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

A museum records the number of visitors (thousands per year) x years after 1990. 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 = −0.856, b = 29.9, c = 84.8

2. (b) The coefficient of determination is R² = 0.99996 (5 d.p.). State whether R² supports a quadratic model and justify your answer.

   • R² = 0.99996 is very 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 a.

   • a is negative, so the graph is a downward parabola and f(x) < 0 for x > 37.6
   • A number of visitors cannot be negative, so the model is not valid for large x

A water tank has a square base of side w m. Its height h m is 4 m more than the side of the base. The tank holds 63 m³ of water when full.

1. (a) Write down an expression for h in terms of w.

   • "4 more than" w: h = w + 4

2. (b) Find w and h.

   • w²(w + 4) = 63, so w³ + 4w² − 63 = 0
   • Polynomial solver, coefficients 1, 4, 0, −63: w = 3 (the other roots are complex)
   • h = 3 + 4 = 7

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

   • 2(3×3 + 3×7 + 3×7)
   • = 102 m²

A biscuit tin is a closed cylinder of radius r cm and height h cm. It holds 280 cm³. Let A cm² be the total surface area of the tin.

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

   • πr²h = 280
   • h = 280/(πr²)

2. (b) Show that \(A=2\pi r^2+\frac{560}{r}\).

   • A = 2πr² + 2πrh
   • = 2πr² + 2πr × 280/(πr²)
   • = 2πr² + 560/r

A pot of face cream is a closed cylinder that holds 360 cm³. Its total surface area is \(A=2\pi r^2+\frac{720}{r}\) cm², where r cm is the radius, r > 0.

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

   • A = 2πr² + 720r⁻¹
   • dA/dr = 4πr − 720r⁻²

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

   • Set dA/dr = 0: 4πr = 720/r²
   • r³ = 720/(4π) = 57.3
   • r = 3.86

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

   • A = 2π(3.86)² + 720/3.86
   • = 280 cm²

An open fish tank has a square base of side x dm and volume 108 dm³. Its glass area is \(A=x^2+\frac{432}{x}\) dm², where x > 0.

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

   • A = x² + 432x⁻¹
   • dA/dx = 2x − 432x⁻²
   • d²A/dx² = 2 + 864x⁻³

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

   • x > 0, so 864/x³ > 0
   • So d²A/dx² > 2 > 0: concave up for all x > 0

3. (c) Given that dA/dx = 0 when x = 6, state what the answer to part (b) tells you about this value of x.

   • d²A/dx² > 0 at x = 6 and dA/dx = 0
   • So x = 6 gives a minimum value of A: A = 108

Consider the curve \(y=3x(1-e^{4x})\).

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

   • u = 3x, v = 1 − e^(4x), u′ = 3, v′ = −4e^(4x)
   • dy/dx = 3(1 − e^(4x)) + 3x(−4e^(4x))
   • = 3 − 3e^(4x) − 12xe^(4x)

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

   • Differentiate −3e^(4x): −12e^(4x). Product rule on −12xe^(4x): −12e^(4x) − 48xe^(4x)
   • d²y/dx² = −12e^(4x) − 12e^(4x) − 48xe^(4x)
   • = −24e^(4x) − 48xe^(4x) = −24(1 + 2x)e^(4x)

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

   • −24(1 + 2a)e^(4a) = 0 and e^(4a) is never 0
   • 1 + 2a = 0, so a = −1/2
   • d²y/dx² changes sign here, so it is an inflexion

A cycling club records its number of members every 5 years. Let x be the number of years after 1995. 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 = −1.41, b = 79.3, c = 295

2. (b) Find f′(30) and interpret your answer in context.

   • f′(x) = 2ax + b = −2.82x + 79.3
   • f′(30) = −5.29
   • At x = 30 the membership is decreasing at 5.29 members per year

3. (c) Use the change in the table between x = 25 and x = 30 to comment on the validity of the model.

   • Table: (1400 − 1390) ÷ 5 = 2 members per year
   • The data are still rising but the model says they are falling
   • The model is not valid at the end of the data

The number of birds on an island, P hundred, n years after they arrive is modelled by \(P=\frac{18}{1+8e^{-0.5n}}\).

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

   • P = 18(1 + 8e^(−0.5n))⁻¹
   • dP/dn = −18(1 + 8e^(−0.5n))⁻² × (−4e^(−0.5n)) = 72e^(−0.5n)/(1 + 8e^(−0.5n))²
   • The top is positive and the bottom is a square, so dP/dn > 0: always increasing

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

   • n = 0: P = 18 ÷ (1 + 8) = 2
   • dP/dn = 72 ÷ 81 = 0.889

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

   • As n → ∞, e^(−0.5n) → 0
   • P → 18 ÷ 1 = 18

4. (d) Hence sketch P against n.

   • Start at (0, 2)
   • Rising all the time, with a horizontal asymptote P = 18

A pet-food company packs 480 cm³ of food 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 = 480
   • h = 480/(πr²)

2. (b) Show that \(A=2\pi r^2+\frac{960}{r}\).

   • A = 2πr² + 2πrh = 2πr² + 2πr × 480/(πr²)
   • = 2πr² + 960/r

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

   • A = 2πr² + 960r⁻¹
   • dA/dr = 4πr − 960r⁻²

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

   • 4πr = 960/r², so r³ = 960/(4π) = 76.4
   • r = 4.24 cm
   • A = 2π(4.24)² + 960/4.24 = 339 cm²

5. (e) A cube-shaped box holds the same 480 cm³. Find its total surface area and say whether the cylinder or the cube needs less material.

   • Cube: s³ = 480, s = 7.83 cm
   • Area = 6s² = 6 × 7.83² = 368 cm²
   • 339 < 368: the cylinder uses less material

Consider the curve \(y=x^{4x}\) for x > 0.

1. (a) Show that \(x^{4x}=e^{4x\ln x}\).

   • x = e^(ln x)
   • x^(4x) = (e^(ln x))^(4x) = e^(4x ln x)

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

   • dy/dx = e^(4x ln x) × (4 ln x + 4)
   • At x = 1: y = 1 and dy/dx = 1 × (0 + 4) = 4
   • y − 1 = 4(x − 1), so y = 4x − 3

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

   • e^(4x ln x) is never 0
   • So 4 ln x + 4 = 0, ln x = −1
   • x = 1/e