Derivatives: exam-style practice Problem set A One question at a time

A language app records its users (thousands) x months after launch. The data are in the table. A quadratic model f(x) = ax² + bx + c is fitted to the data.

1. (a) Use quadratic regression to find a, b and c to 3 significant figures.

   • From the calculator: quadratic regression on the two lists
   • a = −0.650, b = 44.6, c = −21.4

2. (b) The coefficient of determination is R² = 0.987 (3 s.f.). State whether R² supports a quadratic model and justify your answer.

   • R² = 0.987 is close to 1
   • So the quadratic explains nearly all of the variation in the data: it supports the model

3. (c) Comment on the validity of the model with reference to the value of c.

   • f(0) = c = −21.4
   • The model gives a negative number of users at launch (the data show 15 thousand)
   • So the model is not valid close to x = 0

A packing crate has a square base of side l m. Its height d m is 1 m less than twice the side of the base. The crate holds 112 m³.

1. (a) Write down an expression for d in terms of l.

   • "1 less than twice" l: d = 2l − 1

2. (b) Find l and d.

   • l²(2l − 1) = 112, so 2l³ − l² − 112 = 0
   • Polynomial solver, coefficients 2, −1, 0, −112: l = 4 (the other roots are complex)
   • d = 2(4) − 1 = 7

3. (c) Find the total external surface area of the crate.

   • 2(4×4 + 4×7 + 4×7)
   • = 144 m²

A paint tin is a closed cylinder of radius r cm and height h cm. It holds 350 cm³ of paint. Let A cm² be the total surface area of the tin.

1. (a) Find an expression for h in terms of r.

   • πr²h = 350
   • h = 350/(πr²)

2. (b) Show that \(A=2\pi r^2+\frac{700}{r}\).

   • A = 2πr² + 2πrh
   • = 2πr² + 2πr × 350/(πr²)
   • = 2πr² + 700/r

A soup can is a closed cylinder that holds 600 cm³. Its total surface area is \(A=2\pi r^2+\frac{1200}{r}\) cm², where r cm is the radius, r > 0.

1. (a) Find \(\frac{dA}{dr}\).

   • A = 2πr² + 1200r⁻¹
   • dA/dr = 4πr − 1200r⁻²

2. (b) Find the value of r that minimises A.

   • Set dA/dr = 0: 4πr = 1200/r²
   • r³ = 1200/(4π) = 95.5
   • r = 4.57

3. (c) Find the minimum value of A.

   • A = 2π(4.57)² + 1200/4.57
   • = 394 cm²

An open tool box has a square base of side x m and volume 32 m³. Its surface area is \(A=x^2+\frac{128}{x}\) m², where x > 0.

1. (a) Find \(\frac{d^2A}{dx^2}\).

   • A = x² + 128x⁻¹
   • dA/dx = 2x − 128x⁻²
   • d²A/dx² = 2 + 256x⁻³

2. (b) Show that the graph of A is concave up for all x > 0.

   • x > 0, so 256/x³ > 0
   • So d²A/dx² > 2 > 0: concave up for all x > 0

3. (c) Given that dA/dx = 0 when x = 4, state what the answer to part (b) tells you about this value of x.

   • d²A/dx² > 0 at x = 4 and dA/dx = 0
   • So x = 4 gives a minimum value of A: A = 48

Consider the curve \(y=2x(5-e^{3x})\).

1. (a) Find \(\frac{dy}{dx}\).

   • u = 2x, v = 5 − e^(3x), u′ = 2, v′ = −3e^(3x)
   • dy/dx = 2(5 − e^(3x)) + 2x(−3e^(3x))
   • = 10 − 2e^(3x) − 6xe^(3x)

2. (b) Find \(\frac{d^2y}{dx^2}\).

   • Differentiate −2e^(3x): −6e^(3x). Product rule on −6xe^(3x): −6e^(3x) − 18xe^(3x)
   • d²y/dx² = −6e^(3x) − 6e^(3x) − 18xe^(3x)
   • = −12e^(3x) − 18xe^(3x) = −6(2 + 3x)e^(3x)

3. (c) The curve has a point of inflexion at x = a. Find a.

   • −6(2 + 3a)e^(3a) = 0 and e^(3a) is never 0
   • 2 + 3a = 0, so a = −2/3
   • d²y/dx² changes sign here, so it is an inflexion

A conservation group counts a colony of seals (hundreds) every 10 years. Let x be the number of years after 1960. The data are in the table. A quadratic model f(x) = ax² + bx + c is fitted.

1. (a) Use quadratic regression to find a, b and c to 3 significant figures.

   • From the calculator: quadratic regression
   • a = −0.0410, b = 6.49, c = −2.90

2. (b) Find f′(60) and interpret your answer in context.

   • f′(x) = 2ax + b = −0.0819x + 6.49
   • f′(60) = 1.57
   • At x = 60 the population is increasing at 1.57 hundred seals per year

3. (c) Use the change in the table between x = 50 and x = 60 to comment on the validity of the model.

   • Table: (229 − 225) ÷ 10 = 0.4 hundred per year
   • The model's rate 1.57 is much larger than 0.4
   • The data show the population levelling off, which the model does not: the model is not valid for later years

The number of users of a new website, P thousand, after n weeks is modelled by \(P=\frac{15}{1+4e^{-0.3n}}\).

1. (a) Find \(\frac{dP}{dn}\) and explain why the model predicts that the number of users is always increasing.

   • P = 15(1 + 4e^(−0.3n))⁻¹
   • dP/dn = −15(1 + 4e^(−0.3n))⁻² × (−1.2e^(−0.3n)) = 18e^(−0.3n)/(1 + 4e^(−0.3n))²
   • The top is positive and the bottom is a square, so dP/dn > 0: always increasing

2. (b) Find the initial number of users and the initial rate of growth.

   • n = 0: P = 15 ÷ (1 + 4) = 3
   • dP/dn = 18 ÷ 25 = 0.72

3. (c) Find the value of P as n becomes very large.

   • As n → ∞, e^(−0.3n) → 0
   • P → 15 ÷ 1 = 15

4. (d) Hence sketch P against n.

   • Start at (0, 3)
   • Rising all the time, with a horizontal asymptote P = 15

A juice company packs 640 cm³ of juice in a closed cylinder of radius r cm and height h cm. Let A cm² be its total surface area.

1. (a) Find an expression for h in terms of r.

   • πr²h = 640
   • h = 640/(πr²)

2. (b) Show that \(A=2\pi r^2+\frac{1280}{r}\).

   • A = 2πr² + 2πrh = 2πr² + 2πr × 640/(πr²)
   • = 2πr² + 1280/r

3. (c) Find \(\frac{dA}{dr}\).

   • A = 2πr² + 1280r⁻¹
   • dA/dr = 4πr − 1280r⁻²

4. (d) Find the value of r that minimises A, and the minimum value of A.

   • 4πr = 1280/r², so r³ = 1280/(4π) = 102
   • r = 4.67 cm
   • A = 2π(4.67)² + 1280/4.67 = 411 cm²

5. (e) A cube-shaped box holds the same 640 cm³. Find its total surface area and say whether the cylinder or the cube needs less material.

   • Cube: s³ = 640, s = 8.62 cm
   • Area = 6s² = 6 × 8.62² = 446 cm²
   • 411 < 446: the cylinder uses less material

Consider the curve \(y=x^{2x}\) for x > 0.

1. (a) Show that \(x^{2x}=e^{2x\ln x}\).

   • x = e^(ln x)
   • x^(2x) = (e^(ln x))^(2x) = e^(2x ln x)

2. (b) Hence find the equation of the tangent to the curve at x = 1.

   • dy/dx = e^(2x ln x) × (2 ln x + 2)
   • At x = 1: y = 1 and dy/dx = 1 × (0 + 2) = 2
   • y − 1 = 2(x − 1), so y = 2x − 1

3. (c) Find the exact value of x at which \(\frac{dy}{dx}=0\).

   • e^(2x ln x) is never 0
   • So 2 ln x + 2 = 0, ln x = −1
   • x = 1/e