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IG.M10 Direct and Inverse Proportion With Non-Linear Powers

EDEXCEL 4MA1 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer all questions. Full sentences are required in 2 questions (where an explanation is asked). Allow 60 minutes.

Given that y is proportional to x² for a square tile, and when x = 4 cm then y = 36 (area units), write the equation relating y and x and find the value of k.

(Total for Question 1 is 1 mark)

Given that T is proportional to v³ for a model where thrust T increases with the cube of speed v, and T = 54 N when v = 3 m/s, find the constant of proportionality k in T = kv³.

(Total for Question 2 is 1 mark)

Given that y is proportional to 1/√x (y = k / x^1/2) for cooling time y depending inversely on the square root of mass x, and y = 10 when x = 4, find k and state the proportionality equation.

(Total for Question 3 is 2 marks)

A circular pond has water depth y that varies directly as the square of the radius r, y = kr², in an experimental model. Given that y = 18 when r = 3 m, form the equation and hence find y when r = 5 m.

(Total for Question 4 is 2 marks)

Given that the cost C of making a decorative square panel varies as the square of its side length s, C = ks². If a panel of side 1.2 m costs pound 43.20, find k and then find the cost of a panel of side 0.8 m. Give answers to 2 decimal places.

(Total for Question 5 is 2 marks)

Given that y is proportional to x³, y = kx³, and y = 16 when x = 2, find k and then hence find x when y = 432.

(Total for Question 6 is 3 marks)

Show that if the period T of a simple pendulum is proportional to the square root of its length L, T = k L^1/2, and T = 1.8 s when L = 0.81 m, then k = 2. Verify the period when L = 2.25 m.

(Total for Question 7 is 3 marks)

The shaking force F on a small structure varies inversely as the square of the distance d from the source, F = k / d². If F = 0.5 N at d = 2 m, find k and hence find F at d = 5 m, giving your answer to 3 significant figures.

(Total for Question 8 is 3 marks)

A physics experiment models intensity I of radiation as I = k / r³, inversely proportional to the cube of distance r. If I = 0.125 units at r = 2 m, find k and hence find r when I = 0.001 units. Give r to 2 decimal places.

(Total for Question 9 is 4 marks)

A recipe says that the volume V of a loaf varies directly as the 3/2 power of the amount of yeast x used, V = k x^3/2. If using 8 g of yeast gives volume 1440 cm³, find k and then the volume when 2 g of yeast is used. Give the final volume to the nearest whole number.

(Total for Question 10 is 4 marks)

A thin metal plate has heat loss rate H that varies as the square root of its area A, H = k A^1/2. An experiment gives H = 12 W when A = 16 cm². Find k and then find H when A = 100 cm².

(Total for Question 11 is 4 marks)

Show that if the period P of oscillation of a particle varies inversely as the square of amplitude a, P = k / a², and P = 0.5 s when a = 0.4 m, then k = 0.5*(0.4)² = 0.08. Hence find P when a = 0.2 m and explain briefly whether P increases or decreases when amplitude halves.

(Total for Question 12 is 4 marks)

A scale model ship's resistance R varies as the square of its speed v and inversely as the cube root of its length L, R = k v² / L^1/3. In a test tank, R = 0.96 N when v = 3 m/s and L = 0.125 m. Find k and then find R when v = 5 m/s and L = 1.0 m. Give R to 3 significant figures.

(Total for Question 13 is 5 marks)

A charity fundraiser states that the amount raised A varies directly as the square of the number of attendees n, A = k n². At a small event with n = 30 attendees, A = pound 2700. Use this to estimate how many attendees would be needed to raise pound 3840, assuming the same k. Give your answer to the nearest whole person and show your working in words.

(Total for Question 14 is 5 marks)

A lamp's illuminance E varies inversely as the square root of distance d, E = k / d^1/2. At d = 0.25 m, E = 80 lux. A photographer asks: how far must the lamp be moved so that illuminance is reduced to 20 lux? Solve for the new distance d and give the answer to 2 decimal places.

(Total for Question 15 is 4 marks)

Mark scheme · IG.M10 Direct and Inverse Proportion With Non-Linear Powers

Question 1

B1 y = kx² with k = 36/16 = 9/4 or 2.25 stated
Answer: y = (9/4)x² or y = 2.25x²

Question 2

B1 k = 54/27 = 2 stated
Answer: k = 2

Question 3

M1 substitutes values into y = k/x^1/2: 10 = k/2 or equivalent
A1 k = 20 and equation y = 20 / x^1/2 or y = 20 x^-1/2 cao
Answer: k = 20, so y = 20 / x^1/2

Question 4

M1 finds k from 18 = k*3², k = 2
A1 y = 2*5² = 50 stated
Answer: y = 50 (depth units)

Question 5

M1 finds k = 43.20 / (1.2²) = 43.20 / 1.44 = 30
A1 cost at 0.8 m is C = 30 * 0.8² = 30 * 0.64 = 19.20, cao
Answer: k = 30, cost = pound 19.20

Question 6

M1 finds k from 16 = k*(2³) so k = 16/8 = 2
M1 substitutes k into 432 = 2*x³ and rearranges to x³ = 216
A1 x = 6 cao
Answer: k = 2 and x = 6

Question 7

M1 substitutes 1.8 = k*(0.81)^1/2 and finds √0.81 = 0.9 so k = 1.8/0.9 = 2
M1 uses k = 2 to find T = 2*(2.25)^1/2 and evaluates √2.25 = 1.5
A1 T = 2*1.5 = 3.0 s cao
Answer: k = 2 and T = 3.0 s when L = 2.25 m

Question 8

M1 finds k from 0.5 = k/2² so k = 0.5 * 4 = 2
M1 substitutes k to get F = 2 / 5² = 2 / 25 = 0.08
A1 F = 0.0800... so 0.08 N or 0.080 to 3 s.f., accept 0.080 N cao
Answer: k = 2, F = 0.0800... so 0.080 N (3 s.f.)

Question 9

M1 finds k from 0.125 = k / 2³ so k = 0.125 * 8 = 1
M1 uses k = 1 and sets 0.001 = 1 / r³ to get r³ = 1000
M1 takes cube root to find r = 10, or evaluates r = 1000^1/3
A1 r = 10.00 m cao
Answer: k = 1 and r = 10.00 m

Question 10

M1 substitutes 1440 = k*(8)^3/2. Recognises 8^1/2 = 2*√2 or better to compute 8^3/2 = (√8)³ = (2.828427...)³ = 22.627... or 8^3/2= (8¹)*(8¹/2)=8*2.828...
M1 calculates k = 1440 / 8^3/2 = 1440 / 22.627416997 = approx 63.64, or uses exact form k = 1440 / (8*√8) = 45*√2
M1 uses k to find V at x=2: V = k*(2)^3/2 and evaluates
A1 V = 180 cm³ (nearest whole number) cao
Answer: k approx 63.64 (exact 45*√2), V at 2 g = 180 cm³

Question 11

M1 finds k from 12 = k*(16)^1/2 so k = 12/4 = 3
M1 uses k to get H = 3*(100)^1/2 = 3*10
A1 H = 30 W cao
B1 units W stated with final numeric answer
Answer: k = 3, H = 30 W

Question 12

M1 substitutes to find k: 0.5 = k / 0.4² so k = 0.5 * 0.16 = 0.08
M1 uses k to find P at a = 0.2: P = 0.08 / 0.2² = 0.08 / 0.04 = 2
A1 P = 2.0 s cao
B1 brief explanation: when amplitude halves, denominator a² becomes one quarter, so P becomes four times larger, so P increases, oe
Answer: k = 0.08, P = 2.0 s; P increases (becomes 4 times larger) when amplitude halves

Question 13

M1 substitutes 0.96 = k*(3²) / (0.125)^1/3 and recognises 3² = 9 and cube root 0.125 = 0.5
M1 computes k = 0.96 * 0.5 / 9 = (0.48)/9 = 0.053333... or 0.0533333
M1 substitutes k into R = k*v² / L^1/3 with v=5 and L=1.0 giving R = k*25 / 1 = 25k
M1 calculates R = 25 * 0.053333... = 1.333333...
A1 R = 1.33 N to 3 s.f. cao
Answer: k = 0.053333... (16/300), R = 1.33 N (3 s.f.)

Question 14

M1 finds k from 2700 = k*30² so k = 2700/900 = 3
M1 sets 3840 = 3*n² and rearranges to n² = 1280
M1 takes square root to get n = √1280 and evaluates
A1 n approx 35.777... and rounded to nearest whole person gives 36 cao
B1 brief wording: states using the same k, n = √3840/3 approx 35.78, so 36 attendees needed, oe
Answer: n approx 35.78, so 36 attendees (nearest whole person)

Question 15

M1 finds k from 80 = k / 0.25^1/2; √0.25=0.5 so k = 80*0.5 = 40
M1 sets 20 = 40 / d^1/2 and rearranges to d^1/2 = 40/20 = 2
M1 squares to get d = 4 and evaluates
A1 d = 4.00 m cao
Answer: d = 4.00 m

Direct and Inverse Proportion With Non-Linear Powers - Worksheets, Questions and Revision