letter that stands on its own is called the subject of the formula. For example in the formula for circumference of a circle C = 2πr; C is the subject of the formula.
To change the subject of a formula means to rearrange the formula in a way that the relationship between the quantities is still correct.
If M = 2D, then M/2 = D.
There are some simple steps that help when changing the subject of a formula.
Step I: Start from the letters furthest away from the quantity you want to make the subject of the formula.
Step II: Eliminate the unwanted letters by using the reciprocal of their relationship to the letter.
NOTE: Whatever you do to one side must be done to the other side for there to be balance in the equation.
Example 1: Make R the subject of the formula I = PRT
100
Solution:
For I = P x R x T, 100 is the furthest away from the R, so multiply both sides by 100
100
Now 100I = P x R x T so divide both sides by P x T so that R can be alone
This gives (100I)/ (P x T) = P x R x T/ (P x T); Therefore R = 100I/PT
Example 2: In the formula V = U + at, make t the subject of the formula
Solution: U is furthest away from t so we subtract U from both sides
V – U = U + at – U ====> V – U = at
Divide both sides by “a” so we have t alone
(V – U) = at; Thereforet = V – U
a a a
Example 2: Make b the subject of the formula S = 2L + 2B
Solution: Subtract 2L from both sides
S – 2L = 2L + 2B – 2L ======> S – 2L = 2B
Divide both sides by 2
S – 2L = 2B
2 2
B = S – 2L
2
TRY THESE:
1. Make h the subject of the formula A = ½bh [ANS: h = 2A/b]
2. How would the formula T = θ +273 be written if θ is made the subject of the formula [θ = T – 273]
Tests and Exercises
1. Make r the subject of formula V =⅓πhr (a) 3V/h (b) πh/v (c) 3V/πh (d) 3V/r
Guideline: 1st we would clear the fractions by multiplying both sides by 3
3 x V = 3 x (⅓πhr) è 3V = πhr
Divide both sides by πh to isolate r
This gives us h = 3V/πh
2. If a = 2b – c, find the value of b when a = 11 and c = – 3 (a) 4 (b) 7 (c) 8 (d) 14
Guideline: Substituting a = 11 and c = -3 in the equation gives
11 = 2b – (-3) = 2b + 3
11 = 2b + 3, subtract 3 from both sides
11 – 3 = 2b; 8 = 2b, b = 4
3. If y = x + 5, find the value of y when x = 5 (a) – 5 (b) – 2 (c) 5 (d) 7
x – 3
Guideline: x + 5 = 5 + 5 = 10 = 5
x – 3 5 – 3 2
4. Express h in terms of A, a & b if A = 2(a + b) h (a) A (b) ½ (a + b) (c) 2A (d) a + b [ANS: A]
2(a + b) A (a +b) h 2A
Guideline: A = 2 (a + b) h; we can see that 2 (a + b) is what we need to remove from the RHS so we divide both sides by 2 (a + b) to get A__ i.e. option A
2(a + b)
5. The surface area of a sphere is S = 4πr2. If π = 22/7, find the surface area of a sphere of radius 7cm (a) 154cm2 (b) 176cm2 (c) 308cm2 (d) 616cm2
Guideline: Substitute the value of r and π in S = 4πr2
S = 4 x 22/7 x 72 = 4 x 22 x 7 x 7 = 616cm2
7
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