Harder Proportion – Direct and Inverse Relationships | Pearson Edexcel International GCSE Maths
These two examples use proportional relationships involving square roots. In each case, the proportional statement is first converted into an equation containing a constant of proportionality, k. The given values are then used to find k before the required value is calculated.
Question 1 – Direct Proportion and Square Roots
The time t seconds taken for a stone to fall is directly proportional to the square root of the distance d metres.
When t = 4.6, d = 25.
(a) Express t in terms of d.
(b) Find the time taken when d = 42.25.
(a) Express t in terms of d
(b) When d = 42.25
Direct Proportion
If t is directly proportional to √d, then:
t ∝ √d
Replacing the proportionality symbol with an equals sign requires a constant of proportionality:
t = k√d
Question 2 – Inverse Proportion and Square Roots
The speed v km/s of a satellite is inversely proportional to the square root of the radius r km of its orbit.
A satellite travelling at a radius of 6940 km has a speed of 7.6 km/s.
Find the speed when the orbital radius is 42 000 km, giving the answer to 3 significant figures.
Working
Inverse Proportion
If v is inversely proportional to √r, then:
v ∝ 1 / √r
Introducing the constant of proportionality gives:
v = k / √r
Method Summary
1. Write the proportional relationship.
2. Replace ∝ with an equals sign and introduce k.
3. Substitute the known values to find k.
4. Write the resulting formula.
5. Substitute the new value and calculate the required result.
Key Results
Question 1(a):
t = 0.92√d
Question 1(b):
t = 5.98 s
Question 2:
v = 3.09 km/s
Source: Questions adapted from Harry Smith, Revise Pearson Edexcel International GCSE (9–1) Mathematics A – Higher Tier Revision Guide, Pearson.
Solutions and workings: Tiago Hands.